1 −
x
(d)
y p
y = csc–1x
−2 −1 1
p 2 (c)
Domain: x ≤ −1 or x ≥ 1 Range: −p ≤ y ≤ p , y ≠ 0 2 2 y
Domain: x ≤ −1 or x ≥ 1 p Range: 0 ≤ y ≤ p, y ≠ 2 y
y = tan –1x x 1 2
p 2
x
2
p 2
−2 −1
y = cot –1x
1
2
x
(f )
(e)
FIGURE 1.66 Graphs of the six basic inverse trigonometric functions.
DEFINITION y = sin−1 x is the number in 3 p>2, p>24 for which sin y = x. y = cos−1 x is the number in 3 0, p4 for which cos y = x. The “Arc” in Arcsine and Arccosine For a unit circle and radian angles, the arc length equation s = ru becomes s = u, so central angles and the arcs they subtend have the same measure. If x = sin y, then, in addition to being the angle whose sine is x, y is also the length of arc on the unit circle that subtends an angle whose sine is x. So we call y “the arc whose sine is x.”
The graph of y = sin1 x (Figure 1.65b) is symmetric about the origin (it lies along the graph of x = sin y). The arcsine is therefore an odd function: sin1(x) = sin1 x. The graph of y = cos x (Figure 1.67b) has no such symmetry.
EXAMPLE 8
y x2 + y2 = 1
0
Arc whose cosine is x x Angle whose cosine is x
1
Evaluate
(a) sin1 a
23
2
b
and
1 (b) cos1 a b . 2
Solution (a) We see that
Arc whose sine is x
Angle whose sine is x
(2)
1
x
sin1 a
23
2
b =
p 3
because sin (p>3) = 23>2 and p>3 belongs to the range 3 p>2, p>24 of the arcsine function. See Figure 1.68a. (b) We have 2p 1 cos1 a b = 3 2 because cos (2p>3) = 1>2 and 2p>3 belongs to the range 3 0, p4 of the arccosine function. See Figure 1.68b.
50
Chapter 1: Functions
y
Using the same procedure illustrated in Example 8, we can create the following table of common values for the arcsine and arccosine functions.
y = cos x, 0 ≤ x ≤ p Domain: [0, p] Range: [−1, 1]
1
p 2
0 −1
x
x
p
sin1 x
cos1 x p>6 p>4 p>3 2p>3
(a)
22>2
y
1>2 1>2
p>3 p>4 p>6 p>6
 22>2
p>4
3p>4
 23>2
p>3
5p>6
23>2
x = cos y p
y = cos –1 x Domain: [−1, 1] Range: [0, p]
p 2 −1 0
y
x
1
2 p 3 0 1
(b)
FIGURE 1.67 The graphs of (a) y = cos x, 0 … x … p, and (b) its inverse, y = cos1 x. The graph of cos1 x, obtained by reflection across the line y = x, is a portion of the curve x = cos y.
Chicago 179 180 61
12 62 b
St. Louis
Plane position
a
c
y
sin –1" 3 = p 2 3
cos–1a− 1b = 2 p 3 2 2
"3
2p 3
"3 x
sin p = " 3 2 3 (a)
x
−1 0
cos a2 pb = – 1 3 2 (b)
FIGURE 1.68 Values of the arcsine and arccosine functions (Example 8).
EXAMPLE 9 During a 240 mi airplane flight from Chicago to St. Louis, after flying 180 mi the navigator determines that the plane is 12 mi off course, as shown in Figure 1.69. Find the angle a for a course parallel to the original correct course, the angle b, and the drift correction angle c = a + b. Solution From the Pythagorean theorem and given information, we compute an approximate hypothetical flight distance of 179 mi, had the plane been flying along the original correct course (see Figure 1.69). Knowing the flight distance from Chicago to St. Louis, we next calculate the remaining leg of the original course to be 61 mi. Applying the Pythagorean theorem again then gives an approximate distance of 62 mi from the position of the plane to St. Louis. Finally, from Figure 1.69, we see that 180 sin a = 12 and 62 sin b = 12, so
FIGURE 1.69 Diagram for drift correction (Example 9), with distances surrounded to the nearest mile (drawing not to scale).
12 ≈ 0.067 radian ≈ 3.8° 180 12 b = sin1 ≈ 0.195 radian ≈ 11.2° 62 c = a + b ≈ 15°.
a = sin1
y cos–1(−x)
Identities Involving Arcsine and Arccosine cos–1x
−1 −x
0
x
1
As we can see from Figure 1.70, the arccosine of x satisfies the identity
x
cos1 x + cos1(x) = p,
(3)
cos1 (x) = p  cos1 x.
(4)
or
FIGURE 1.70 cos1 x and cos1( x) are supplementary angles (so their sum is p).
Also, we can see from the triangle in Figure 1.71 that for x 7 0, sin1 x + cos1 x = p>2.
(5)
51
1.6 Inverse Functions and Logarithms
cos–1x
1
x
sin–1x
FIGURE 1.71 sin1 x and cos1 x are complementary angles (so their sum is p>2).
Equation (5) holds for the other values of x in 3 1, 1] as well, but we cannot conclude this from the triangle in Figure 1.71. It is, however, a consequence of Equations (2) and (4) (Exercise 76). The arctangent, arccotangent, arcsecant, and arccosecant functions are defined in Section 3.9. There we develop additional properties of the inverse trigonometric functions in a calculus setting using the identities discussed here.
1.6
Exercises
Identifying OnetoOne Functions Graphically Which of the functions graphed in Exercises 1–6 are onetoone, and which are not? y
1.
y
2. y = −3x 3
Graphing Inverse Functions Each of Exercises 11–16 shows the graph of a function y = ƒ(x). Copy the graph and draw in the line y = x. Then use symmetry with respect to the line y = x to add the graph of ƒ 1 to your sketch. (It is not necessary to find a formula for ƒ 1.) Identify the domain and range of ƒ 1.
11. x
0
−1
0 1 y = x4 − x2
12. y
x
y y = f (x) = 2 1 , x ≥ 0 x +1
1
3.
4.
y
1 0
y = f (x) = 1 − 1x , x > 0 x
1
y y = int x 0
x
1
y = 20 x 0
13. x
y
5.
y
y = f (x) = sin x, p 1 p − ≤x≤ 2 2 y
6. y = 1x
14. y
−
y = x13
p 2
y = f (x) = tan x, p p − x. What symmetry does the graph have? b. Show that ƒ is its own inverse.
52
Chapter 1: Functions
Formulas for Inverse Functions Each of Exercises 19–24 gives a formula for a function y = ƒ(x) and shows the graphs of ƒ and ƒ 1. Find a formula for ƒ 1 in each case.
19. ƒ(x) = x2 + 1, x Ú 0
20. ƒ(x) = x2, x … 0
y
b. What can you conclude about the inverse of a function y = ƒ(x) whose graph is a line through the origin with a nonzero slope m?
y y = f (x)
38. Show that the graph of the inverse of ƒ(x) = mx + b, where m and b are constants and m ≠ 0, is a line with slope 1>m and yintercept b>m.
y = f (x)
1
y = f –1(x)
1 0
0 x
1
21. ƒ(x) = x3  1
x
1
22. ƒ(x) = x2  2x + 1, x Ú 1 y = f (x)
b. Find the inverse of ƒ(x) = x + b (b constant). What angle does the line y = x + b make with the line y = x?
1 −1
x
1
−1
y = f (x)
1 0
c. What can you conclude about the inverses of functions whose graphs are lines parallel to the line y = x? 40. a. Find the inverse of ƒ(x) = x + 1. Graph the line y = x + 1 together with the line y = x. At what angle do the lines intersect?
y = f –1(x)
y = f –1(x)
39. a. Find the inverse of ƒ(x) = x + 1. Graph ƒ and its inverse together. Add the line y = x to your sketch, drawing it with dashes or dots for contrast. b. Find the inverse of ƒ(x) = x + b (b constant). How is the graph of ƒ 1 related to the graph of ƒ?
y = f –1(x)
y
y
Inverses of Lines 37. a. Find the inverse of the function ƒ(x) = mx, where m is a constant different from zero.
x
1
c. What can you conclude about the inverses of functions whose graphs are lines perpendicular to the line y = x? Logarithms and Exponentials 41. Express the following logarithms in terms of ln 2 and ln 3.
23. ƒ(x) = (x + 1)2, x Ú  1 24. ƒ(x) = x2>3, x Ú 0 y
y y = f (x)
−1
0
1
1
−1 0
x
1
Each of Exercises 25–36 gives a formula for a function y = ƒ(x). In each case, find ƒ 1(x) and identify the domain and range of ƒ 1. As a check, show that ƒ(ƒ 1(x)) = ƒ 1(ƒ(x)) = x. 25. ƒ(x) = x5
26. ƒ(x) = x4, x Ú 0
27. ƒ(x) = x3 + 1
28. ƒ(x) = (1>2)x  7>2
29. ƒ(x) = 1>x2, x 7 0
30. ƒ(x) = 1>x3, x ≠ 0
x + 3 31. ƒ(x) = x  2
32. ƒ(x) =
2x 2x  3
34. ƒ(x) = (2x3 + 1)1>5 33. ƒ(x) = x2  2x, x … 1 (Hint: Complete the square.) 35. ƒ(x) =
x + b , b 7  2 and constant x  2
36. ƒ(x) = x  2bx, b 7 0 and constant, x … b 2
3 d. ln 2 9
f. ln 213.5
42. Express the following logarithms in terms of ln 5 and ln 7.
y = f (x) x
b. ln (4>9)
c. ln (1>2) e. ln 3 22
y = f –1(x)
y = f –1(x)
1
a. ln 0.75
a. ln (1>125)
b. ln 9.8
c. ln 7 27
d. ln 1225
e. ln 0.056
f. (ln 35 + ln (1>7))>(ln 25)
Use the properties of logarithms to write the expressions in Exercises 43 and 44 as a single term. sin u 1 43. a. ln sin u  ln a b b. ln (3x2  9x) + ln a b 3x 5 c.
1 ln (4t 4)  ln b 2
44. a. ln sec u + ln cos u
b. ln (8x + 4)  2 ln c
3 c. 3 ln 2t 2  1  ln (t + 1)
Find simpler expressions for the quantities in Exercises 45–48. b. eln x
2
46. a. eln (x
+y ) 2
47. a. 2 ln 2e sec u
48. a. ln (e
c. eln x  ln y
2
45. a. eln 7.2
b. eln 0.3
c. eln px  ln 2 e
b. ln (ln e ) )
x
(e )
b. ln (e )
c. ln (ex
2
 y2
2 ln x
c. ln (e
)
In Exercises 49–54, solve for y in terms of t or x, as appropriate. 49. ln y = 2t + 4
50. ln y = t + 5
51. ln (y  b) = 5t
52. ln (c  2y) = t
53. ln (y  1)  ln 2 = x + ln x 54. ln (y2  1)  ln (y + 1) = ln (sin x)
)
1.6 Inverse Functions and Logarithms
74. If a composite ƒ ∘ g is onetoone, must g be onetoone? Give reasons for your answer.
In Exercises 55 and 56, solve for k. 55. a. e2k = 4 56. a. e5k =
1 4
b. 100e10k = 200
c. ek>1000 = a
b. 80ek = 1
c. e(ln 0.8)k = 0.8
57. a. e0.3t = 27
b. ekt =
1 2
c. e(ln 0.2)t = 0.4
58. a. e0.01t = 1000
b. ekt =
1 10
c. e(ln 2)t =
1 2
60. e(x )e(2x + 1) = et 2
log8 22
61. a. 5
d. log4 16 62. a. 2log2 3 d. log11 121 63. a. 2log4 x 2
64. a. 25log5 (3x )
log1.3 75
b. 8
c. 1.3
e. log3 23
1 f. log4 a b 4
b. 10log10 (1>2)
c. plogp 7
c. shifting left 1, up 3 units.
c. log2 (e(ln 2)(sin x))
b. loge (ex)
c. log4 (2e sin x)
e. reflecting about the yaxis.
x
Express the ratios in Exercises 65 and 66 as ratios of natural logarithms and simplify. log2 x log8 x
c.
logx a logx2 a
66. a.
log9 x log3 x
b.
log 210 x log 22 x
c.
loga b logb a
1 b 2
b. sin1 a
1 22
1 68. a. cos1 a b 2
b. cos1 a
69. a. arccos ( 1)
b. arccos (0)
70. a. arcsin ( 1)
b. arcsin a
1 22
f. reflecting about the line y = x. 78. Start with the graph of y = ln x. Find an equation of the graph that results from a. vertical stretching by a factor of 2. b. horizontal stretching by a factor of 3. c. vertical compression by a factor of 4. d. horizontal compression by a factor of 2. T 79. The equation x2 = 2x has three solutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by graphing.
Arcsine and Arccosine In Exercises 67–70, find the exact value of each expression.
67. a. sin1 a
76. The identity sin1 x + cos1 x = P>2 Figure 1.71 establishes the identity for 0 6 x 6 1. To establish it for the rest of 3 1, 1], verify by direct calculation that it holds for x = 1, 0, and  1. Then, for values of x in (1, 0), let x = a, a 7 0, and apply Eqs. (3) and (5) to the sum sin1 ( a) + cos1 ( a).
d. shifting down 4, right 2 units.
b. 9log3 x
b.
50 1 + 1.1x
b. shifting right 1 unit.
1 f. log3 a b 9
log2 x log3 x
b. ƒ(x) =
a. shifting down 3 units.
e. log121 11
65. a.
100 1 + 2x
77. Start with the graph of y = ln x. Find an equation of the graph that results from
Simplify the expressions in Exercises 61–64. log5 7
75. Find a formula for the inverse function ƒ 1 and verify that (ƒ ∘ ƒ 1)(x) = (ƒ 1 ∘ ƒ)(x) = x. a. ƒ(x) =
In Exercises 57–60, solve for t.
59. e2t = x2
53
 23 b 2
b
c. sin1 a
b
c. cos1 a
23
2
b
T 80. Could xln 2 possibly be the same as 2ln x for x 7 0? Graph the two functions and explain what you see. 81. Radioactive decay The halflife of a certain radioactive substance is 12 hours. There are 8 grams present initially. a. Express the amount of substance remaining as a function of time t. b. When will there be 1 gram remaining?
1 22
b
Theory and Examples 71. If ƒ(x) is onetoone, can anything be said about g(x) = ƒ(x)? Is it also onetoone? Give reasons for your answer.
72. If ƒ(x) is onetoone and ƒ(x) is never zero, can anything be said about h(x) = 1>ƒ(x)? Is it also onetoone? Give reasons for your answer. 73. Suppose that the range of g lies in the domain of ƒ so that the composite ƒ ∘ g is defined. If ƒ and g are onetoone, can anything be said about ƒ ∘ g? Give reasons for your answer.
82. Doubling your money Determine how much time is required for a $500 investment to double in value if interest is earned at the rate of 4.75% compounded annually. 83. Population growth The population of Glenbrook is 375,000 and is increasing at the rate of 2.25% per year. Predict when the population will be 1 million. 84. Radon222 The decay equation for radon222 gas is known to be y = y0 e0.18t, with t in days. About how long will it take the radon in a sealed sample of air to fall to 90% of its original value?
54
Chapter 1: Functions
Chapter
1
Questions to Guide Your Review
1. What is a function? What is its domain? Its range? What is an arrow diagram for a function? Give examples. 2. What is the graph of a realvalued function of a real variable? What is the vertical line test? 3. What is a piecewisedefined function? Give examples. 4. What are the important types of functions frequently encountered in calculus? Give an example of each type. 5. What is meant by an increasing function? A decreasing function? Give an example of each. 6. What is an even function? An odd function? What symmetry properties do the graphs of such functions have? What advantage can we take of this? Give an example of a function that is neither even nor odd. 7. If ƒ and g are realvalued functions, how are the domains of ƒ + g, ƒ  g, ƒg, and ƒ>g related to the domains of ƒ and g? Give examples. 8. When is it possible to compose one function with another? Give examples of composites and their values at various points. Does the order in which functions are composed ever matter? 9. How do you change the equation y = ƒ(x) to shift its graph vertically up or down by k units? Horizontally to the left or right? Give examples. 10. How do you change the equation y = ƒ(x) to compress or stretch the graph by a factor c 7 1? Reflect the graph across a coordinate axis? Give examples. 11. What is radian measure? How do you convert from radians to degrees? Degrees to radians? 12. Graph the six basic trigonometric functions. What symmetries do the graphs have? 13. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions?
15. How does the formula for the general sine function ƒ(x) = A sin ((2p>B)(x  C)) + D relate to the shifting, stretching, compressing, and reflection of its graph? Give examples. Graph the general sine curve and identify the constants A, B, C, and D. 16. Name three issues that arise when functions are graphed using a calculator or computer with graphing software. Give examples. 17. What is an exponential function? Give examples. What laws of exponents does it obey? How does it differ from a simple power function like ƒ(x) = xn ? What kind of realworld phenomena are modeled by exponential functions? 18. What is the number e, and how is it defined? What are the domain and range of ƒ(x) = ex ? What does its graph look like? How do the values of ex relate to x2, x3, and so on? 19. What functions have inverses? How do you know if two functions ƒ and g are inverses of one another? Give examples of functions that are (are not) inverses of one another. 20. How are the domains, ranges, and graphs of functions and their inverses related? Give an example. 21. What procedure can you sometimes use to express the inverse of a function of x as a function of x? 22. What is a logarithmic function? What properties does it satisfy? What is the natural logarithm function? What are the domain and range of y = ln x? What does its graph look like? 23. How is the graph of loga x related to the graph of ln x? What truth is in the statement that there is really only one exponential function and one logarithmic function? 24. How are the inverse trigonometric functions defined? How can you sometimes use right triangles to find values of these functions? Give examples.
14. Starting with the identity sin2 u + cos2 u = 1 and the formulas for cos (A + B) and sin (A + B), show how a variety of other trigonometric identities may be derived.
Chapter
1
Practice Exercises
Functions and Graphs 1. Express the area and circumference of a circle as functions of the circle’s radius. Then express the area as a function of the circumference.
4. A hotair balloon rising straight up from a level field is tracked by a range finder located 500 ft from the point of liftoff. Express the balloon’s height as a function of the angle the line from the range finder to the balloon makes with the ground.
2. Express the radius of a sphere as a function of the sphere’s surface area. Then express the surface area as a function of the volume.
In Exercises 5–8, determine whether the graph of the function is symmetric about the yaxis, the origin, or neither.
3. A point P in the first quadrant lies on the parabola y = x2. Express the coordinates of P as functions of the angle of inclination of the line joining P to the origin.
5. y = x1>5
6. y = x2>5
7. y = x  2x  1
8. y = ex
2
2
Chapter 1 Practice Exercises
In Exercises 9–16, determine whether the function is even, odd, or neither. 9. y = x2 + 1
10. y = x5  x3  x
1 35. ƒ(x) = x ,
55
1
g(x) =
2x + 2
3 g(x) = 2 x + 1
36. ƒ(x) = 2  x,
11. y = 1  cos x
12. y = sec x tan x
x4 + 1 13. y = 3 x  2x
14. y = x  sin x
In Exercises 37 and 38, (a) write formulas for ƒ ∘ g and g ∘ ƒ and find the (b) domain and (c) range of each.
15. y = x + cos x
16. y = x cos x
38. ƒ(x) = 2x,
g(x) = 2x + 2
37. ƒ(x) = 2  x2,
17. Suppose that ƒ and g are both odd functions defined on the entire real line. Which of the following (where defined) are even? odd? e. 0 g 0
g(x) = 21  x
For Exercises 39 and 40, sketch the graphs of ƒ and ƒ ∘ ƒ.
18. If ƒ(a  x) = ƒ(a + x), show that g(x) = ƒ(x + a) is an even function.
x  2, 39. ƒ(x) = c  1, x  2,
In Exercises 19–28, find the (a) domain and (b) range.
40. ƒ(x) = b
b. ƒ3
a. ƒg
c. ƒ(sin x)
d. g(sec x)
19. y = x  2
20. y =  2 + 21  x
21. y = 216  x2
22. y = 32  x + 1
23. y = 2ex  3
24. y = tan (2x  p)
25. y = 2 sin (3x + p)  1
26. y = x2>5
27. y = ln (x  3) + 1
3 28. y =  1 + 2 2  x
ƒ1(x)
a. Volume of a sphere as a function of its radius c. Height above Earth’s sea level as a function of atmospheric pressure (assumed nonzero) d. Kinetic energy as a function of a particle’s velocity 30. Find the largest interval on which the given function is increasing.
44. x2 + x 45. 4  x2 1 46. x
a. ƒ(x) = 0 x  2 0 + 1
b. ƒ(x) = (x + 1)4
47. 2x
c. g(x) = (3x  1)
d. R(x) = 22x  1
48. sin x
2 x, 2x,
4 … x … 0 0 6 x … 4
x  2, 32. y = c x, x + 2,
2 … x …  1 1 6 x … 1 1 6 x … 2
a. Up
34.
0
1
0x0
20x0
sin 0 x 0
1 unit, right 3 2 2 3
c. Reflect about the yaxis d. Reflect about the xaxis
y 5
1
0x0 0x02 0 x3 0 0 x2 + x 0 0 4  x2 0
b. Down 2 units, left
In Exercises 33 and 34, write a piecewise formula for the function. y
ƒ2(x)
Shifting and Scaling Graphs 49. Suppose the graph of g is given. Write equations for the graphs that are obtained from the graph of g by shifting, scaling, or reflecting, as indicated.
PiecewiseDefined Functions In Exercises 31 and 32, find the (a) domain and (b) range.
33.
41. x 42. x2 43. x3
b. Greatest integer function
31. y = e
2 … x 6 0 0 … x … 2
Composition with absolute values In Exercises 41–48, graph ƒ1 and ƒ2 together. Then describe how applying the absolute value function in ƒ2 affects the graph of ƒ1 .
29. State whether each function is increasing, decreasing, or neither.
1>3
x + 1, x  1,
 4 … x … 1 1 6 x … 1 1 6 x … 2
(2, 5)
e. Stretch vertically by a factor of 5 f. Compress horizontally by a factor of 5
1
2
50. Describe how each graph is obtained from the graph of y = ƒ(x).
x
0
Composition of Functions In Exercises 35 and 36, find
a. (ƒ ∘ g) ( 1).
b. (g ∘ ƒ) (2).
c. (ƒ ∘ ƒ) (x).
d. (g ∘ g) (x).
4
x
a. y = ƒ(x  5)
b. y = ƒ(4x)
c. y = ƒ(3x)
d. y = ƒ(2x + 1)
x e. y = ƒa b  4 3
f. y = 3ƒ(x) +
1 4
56
Chapter 1: Functions
In Exercises 51–54, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.15–1.17, and applying an appropriate transformation. 51. y = 
A
1 +
1 53. y = 2 + 1 2x
x 2
52. y = 1 
x 3
54. y = ( 5x)
1>3
Trigonometry In Exercises 55–58, sketch the graph of the given function. What is the period of the function? x 55. y = cos 2x 56. y = sin 2
57. y = sin px
58. y = cos
px 2
p 59. Sketch the graph y = 2 cos ax  b . 3 60. Sketch the graph y = 1 + sin ax +
p b. 4
In Exercises 61–64, ABC is a right triangle with the right angle at C. The sides opposite angles A, B, and C are a, b, and c, respectively. 61. a. Find a and b if c = 2, B = p>3. b. Find a and c if b = 2, B = p>3. 62. a. Express a in terms of A and c. b. Express a in terms of A and b. 63. a. Express a in terms of B and b. b. Express c in terms of A and a. 64. a. Express sin A in terms of a and c. b. Express sin A in terms of b and c. 65. Height of a pole Two wires stretch from the top T of a vertical pole to points B and C on the ground, where C is 10 m closer to the base of the pole than is B. If wire BT makes an angle of 35° with the horizontal and wire CT makes an angle of 50° with the horizontal, how high is the pole? 66. Height of a weather balloon Observers at positions A and B 2 km apart simultaneously measure the angle of elevation of a weather balloon to be 40° and 70°, respectively. If the balloon is directly above a point on the line segment between A and B, find the height of the balloon. T 67. a. Graph the function ƒ(x) = sin x + cos(x>2). b. What appears to be the period of this function? c. Confirm your finding in part (b) algebraically. T 68. a. Graph ƒ(x) = sin (1>x). b. What are the domain and range of ƒ? c. Is ƒ periodic? Give reasons for your answer.
Transcendental Functions In Exercises 69–72, find the domain of each function.
69. a. ƒ(x) = 1 + esin x
b. g(x) = ex + ln 2x
70. a. ƒ(x) = e1>x
b. g(x) = ln 0 4  x2 0
x 71. a. h(x) = sin1 a b 3
b. ƒ(x) = cos1 ( 2x  1)
72. a. h(x) = ln (cos1 x)
b. ƒ(x) = 2p  sin1x
2
73. If ƒ(x) = ln x and g(x) = 4  x2, find ƒ ∘ g, g ∘ ƒ, ƒ ∘ ƒ, g ∘ g, and their domains.
the
functions
74. Determine whether ƒ is even, odd, or neither. a. ƒ(x) = ex
b. ƒ(x) = 1 + sin1( x)
2
c. ƒ(x) = 0 ex 0
d. ƒ(x) = eln x + 1
T 75. Graph ln x, ln 2x, ln 4x, ln 8x, and ln 16x (as many as you can) together for 0 6 x … 10. What is going on? Explain. T 76. Graph y = ln (x2 + c) for c = 4, 2, 0, 3, and 5. How does the graph change when c changes? T 77. Graph y = ln sin x in the window 0 … x … 22,  2 … y … 0. Explain what you see. How could you change the formula to turn the arches upside down? T 78. Graph the three functions y = xa, y = ax, and y = loga x together on the same screen for a = 2, 10, and 20. For large values of x, which of these functions has the largest values and which has the smallest values? Theory and Examples In Exercises 79 and 80, find the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers and comment on any differences you see.
79. a. y = sin1(sin x) 1
80. a. y = cos (cos x)
b. y = sin (sin1 x) b. y = cos (cos1 x)
81. Use a graph to decide whether ƒ is onetoone. a. ƒ(x) = x3 
x 2
b. ƒ(x) = x3 +
x 2
T 82. Use a graph to find to 3 decimal places the values of x for which ex 7 10,000,000. 3 83. a. Show that ƒ(x) = x3 and g(x) = 2x are inverses of one another.
T b. Graph ƒ and g over an xinterval large enough to show the graphs intersecting at (1, 1) and (1,  1). Be sure the picture shows the required symmetry in the line y = x.
84. a. Show that h(x) = x3 >4 and k(x) = (4x)1>3 are inverses of one another. T b. Graph h and k over an xinterval large enough to show the graphs intersecting at (2, 2) and (2,  2). Be sure the picture shows the required symmetry in the line y = x.
Chapter 1 Additional and Advanced Exercises
Chapter
1
57
Additional and Advanced Exercises
Functions and Graphs 1. Are there two functions ƒ and g such that ƒ ∘ g = g ∘ ƒ? Give reasons for your answer.
2. Are there two functions ƒ and g with the following property? The graphs of ƒ and g are not straight lines but the graph of ƒ ∘ g is a straight line. Give reasons for your answer. 3. If ƒ(x) is odd, can anything be said of g(x) = ƒ(x)  2? What if ƒ is even instead? Give reasons for your answer. 4. If g(x) is an odd function defined for all values of x, can anything be said about g(0)? Give reasons for your answer. 5. Graph the equation 0 x 0 + 0 y 0 = 1 + x.
Effects of Parameters on Graphs T 13. What happens to the graph of y = ax2 + bx + c as a. a changes while b and c remain fixed?
b. b changes (a and c fixed, a ≠ 0)? c. c changes (a and b fixed, a ≠ 0)? T 14. What happens to the graph of y = a(x + b)3 + c as a. a changes while b and c remain fixed?
6. Graph the equation y + y = x + x . Derivations and Proofs 7. Prove the following identities.
1  cos x sin x = a. 1 + cos x sin x
b. Uniqueness Show that there is only one way to write ƒ as the sum of an even and an odd function. (Hint: One way is given in part (a). If also ƒ(x) = E1(x) + O1(x) where E1 is even and O1 is odd, show that E  E1 = O1  O. Then use Exercise 11 to show that E = E1 and O = O1.)
b. b changes (a and c fixed, a ≠ 0)? x 1  cos x b. = tan2 1 + cos x 2
8. Explain the following “proof without words” of the law of cosines. (Source: Kung, Sidney H., “Proof Without Words: The Law of Cosines,” Mathematics Magazine, Vol. 63, no. 5, Dec. 1990, p. 342.) 2a cos u − b a−c c a
a
b u
c. c changes (a and b fixed, a ≠ 0)? Geometry 15. An object’s center of mass moves at a constant velocity y along a straight line past the origin. The accompanying figure shows the coordinate system and the line of motion. The dots show positions that are 1 sec apart. Why are the areas A1, A2, c, A5 in the figure all equal? As in Kepler’s equal area law (see Section 13.6), the line that joins the object’s center of mass to the origin sweeps out equal areas in equal times. y
a t=6
10
9. Show that the area of triangle ABC is given by (1>2)ab sin C = (1>2)bc sin A = (1>2)ca sin B. C
Kilometers
t=5 yΔt
A5 A4
5
yΔt
A3
t=2
A2
t=1
A1 b
a 0
A
c
5
10 Kilometers
15
x
B
10. Show that the area of triangle ABC is given by 2s(s  a)(s  b)(s  c) where s = (a + b + c)>2 is the semiperimeter of the triangle. 11. Show that if ƒ is both even and odd, then ƒ(x) = 0 for every x in the domain of ƒ.
16. a. Find the slope of the line from the origin to the midpoint P of side AB in the triangle in the accompanying figure (a, b 7 0). y
B(0, b)
12. a. Evenodd decompositions Let ƒ be a function whose domain is symmetric about the origin, that is,  x belongs to the domain whenever x does. Show that ƒ is the sum of an even function and an odd function: ƒ(x) = E(x) + O(x), where E is an even function and O is an odd function. (Hint: Let E(x) = (ƒ(x) + ƒ( x))>2. Show that E( x) = E(x), so that E is even. Then show that O(x) = ƒ(x)  E(x) is odd.)
P
O
b. When is OP perpendicular to AB?
A(a, 0)
x
58
Chapter 1: Functions
17. Consider the quartercircle of radius 1 and right triangles ABE and ACD given in the accompanying figure. Use standard area formulas to conclude that u 1 sin u 1 . sin u cos u 6 6 2 2 2 cos u
22. Drug absorption A drug is administered intravenously for pain. The function ƒ(t) = 90  52 ln (1 + t),
0 … t … 4
gives the number of units of the drug remaining in the body after t hours. a. What was the initial number of units of the drug administered?
y
b. How much is present after 2 hours? (0, 1)
C
c. Draw the graph of ƒ.
B
23. Finding investment time If Juanita invests $1500 in a retirement account that earns 8% compounded annually, how long will it take this single payment to grow to $5000?
1 u E
A
D (1, 0)
x
18. Let ƒ(x) = ax + b and g(x) = cx + d. What condition must be satisfied by the constants a, b, c, d in order that (ƒ ∘ g)(x) = (g ∘ ƒ)(x) for every value of x? Theory and Examples 19. Domain and range Suppose that a ≠ 0, b ≠ 1, and b 7 0. Determine the domain and range of the function.
a. y = a(bc  x) + d
b. y = a logb(x  c) + d
20. Inverse functions Let ƒ(x) =
ax + b , cx + d
c ≠ 0,
b. Find a formula for the inverse of ƒ. 21. Depreciation Smith Hauling purchased an 18wheel truck for $100,000. The truck depreciates at the constant rate of $10,000 per year for 10 years. a. Write an expression that gives the value y after x years. b. When is the value of the truck $55,000?
1
25. For what x 7 0 does x(x ) = (xx)x ? Give reasons for your answer. x
T 26. a. If (ln x)>x = (ln 2)>2, must x = 2? b. If (ln x)>x = 2 ln 2, must x = 1>2?
ad  bc ≠ 0.
a. Give a convincing argument that ƒ is onetoone.
Chapter
24. The rule of 70 If you use the approximation ln 2 ≈ 0.70 (in place of 0.69314 c), you can derive a rule of thumb that says, “To estimate how many years it will take an amount of money to double when invested at r percent compounded continuously, divide r into 70.” For instance, an amount of money invested at 5% will double in about 70>5 = 14 years. If you want it to double in 10 years instead, you have to invest it at 70>10 = 7%. Show how the rule of 70 is derived. (A similar “rule of 72” uses 72 instead of 70, because 72 has more integer factors.)
Give reasons for your answers. 27. The quotient (log4 x)>(log2 x) has a constant value. What value? Give reasons for your answer. T 28. logx (2) vs. log2 (x) How does ƒ(x) = logx (2) compare with g(x) = log2 (x)? Here is one way to find out. a. Use the equation loga b = (ln b)>(ln a) to express ƒ(x) and g(x) in terms of natural logarithms. b. Graph ƒ and g together. Comment on the behavior of ƒ in relation to the signs and values of g.
Technology Application Projects
An Overview of Mathematica An overview of Mathematica sufficient to complete the Mathematica modules appearing on the Web site.
Mathematica/Maple Module: Modeling Change: Springs, Driving Safety, Radioactivity, Trees, Fish, and Mammals Construct and interpret mathematical models, analyze and improve them, and make predictions using them.
2 Limits and Continuity Overview Mathematicians of the seventeenth century were keenly interested in the
study of motion for objects on or near the earth and the motion of planets and stars. This study involved both the speed of the object and its direction of motion at any instant, and they knew the direction at a given instant was along a line tangent to the path of motion. The concept of a limit is fundamental to finding the velocity of a moving object and the tangent to a curve. In this chapter we develop the limit, first intuitively and then formally. We use limits to describe the way a function varies. Some functions vary continuously; small changes in x produce only small changes in ƒ(x). Other functions can have values that jump, vary erratically, or tend to increase or decrease without bound. The notion of limit gives a precise way to distinguish between these behaviors.
2.1 Rates of Change and Tangents to Curves Calculus is a tool that helps us understand how a change in one quantity is related to a change in another. How does the speed of a falling object change as a function of time? How does the level of water in a barrel change as a function of the amount of liquid poured into it? We see change occurring in nearly everything we observe in the world and universe, and powerful modern instruments help us see more and more. In this section we introduce the ideas of average and instantaneous rates of change, and show that they are closely related to the slope of a curve at a point P on the curve. We give precise developments of these important concepts in the next chapter, but for now we use an informal approach so you will see how they lead naturally to the main idea of this chapter, the limit. The idea of a limit plays a foundational role throughout calculus.
Average and Instantaneous Speed HISTORICAL BIOGRAPHY* Galileo Galilei (1564–1642)
In the late sixteenth century, Galileo discovered that a solid object dropped from rest (not moving) near the surface of the earth and allowed to fall freely will fall a distance proportional to the square of the time it has been falling. This type of motion is called free fall. It assumes negligible air resistance to slow the object down, and that gravity is the only force acting on the falling object. If y denotes the distance fallen in feet after t seconds, then Galileo’s law is y = 16t 2, where 16 is the (approximate) constant of proportionality. (If y is measured in meters, the constant is 4.9.) A moving object’s average speed during an interval of time is found by dividing the distance covered by the time elapsed. The unit of measure is length per unit time: kilometers per hour, feet (or meters) per second, or whatever is appropriate to the problem at hand. *To learn more about the historical figures mentioned in the text and the development of many major elements and topics of calculus, visit www.aw.com/thomas.
59
60
Chapter 2: Limits and Continuity
Example 1 A rock breaks loose from the top of a tall cliff. What is its average speed (a) during the first 2 sec of fall? (b) during the 1sec interval between second 1 and second 2?
Solution The average speed of the rock during a given time interval is the change in distance, ∆y, divided by the length of the time interval, ∆t. (Increments like ∆y and ∆t are reviewed in Appendix 3, and pronounced “delta y” and “delta t.”) Measuring distance in feet and time in seconds, we have the following calculations: ∆y 16(2)2 = 2 ∆t ∆y 16(2)2 (b) From sec 1 to sec 2: = 2 ∆t (a) For the first 2 sec:
16(0)2 ft = 32 sec 0 16(1)2 ft = 48 sec 1
We want a way to determine the speed of a falling object at a single instant t0, instead of using its average speed over an interval of time. To do this, we examine what happens when we calculate the average speed over shorter and shorter time intervals starting at t0 . The next example illustrates this process. Our discussion is informal here, but it will be made precise in Chapter 3.
Example 2 Find the speed of the falling rock in Example 1 at t = 1 and t = 2 sec. Solution We can calculate the average speed of the rock over a time interval 3 t0, t0 + h4 , having length ∆t = h, as
∆y 16(t0 + h)2  16t0 2 = . (1) h ∆t
We cannot use this formula to calculate the “instantaneous” speed at the exact moment t0 by simply substituting h = 0, because we cannot divide by zero. But we can use it to calculate average speeds over increasingly short time intervals starting at t0 = 1 and t0 = 2. When we do so, by taking smaller and smaller values of h, we see a pattern (Table 2.1).
Table 2.1 Average speeds over short time intervals 3 t0, t0 + h4 Average speed:
Length of time interval h
∆y 16(t0 + h)2  16t0 2 = h ∆t
Average speed over interval of length h starting at t0 = 1
1 48 0.1 33.6 0.01 32.16 0.001 32.016 0.0001 32.0016
Average speed over interval of length h starting at t0 = 2 80 65.6 64.16 64.016 64.0016
The average speed on intervals starting at t0 = 1 seems to approach a limiting value of 32 as the length of the interval decreases. This suggests that the rock is falling at a speed of 32 ft > sec at t0 = 1 sec. Let’s confirm this algebraically.
2.1 Rates of Change and Tangents to Curves
that
61
If we set t0 = 1 and then expand the numerator in Equation (1) and simplify, we find ∆y 16(1 + h)2  16(1)2 16(1 + 2h + h2)  16 = = h h ∆t =
32h + 16h2 = 32 + 16h. h
For values of h different from 0, the expressions on the right and left are equivalent and the average speed is 32 + 16h ft>sec. We can now see why the average speed has the limiting value 32 + 16(0) = 32 ft>sec as h approaches 0. Similarly, setting t0 = 2 in Equation (1), the procedure yields ∆y = 64 + 16h ∆t for values of h different from 0. As h gets closer and closer to 0, the average speed has the limiting value 64 ft > sec when t0 = 2 sec, as suggested by Table 2.1.
The average speed of a falling object is an example of a more general idea which we discuss next.
Average Rates of Change and Secant Lines Given any function y = ƒ(x), we calculate the average rate of change of y with respect to x over the interval [x1, x2] by dividing the change in the value of y, ∆y = ƒ(x2)  ƒ(x1), by the length ∆x = x2  x1 = h of the interval over which the change occurs. (We use the symbol h for ∆x to simplify the notation here and later on.)
y y = f (x) Q(x 2, f (x 2 ))
Secant
DEFINITION The average rate of change of y = ƒ(x) with respect to x over the
Δy
P(x1, f (x1))
interval [x1, x2] is
Δx = h x2
x1
0
x
Figure 2.1 A secant to the graph y = ƒ(x). Its slope is ∆y> ∆x, the average rate of change of ƒ over the interval [x1, x2].
∆y ƒ(x2)  ƒ(x1) ƒ(x1 + h)  ƒ(x1) = = , x2  x1 h ∆x
h ≠ 0.
Geometrically, the rate of change of ƒ over [x1, x2] is the slope of the line through the points P(x1, ƒ(x1)) and Q(x2, ƒ(x2)) (Figure 2.1). In geometry, a line joining two points of a curve is a secant to the curve. Thus, the average rate of change of ƒ from x1 to x2 is identical with the slope of secant PQ. Let’s consider what happens as the point Q approaches the point P along the curve, so the length h of the interval over which the change occurs approaches zero. We will see that this procedure leads to defining the slope of a curve at a point.
Defining the Slope of a Curve P L O
Figure 2.2 L is tangent to the circle at P if it passes through P perpendicular to radius OP.
We know what is meant by the slope of a straight line, which tells us the rate at which it rises or falls—its rate of change as a linear function. But what is meant by the slope of a curve at a point P on the curve? If there is a tangent line to the curve at P—a line that just touches the curve like the tangent to a circle—it would be reasonable to identify the slope of the tangent as the slope of the curve at P. So we need a precise meaning for the tangent at a point on a curve. For circles, tangency is straightforward. A line L is tangent to a circle at a point P if L passes through P perpendicular to the radius at P (Figure 2.2). Such a line just touches the circle. But what does it mean to say that a line L is tangent to some other curve C at a point P?
62
Chapter 2: Limits and Continuity
To define tangency for general curves, we need an approach that takes into account the behavior of the secants through P and nearby points Q as Q moves toward P along the curve (Figure 2.3). Here is the idea: 1. Start with what we can calculate, namely the slope of the secant PQ. 2. Investigate the limiting value of the secant slope as Q approaches P along the curve. (We clarify the limit idea in the next section.) 3. If the limit exists, take it to be the slope of the curve at P and define the tangent to the curve at P to be the line through P with this slope. HISTORICAL BIOGRAPHY Pierre de Fermat (1601–1665)
This procedure is what we were doing in the fallingrock problem discussed in Example 2. The next example illustrates the geometric idea for the tangent to a curve.
Secants
Tangent
P
P Q
Tangent
Secants
Q
Figure 2.3 The tangent to the curve at P is the line through P whose slope is the limit of the secant slopes as Q S P from either side.
Example 3 Find the slope of the parabola y = x2 at the point P(2, 4). Write an equation for the tangent to the parabola at this point.
Solution We begin with a secant line through P(2, 4) and Q(2 + h, (2 + h)2) nearby. We then write an expression for the slope of the secant PQ and investigate what happens to the slope as Q approaches P along the curve: Secant slope =
∆y (2 + h)2  22 h2 + 4h + 4  4 = = h h ∆x h2 + 4h = h + 4. h
=
If h 7 0, then Q lies above and to the right of P, as in Figure 2.4. If h 6 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope h + 4 approaches 4. We take 4 to be the parabola’s slope at P. y y = x2
Secant slope is
(2 + h) 2 − 4 = h + 4. h
Q(2 + h, (2 + h) 2) Tangent slope = 4 Δy = (2 + h)2 − 4 P(2, 4) Δx = h 0
2
2+h
x
NOT TO SCALE
Figure 2.4 Finding the slope of the parabola y = x2 at the point P(2, 4) as the limit of secant slopes (Example 3).
2.1 Rates of Change and Tangents to Curves
63
The tangent to the parabola at P is the line through P with slope 4: y = 4 + 4(x  2)
Pointslope equation
y = 4x  4.
Instantaneous Rates of Change and Tangent Lines The rates at which the rock in Example 2 was falling at the instants t = 1 and t = 2 are called instantaneous rates of change. Instantaneous rates and slopes of tangent lines are closely connected, as we see in the following examples.
Example 4 Figure 2.5 shows how a population p of fruit flies (Drosophila) grew in a 50day experiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time t, and the points joined by a smooth curve (colored blue in Figure 2.5). Find the average growth rate from day 23 to day 45. Solution There were 150 flies on day 23 and 340 flies on day 45. Thus the number of flies increased by 340  150 = 190 in 45  23 = 22 days. The average rate of change of the population from day 23 to day 45 was Average rate of change:
∆p 340  150 190 = = ≈ 8.6 flies>day. 22 45  23 ∆t
p
Number of flies
350
Q(45, 340)
300
Δp = 190
250 200
P(23, 150)
150
Δp ≈ 8.6 fliesday Δt Δt = 22
100 50 0
10
20 30 Time (days)
40
50
t
Figure 2.5 Growth of a fruit fly population in a controlled experiment. The average rate of change over 22 days is the slope ∆p> ∆t of the secant line (Example 4).
This average is the slope of the secant through the points P and Q on the graph in Figure 2.5. The average rate of change from day 23 to day 45 calculated in Example 4 does not tell us how fast the population was changing on day 23 itself. For that we need to examine time intervals closer to the day in question.
Example 5 How fast was the number of flies in the population of Example 4 growing on day 23? Solution To answer this question, we examine the average rates of change over increasingly short time intervals starting at day 23. In geometric terms, we find these rates by calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure 2.6).
Chapter 2: Limits and Continuity
Q
Slope of PQ = 𝚫p , 𝚫t (flies , day)
(45, 340)
340  150 ≈ 8.6 45  23
(40, 330)
330  150 ≈ 10.6 40  23
(35, 310)
310  150 ≈ 13.3 35  23
(30, 265)
265  150 ≈ 16.4 30  23
p B(35, 350)
350
Q(45, 340)
300 Number of flies
64
250 200
P(23, 150)
150 100 50 0
10 20 30 A(14, 0) Time (days)
40
50
t
Figure 2.6 The positions and slopes of four secants through the point P on the fruit fly graph (Example 5).
The values in the table show that the secant slopes rise from 8.6 to 16.4 as the tcoordinate of Q decreases from 45 to 30, and we would expect the slopes to rise slightly higher as t continued on toward 23. Geometrically, the secants rotate counterclockwise about P and seem to approach the red tangent line in the figure. Since the line appears to pass through the points (14, 0) and (35, 350), it has slope 350  0 = 16.7 flies>day (approximately). 35  14 On day 23 the population was increasing at a rate of about 16.7 flies > day.
The instantaneous rates in Example 2 were found to be the values of the average speeds, or average rates of change, as the time interval of length h approached 0. That is, the instantaneous rate is the value the average rate approaches as the length h of the interval over which the change occurs approaches zero. The average rate of change corresponds to the slope of a secant line; the instantaneous rate corresponds to the slope of the tangent line as the independent variable approaches a fixed value. In Example 2, the independent variable t approached the values t = 1 and t = 2. In Example 3, the independent variable x approached the value x = 2. So we see that instantaneous rates and slopes of tangent lines are closely connected. We investigate this connection thoroughly in the next chapter, but to do so we need the concept of a limit.
Exercises 2.1 Average Rates of Change In Exercises 1–6, find the average rate of change of the function over the given interval or intervals.
5. R(u) = 24u + 1;
1. ƒ(x) = x3 + 1
Slope of a Curve at a Point In Exercises 7–14, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.
3 1, 14 a. 32, 34 b.
2. g(x) = x2  2x
3 2, 44 a. 31, 34 b. 3. h(t) = cot t
a. 3p>4, 3p>44 b. 3p>6, p>24 4. g(t) = 2 + cos t
a. 30, p4 b. 3 p, p4
3
2
30, 24
6. P(u) = u  4u + 5u;
31, 24
7. y = x2  5, P(2,  1) 8. y = 7  x2, P(2, 3) 9. y = x2  2x  3, P(2,  3) 10. y = x2  4x, P(1,  3) 11. y = x3, P(2, 8)
2.1 Rates of Change and Tangents to Curves
b. What is the average rate of increase of the profits between 2012 and 2014?
12. y = 2  x3, P(1, 1) 3
13. y = x  12x, P(1,  11)
c. Use your graph to estimate the rate at which the profits were changing in 2012.
14. y = x3  3x2 + 4, P(2, 0) Instantaneous Rates of Change 15. Speed of a car The accompanying figure shows the timetodistance graph for a sports car accelerating from a standstill. P
Q3
T 19. Let g(x) = 2x for x Ú 0.
Q2
a. Find the average rate of change of g(x) with respect to x over the intervals 31, 24, 31, 1.54 and 31, 1 + h4.
300 Q1
200
b. Make a table of values of the average rate of change of g with respect to x over the interval 31, 1 + h4 for some values of h approaching zero, say h = 0.1, 0.01, 0.001, 0.0001, 0.00001, and 0.000001.
100 0
5 10 15 20 Elapsed time (sec)
t
c. What does your table indicate is the rate of change of g(x) with respect to x at x = 1?
a. Estimate the slopes of secants PQ1, PQ2, PQ3, and PQ4, arranging them in order in a table like the one in Figure 2.6. What are the appropriate units for these slopes? b. Then estimate the car’s speed at time t = 20 sec. 16. The accompanying figure shows the plot of distance fallen versus time for an object that fell from the lunar landing module a distance 80 m to the surface of the moon. a. Estimate the slopes of the secants PQ1, PQ2, PQ3, and PQ4, arranging them in a table like the one in Figure 2.6. b. About how fast was the object going when it hit the surface? y
Distance fallen (m)
80 Q3
60
a. Find the average rate of change of ƒ with respect to t over the intervals (i) from t = 2 to t = 3, and (ii) from t = 2 to t = T. b. Make a table of values of the average rate of change of ƒ with respect to t over the interval 32, T4 , for some values of T approaching 2, say T = 2.1, 2.01, 2.001, 2.0001, 2.00001, and 2.000001. c. What does your table indicate is the rate of change of ƒ with respect to t at t = 2?
21. The accompanying graph shows the total distance s traveled by a bicyclist after t hours.
Q2
40
T 20. Let ƒ(t) = 1>t for t ≠ 0.
d. Calculate the limit as T approaches 2 of the average rate of change of ƒ with respect to t over the interval from 2 to T. You will have to do some algebra before you can substitute T = 2.
P
Q4
d. Calculate the limit as h approaches zero of the average rate of change of g(x) with respect to x over the interval 31, 1 + h4.
Q1
20
5
0
s 10
t
Elapsed time (sec)
T 17. The profits of a small company for each of the first five years of its operation are given in the following table: Year
Profit in $1000s
2010 6 2011 27 2012 62 2013 111 2014 174
a. Plot points representing the profit as a function of year, and join them by as smooth a curve as you can.
Distance traveled (mi)
Distance (m)
b. Extending the table if necessary, try to determine the rate of change of F(x) at x = 1.
Q4
500
T 18. Make a table of values for the function F(x) = (x + 2)>(x  2) at the points x = 1.2, x = 11>10, x = 101>100, x = 1001>1000, x = 10001>10000, and x = 1. a. Find the average rate of change of F(x) over the intervals 31, x4 for each x ≠ 1 in your table.
s 650 600
400
65
40 30 20 10 0
1
2 3 Elapsed time (hr)
4
t
a. Estimate the bicyclist’s average speed over the time intervals 30, 14, 31, 2.54 , and 32.5, 3.54 .
b. Estimate the bicyclist’s instantaneous speed at the times t = 12, t = 2, and t = 3. c. Estimate the bicyclist’s maximum speed and the specific time at which it occurs.
66
Chapter 2: Limits and Continuity
22. The accompanying graph shows the total amount of gasoline A in the gas tank of an automobile after being driven for t days.
a. Estimate the average rate of gasoline consumption over the time intervals 30, 34, 30, 54, and 37, 104 . b. Estimate the instantaneous rate of gasoline consumption at the times t = 1, t = 4, and t = 8.
Remaining amount (gal)
A 16
c. Estimate the maximum rate of gasoline consumption and the specific time at which it occurs.
12 8 4 0
1
2
3 4 5 6 7 8 Elapsed time (days)
9 10
t
2.2 Limit of a Function and Limit Laws In Section 2.1 we saw that limits arise when finding the instantaneous rate of change of a function or the tangent to a curve. Here we begin with an informal definition of limit and show how we can calculate the values of limits. A precise definition is presented in the next section. HISTORICAL ESSAY Limits
Limits of Function Values Frequently when studying a function y = ƒ(x), we find ourselves interested in the function’s behavior near a particular point c, but not at c. This might be the case, for instance, if c is an irrational number, like p or 22, whose values can only be approximated by “close” rational numbers at which we actually evaluate the function instead. Another situation occurs when trying to evaluate a function at c leads to division by zero, which is undefined. We encountered this last circumstance when seeking the instantaneous rate of change in y by considering the quotient function ∆y>h for h closer and closer to zero. Here’s a specific example in which we explore numerically how a function behaves near a particular point at which we cannot directly evaluate the function.
y
2 2 y = f (x) = x − 1 x− 1
1
−1
0
1
x
Example 1 How does the function ƒ(x) =
y
x2  1 x  1
behave near x = 1? 2
Solution The given formula defines ƒ for all real numbers x except x = 1 (we cannot divide by zero). For any x ≠ 1, we can simplify the formula by factoring the numerator and canceling common factors:
y=x+1 1
−1
0
1
Figure 2.7 The graph of ƒ is identical with the line y = x + 1 except at x = 1, where ƒ is not defined (Example 1).
x
ƒ(x) =
(x  1)(x + 1) = x + 1 x  1
for
x ≠ 1.
The graph of ƒ is the line y = x + 1 with the point (1, 2) removed. This removed point is shown as a “hole” in Figure 2.7. Even though ƒ(1) is not defined, it is clear that we can make the value of ƒ(x) as close as we want to 2 by choosing x close enough to 1 (Table 2.2).
2.2 Limit of a Function and Limit Laws
Table 2.2 As x gets closer to 1, ƒ(x ) gets closer to 2.
ƒ(x) =
x
x2 −1 x−1
0.9 1.9 1.1 2.1 0.99 1.99 1.01 2.01 0.999 1.999 1.001 2.001 0.999999 1.999999 1.000001 2.000001
67
Generalizing the idea illustrated in Example 1, suppose ƒ(x) is defined on an open interval about c, except possibly at c itself. If ƒ(x) is arbitrarily close to the number L (as close to L as we like) for all x sufficiently close to c, we say that ƒ approaches the limit L as x approaches c, and write lim ƒ(x) = L,
xSc
which is read “the limit of ƒ(x) as x approaches c is L.” For instance, in Example 1 we would say that ƒ(x) approaches the limit 2 as x approaches 1, and write lim ƒ(x) = 2,
x2  1 = 2. xS1 x  1
or
xS1
lim
Essentially, the definition says that the values of ƒ(x) are close to the number L whenever x is close to c (on either side of c). Our definition here is “informal” because phrases like arbitrarily close and sufficiently close are imprecise; their meaning depends on the context. (To a machinist manufacturing a piston, close may mean within a few thousandths of an inch. To an astronomer studying distant galaxies, close may mean within a few thousand lightyears.) Nevertheless, the definition is clear enough to enable us to recognize and evaluate limits of many specific functions. We will need the precise definition given in Section 2.3, however, when we set out to prove theorems about limits or study complicated functions. Here are several more examples exploring the idea of limits.
Example 2 The limit value of a function does not depend on how the function is defined at the point being approached. Consider the three functions in Figure 2.8. The function ƒ has limit 2 as x S 1 even though ƒ is not defined at x = 1. The function g has limit 2 as x S 1 even though 2 ≠ g(1). The function h is the only one of the three functions in Figure 2.8 whose limit as x S 1 equals its value at x = 1. For h, we have limx S 1 h(x) = h(1). This equality of limit and function value is of special importance, and we return to it in Section 2.5. y y
y
y
2
2
2
1
1
1
y=x c −1
0
1
x
0
−1
x
c
2 (a) f (x) = x − 1 x− 1
(b) g(x) =
(a) Identity function
1
x
x2 − 1 , x ≠ 1 x− 1 1,
−1
0
1
x
(c) h(x) = x + 1
x=1
Figure 2.8 The limits of ƒ(x), g(x), and h(x) all equal 2 as x approaches 1. However, only h(x) has the same function value as its limit at x = 1 (Example 2).
y
y=k
k
Example 3 (a) If ƒ is the identity function ƒ(x) = x, then for any value of c (Figure 2.9a),
0
c
lim ƒ(x) = lim x = c.
x
(b) Constant function
Figure 2.9 The functions in Example 3 have limits at all points c.
xSc
xSc
(b) If ƒ is the constant function ƒ(x) = k (function with the constant value k), then for
any value of c (Figure 2.9b),
lim ƒ(x) = lim k = k.
xSc
xSc
68
Chapter 2: Limits and Continuity
For instances of each of these rules we have lim x = 3
xS3
lim (4) = lim (4) = 4.
and
x S 7
xS2
We prove these rules in Example 3 in Section 2.3. A function may not have a limit at a particular point. Some ways that limits can fail to exist are illustrated in Figure 2.10 and described in the next example.
y
y y=
0, x < 0
y y=
1, x ≥ 0
1
0, x = 0
1
0
1, x≠0 x
x
x
0
x
0 y=
0,
x≤0
sin 1x , x > 0
–1 (a) Unit step function U(x)
(b) g(x)
(c) f (x)
Figure 2.10 None of these functions has a limit as x approaches 0 (Example 4).
Example 4 Discuss the behavior of the following functions, explaining why they have no limit as x S 0. 0, 1,
x 6 0 x Ú 0
1 x, (b) g(x) = • 0,
x≠0
0,
x … 0
1 sin x ,
x 7 0
(a) U(x) = e
(c) ƒ(x) = c
x = 0
Solution
(a) It jumps: The unit step function U(x) has no limit as x S 0 because its values jump
at x = 0. For negative values of x arbitrarily close to zero, U(x) = 0. For positive values of x arbitrarily close to zero, U(x) = 1. There is no single value L approached by U(x) as x S 0 (Figure 2.10a). (b) It grows too “large” to have a limit: g(x) has no limit as x S 0 because the values of g grow arbitrarily large in absolute value as x S 0 and do not stay close to any fixed real number (Figure 2.10b). We say the function is not bounded. (c) It oscillates too much to have a limit: ƒ(x) has no limit as x S 0 because the function’s values oscillate between +1 and 1 in every open interval containing 0. The values do not stay close to any one number as x S 0 (Figure 2.10c).
2.2 Limit of a Function and Limit Laws
69
The Limit Laws To calculate limits of functions that are arithmetic combinations of functions having known limits, we can use several fundamental rules.
THEOREM 1—Limit Laws If L, M, c, and k are real numbers and lim ƒ(x) = L
xSc
1. Sum Rule: 2. Difference Rule: 3. Constant Multiple Rule: 4. Product Rule: 5. Quotient Rule: 6. Power Rule: 7. Root Rule:
lim g(x) = M, then
and
xSc
lim (ƒ(x) + g(x)) = L + M
xSc
lim (ƒ(x)  g(x)) = L  M
xSc
lim (k # ƒ(x)) = k # L
xSc
lim (ƒ(x) # g(x)) = L # M
xSc
ƒ(x) L = , M ≠ 0 M x S c g(x) lim
lim 3 ƒ(x)4 n = L n, n a positive integer
xSc
n
n
lim 2ƒ(x) = 2L = L 1>n, n a positive integer
xSc
(If n is even, we assume that lim ƒ(x) = L 7 0.) xSc
In words, the Sum Rule says that the limit of a sum is the sum of the limits. Similarly, the next rules say that the limit of a difference is the difference of the limits; the limit of a constant times a function is the constant times the limit of the function; the limit of a product is the product of the limits; the limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0); the limit of a positive integer power (or root) of a function is the integer power (or root) of the limit (provided that the root of the limit is a real number). It is reasonable that the properties in Theorem 1 are true (although these intuitive arguments do not constitute proofs). If x is sufficiently close to c, then ƒ(x) is close to L and g(x) is close to M, from our informal definition of a limit. It is then reasonable that ƒ(x) + g(x) is close to L + M; ƒ(x)  g(x) is close to L  M; kƒ(x) is close to kL; ƒ(x)g(x) is close to LM; and ƒ(x)>g(x) is close to L>M if M is not zero. We prove the Sum Rule in Section 2.3, based on a precise definition of limit. Rules 2–5 are proved in Appendix 4. Rule 6 is obtained by applying Rule 4 repeatedly. Rule 7 is proved in more advanced texts. The Sum, Difference, and Product Rules can be extended to any number of functions, not just two.
Example 5 Use the observations limx S c k = k and limx S c x = c (Example 3) and the fundamental rules of limits to find the following limits. (a) lim ( x3 + 4x2  3 ) xSc
x4 + x2  1 xSc x2 + 5
(b) lim (c)
lim 24x2  3
x S 2
70
Chapter 2: Limits and Continuity
Solution (a) lim ( x3 + 4x2  3 ) = lim x3 + lim 4x2  lim 3
Sum and Difference Rules
= c3 + 4c2  3
Power and Multiple Rules
xSc
xSc
xSc
lim ( x4 + x2  1 )
x4 + x2  1 = (b) lim S x c x2 + 5
xSc
=
xSc
lim ( x2 + 5 )
(c)
Quotient Rule
xSc
lim x4 + lim x2  lim 1 xSc
xSc
lim x2 + lim 5
xSc
=
xSc
c4 + c2  1 c2 + 5
Power or Product Rule
lim 24x2  3 = 2 lim ( 4x2  3 )
x S 2
Sum and Difference Rules
xSc
x S 2
= 2 lim 4x2  lim 3 x S 2
x S 2
= 24(2)2  3
Root Rule with n = 2 Difference Rule Product and Multiple Rules
= 216  3 = 213
Theorem 1 simplifies the task of calculating limits of polynomials and rational functions. To evaluate the limit of a polynomial function as x approaches c, merely substitute c for x in the formula for the function. To evaluate the limit of a rational function as x approaches a point c at which the denominator is not zero, substitute c for x in the formula for the function. (See Examples 5a and 5b.) We state these results formally as theorems.
THEOREM 2—Limits of Polynomials If P(x) = an xn + an  1 xn  1 + g + a0, then lim P(x) = P(c) = an cn + an  1 cn  1 + g + a0.
xSc
THEOREM 3—Limits of Rational Functions If P(x) and Q(x) are polynomials and Q(c) ≠ 0, then P(x) P(c) = . Q(c) x S c Q(x) lim
Example 6 The following calculation illustrates Theorems 2 and 3: Identifying Common Factors It can be shown that if Q(x) is a polynomial and Q(c) = 0, then (x  c) is a factor of Q(x). Thus, if the numerator and denominator of a rational function of x are both zero at x = c, they have (x  c) as a common factor.
3 2 x3 + 4x2  3 (1) + 4(1)  3 0 = = = 0 6 x S 1 x2 + 5 (1)2 + 5
lim
Eliminating Common Factors from Zero Denominators Theorem 3 applies only if the denominator of the rational function is not zero at the limit point c. If the denominator is zero, canceling common factors in the numerator and denominator may reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can find the limit by substitution in the simplified fraction.
2.2 Limit of a Function and Limit Laws
Example 7 Evaluate
y 2 x−2 y= x + x2 − x (1, 3)
3
−2
0
x2 + x  2 . xS1 x2  x lim
x
1
Solution We cannot substitute x = 1 because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = 1. It is, so it has a factor of (x  1) in common with the denominator. Canceling this common factor gives a simpler fraction with the same values as the original for x ≠ 1:
(a)
x2 + x  2 (x  1)(x + 2) x + 2 = = x , x(x  1) x2  x
y y=x+2 x (1, 3)
3
71
if x ≠ 1.
Using the simpler fraction, we find the limit of these values as x S 1 by Theorem 3: x2 + x  2 x + 2 1 + 2 = lim x = = 3. 1 xS1 xS1 x2  x lim
−2
0
1
x
See Figure 2.11.
Using Calculators and Computers to Estimate Limits (b)
Figure 2.11 The graph of ƒ(x) = ( x2 + x  2 ) > ( x2  x ) in part (a) is the same as the graph of g(x) = (x + 2)>x in part (b) except at x = 1, where ƒ is undefined. The functions have the same limit as x S 1 (Example 7).
When we cannot use the Quotient Rule in Theorem 1 because the limit of the denominator is zero, we can try using a calculator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example 1, but calculators and computers can sometimes give false values and misleading impressions for functions that are undefined at a point or fail to have a limit there. Usually the problem is associated with rounding errors, as we now illustrate.
Example 8 Estimate the value of lim
xS0
2x2 + 100  10
x2
.
Solution Table 2.3 lists values of the function obtained on a calculator for several points approaching x = 0. As x approaches 0 through the points {1, {0.5, {0.10, and {0.01, the function seems to approach the number 0.05. As we take even smaller values of x, {0.0005, {0.0001, {0.00001, and {0.000001, the function appears to approach the number 0. Is the answer 0.05 or 0, or some other value? We resolve this question in the next example.
Table 2.3 Computed values of ƒ(x) =
2x 2 + 100  10
x
ƒ(x)
{1 {0.5 {0.1 {0.01
0.049876 0.049969 t approaches 0.05? 0.049999 0.050000
{0.0005 {0.0001 {0.00001 {0.000001
0.050000 0.000000 t approaches 0? 0.000000 0.000000
x2
near x = 0
72
Chapter 2: Limits and Continuity
Using a computer or calculator may give ambiguous results, as in the last example. The calculator could not keep track of enough digits to avoid rounding errors in computing the values of ƒ(x) when x is very small. We cannot substitute x = 0 in the problem, and the numerator and denominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.
EXAMPLE 9
Evaluate lim
2x2 + 100  10
x2
xS0
.
Solution This is the limit we considered in Example 8. We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression 2x2 + 100 + 10 (obtained by changing the sign after the square root). The preliminary algebra rationalizes the numerator: 2x2 + 100  10
x
2
=
#
2x2 + 100  10 2x2 + 100 + 10
x2
2x2 + 100 + 10
=
x2 + 100  100 x2 ( 2x2 + 100 + 10 )
=
x2 x2( 2x2 + 100 + 10)
Common factor x2
=
1 . 2x + 100 + 10
Cancel x2 for x ≠ 0.
2
Therefore, lim
xS0
2x2 + 100  10
x2
= lim
xS0
1 2x + 100 + 10 2
=
1 2 20 + 100 + 10
=
1 = 0.05. 20
Denominator not 0 at x = 0; substitute.
This calculation provides the correct answer, in contrast to the ambiguous computer results in Example 8. We cannot always algebraically resolve the problem of finding the limit of a quotient where the denominator becomes zero. In some cases the limit might then be found with the aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4), or through methods of calculus (illustrated in Section 4.5). The next theorems give helpful tools by using function comparisons.
y h f
L
The Sandwich Theorem
g
0
c
FIGURE 2.12 The graph of ƒ is sandwiched between the graphs of g and h.
x
The following theorem enables us to calculate a variety of limits. It is called the Sandwich Theorem because it refers to a function ƒ whose values are sandwiched between the values of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of ƒ must also approach L (Figure 2.12). You will find a proof in Appendix 4.
2.2 Limit of a Function and Limit Laws
73
THEOREM 4—The Sandwich Theorem Suppose that g(x) … ƒ(x) … h(x) for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim g(x) = lim h(x) = L. xSc
xSc
Then limx S c ƒ(x) = L. y
Example 10 Given that
2
y = u(x)
1 
1
2 y=1− x 4
0
−1
The Sandwich Theorem is also called the Squeeze Theorem or the Pinching Theorem.
2 y=1+ x 2
x
1
1
xS0
u
−1
0
xSc
Solution (a) In Section 1.3 we established that  0 u 0 … sin u … 0 u 0 for all u (see Figure 2.14a). Since limu S 0 (  0 u 0 ) = limu S 0 0 u 0 = 0, we have lim sin u = 0.
uS0
(b) From Section 1.3, 0 … 1  cos u … 0 u 0 for all u (see Figure 2.14b), and we have
limu S 0 (1  cos u) = 0 or
y
−2
uS0
xSc
(a)
1
(b) lim cos u = 1
(c) For any function ƒ, lim 0 ƒ(x) 0 = 0 implies lim ƒ(x) = 0.
y = −0 u 0
−1
xS0
Example 11 The Sandwich Theorem helps us establish several important limit rules:
y = sin u
p
lim ( 1 + ( x2 >2 ) ) = 1,
and
the Sandwich Theorem implies that limx S 0 u(x) = 1 (Figure 2.13).
uS0
−p
2
Solution Since
(a) lim sin u = 0 y = 0u0
for all x ≠ 0,
find limx S 0 u(x), no matter how complicated u is. lim ( 1  ( x2 >4 ) ) = 1
Figure 2.13 Any function u(x) whose graph lies in the region between y = 1 + (x2 >2) and y = 1  (x2 >4) has limit 1 as x S 0 (Example 10).
y
x2 x2 … u(x) … 1 + 4 2
lim cos u = 1.
uS0
y = 0u0 y = 1 − cos u 1
2
u
(b)
Figure 2.14 The Sandwich Theorem confirms the limits in Example 11.
(c) Since  0 ƒ(x) 0 … ƒ(x) … 0 ƒ(x) 0 and  0 ƒ(x) 0 and 0 ƒ(x) 0 have limit 0 as x S c, it
follows that limx S c ƒ(x) = 0.
Another important property of limits is given by the next theorem. A proof is given in the next section. THEOREM 5 If ƒ(x) … g(x) for all x in some open interval containing c, except possibly at x = c itself, and the limits of ƒ and g both exist as x approaches c, then lim ƒ(x) … lim g(x).
xSc
xSc
Caution The assertion resulting from replacing the less than or equal to (… ) inequality by the strict less than (6 ) inequality in Theorem 5 is false. Figure 2.14a shows that for u ≠ 0,  0 u 0 6 sin u 6 0 u 0 . So limu S 0 sin u = 0 = limu S 0 0 u 0 , not limu S 0 sin u 6 limu S 0 0 u 0 .
74
Chapter 2: Limits and Continuity
Exercises 2.2 Limits from Graphs 1. For the function g(x) graphed here, find the following limits or explain why they do not exist.
a. lim g(x) b. lim g(x) c. lim g(x) d. lim g(x) xS1
xS2
xS3
x S 2.5
d. lim ƒ(x) exists at every point c in (1, 1). xSc
e. lim ƒ(x) exists at every point c in (1, 3). xSc
y
y
1 1
−1
y = g(x) 1
2
x
3
−1 1
2
a. lim ƒ(t) b. lim ƒ(t) c. lim ƒ(t) d. lim ƒ(t) t S 1
tS0
t S 0.5
−1
−2
Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. x 1 6. lim 5. lim xS0 0 x 0 xS1 x  1
7. Suppose that a function ƒ(x) is defined for all real values of x except x = c. Can anything be said about the existence of limx S c ƒ(x)? Give reasons for your answer.
s
s = f (t)
−2
x
3
2. For the function ƒ(t) graphed here, find the following limits or explain why they do not exist. t S 2
y = f (x)
1 0
1
t
8. Suppose that a function ƒ(x) is defined for all x in 3 1, 1]. Can anything be said about the existence of limx S 0 ƒ(x)? Give reasons for your answer. 9. If limx S 1 ƒ(x) = 5, must ƒ be defined at x = 1? If it is, must ƒ(1) = 5? Can we conclude anything about the values of ƒ at x = 1? Explain.
−1
3. Which of the following statements about the function y = ƒ(x) graphed here are true, and which are false? a. lim ƒ(x) exists. xS0
10. If ƒ(1) = 5, must limx S 1 ƒ(x) exist? If it does, then must limx S 1 ƒ(x) = 5? Can we conclude anything about limx S 1 ƒ(x)? Explain.
xS0
Calculating Limits Find the limits in Exercises 11–22.
xS0
11. lim ( x2  13 )
12. lim (  x2 + 5x  2 )
e. lim ƒ(x) = 0
13. lim 8(t  5)(t  7)
14. lim ( x3  2x2 + 4x + 8 )
f. lim ƒ(x) exists at every point c in ( 1, 1).
2x + 5 15. lim x S 2 11  x 3
16. lim (8  3s)(2s  1)
17. lim 4x(3x + 4)2
18. lim
19. lim (5  y)4>3
20. lim 2z2  10
b. lim ƒ(x) = 0 c. lim ƒ(x) = 1 d. lim ƒ(x) = 1
x S 3
xS1
tS6
xS1 xSc
g. lim ƒ(x) does not exist. xS1
y = f (x)
1
−1
2
−1
4. Which of the following statements about the function y = ƒ(x) graphed here are true, and which are false? a. lim ƒ(x) does not exist. xS2
b. lim ƒ(x) = 2 c. lim ƒ(x) does not exist. xS1
21. lim
hS0
3 23h + 1 + 1
yS2
y + 2 y2 + 5y + 6
zS4
y S 3
x
x S 2
s S 2>3
x S 1>2
y 1
xS2
xS2
22. lim
hS0
25h + 4  2
h
Limits of quotients Find the limits in Exercises 23–42. x  5 x + 3 24. lim 2 23. lim 2 x S 5 x  25 x S 3 x + 4x + 3 x2 + 3x  10 x2  7x + 10 25. lim 26. lim x  2 x + 5 x S 5 xS2 t2 + t  2 t 2 + 3t + 2 27. lim 2 28. lim 2 tS1 t  1 t S 1 t  t  2 5y3 + 8y2 2x  4 29. lim 3 30. lim x S 2 x + 2x 2 y S 0 3y4  16y2
2.2 Limit of a Function and Limit Laws
x1  1 xS1 x  1
31. lim
32. lim
39. lim
xS2
2x2 + 12  4
x  2
+
53. Suppose limx S c ƒ(x) = 5 and limx S c g(x) = 2. Find a. lim ƒ(x)g(x) b. lim 2ƒ(x)g(x)
1 x + 1
x y3  8 34. lim 4 y S 2 y  16 4x  x2 36. lim x S 4 2  2x xS0
u4  1 33. lim 3 uS1 u  1 2x  3 35. lim xS9 x  9 x  1 37. lim x S 1 2x + 3  2
1 x  1
x S 1
40. lim
x S 2
2  2x2  5 41. lim x + 3 x S 3
xSc
54. Suppose limx S 4 ƒ(x) = 0 and limx S 4 g(x) = 3. Find lim xƒ(x) a. lim (g(x) + 3) b. xS4
g(x) c. lim (g(x))2 d. lim xS4 x S 4 ƒ(x)  1
x + 2
55. Suppose limx S b ƒ(x) = 7 and limx S b g(x) = 3. Find lim ƒ(x) # g(x) a. lim (ƒ(x) + g(x)) b.
2x2 + 5  3
43. lim (2 sin x  1)
44. lim sin2 x
45. lim sec x
46. lim tan x
47. lim
xS0
48. lim (x2  1)(2  cos x)
=
lim (2ƒ(x)  g(x))
xS0
lim (ƒ(x) + 7)2>3
(a)
lim 2ƒ(x)  lim g(x)
xS0
xS0
a lim ( ƒ(x) + 7 ) b
2>3
(b)
xS0
=
2 lim ƒ(x)  lim g(x) xS0
xS0
a lim ƒ(x) + lim 7b xS0
=
2>3
(c)
(1 + 7)2>3
=
xS1
p(x)(4  r(x))
=
lim 25h(x)
xS1
lim (p(x)(4  r(x)))
(a)
5h(x) 4xlim S1 xS1
5lim h(x) 4 xS1
=
a lim p(x)b a lim 4  lim r(x)b xS1
=
xS1
2(5)(5) 5 = (1)(4  2) 2
57. ƒ(x) = x2, x = 1 59. ƒ(x) = 3x  4, x = 2 60. ƒ(x) = 1>x, x =  2 62. ƒ(x) = 23x + 1, x = 0
64. If 2  x2 … g(x) … 2 cos x for all x, find limx S 0 g(x). 65. a. It can be shown that the inequalities 1 
xS1
lim
xS0
(c)
x2 x sin x 6 6 1 2  2 cos x 6
hold for all values of x close to zero. What, if anything, does this tell you about
(b)
a lim p(x)b a lim ( 4  r(x) ) b xS1
occur frequently in calculus. In Exercises 57–62, evaluate this limit for the given value of x and function ƒ.
63. If 25  2x2 … ƒ(x) … 25  x2 for  1 … x … 1, find limx S 0 ƒ(x).
7 4
xS1
=
ƒ(x + h)  ƒ(x) h
Using the Sandwich Theorem
xS0
(2)(1)  ( 5)
lim
hS0
61. ƒ(x) = 2x, x = 7
52. Let limx S 1 h(x) = 5, limx S 1 p(x) = 1, and limx S 1 r(x) = 2. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. 25h(x)
x S 2
58. ƒ(x) = x2, x =  2
xS0
=
lim
c. lim ( 4p(x) + 5r(x))>s(x) Limits of Average Rates of Change Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form
Using Limit Rules 51. Suppose limx S 0 ƒ(x) = 1 and limx S 0 g(x) =  5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.
(ƒ(x) + 7)2>3
a. lim (p(x) + r(x) + s(x))
xS0 xS0
2ƒ(x)  g(x)
xSb
x S 2
x S p>3
x S p
xS0
xSb
56. Suppose that limx S 2 p(x) = 4, limx S 2 r(x) = 0, and limx S 2 s(x) = 3. Find x S 2
49. lim 2x + 4 cos (x + p) 50. lim 27 + sec2 x
lim
xSb
b. lim p(x) # r(x) # s(x)
x S p>4
1 + x + sin x 3 cos x
xSb
c. lim 4g(x) d. lim ƒ(x)>g(x)
Limits with trigonometric functions Find the limits in Exercises 43–50.
xS0
xS4
x + 1
4  x 42. lim x S 4 5  2x 2 + 9
xS0
xSc
ƒ(x) c. lim (ƒ(x) + 3g(x)) d. lim xSc x S c ƒ(x)  g(x)
2x2 + 8  3
38. lim
75
x sin x ? 2  2 cos x
Give reasons for your answer.
T b. Graph y = 1  ( x2 >6 ) , y = (x sin x)>(2  2 cos x), and y = 1 together for 2 … x … 2. Comment on the behavior of the graphs as x S 0.
76
Chapter 2: Limits and Continuity
66. a. Suppose that the inequalities x2 1  cos x 1 1 6 6 2 24 2 x2 hold for values of x close to zero. (They do, as you will see in Section 9.9.) What, if anything, does this tell you about 1  cos x lim ? xS0 x2 Give reasons for your answer. T b. G raph the equations y = (1>2)  ( x2 >24 ) , y = (1  cos x)>x2, and y = 1>2 together for 2 … x … 2. Comment on the behavior of the graphs as x S 0. Estimating Limits T You will find a graphing calculator useful for Exercises 67–76.
67. Let ƒ(x) = (x2  9)>(x + 3). a. M ake a table of the values of ƒ at the points x = 3.1,  3.01, 3.001, and so on as far as your calculator can go. Then estimate limx S 3 ƒ(x). What estimate do you arrive at if you evaluate ƒ at x =  2.9,  2.99,  2.999, c instead? b. S upport your conclusions in part (a) by graphing ƒ near c = 3 and using Zoom and Trace to estimate yvalues on the graph as x S  3. c. Find limx S 3 ƒ(x) algebraically, as in Example 7.
68. Let g(x) = ( x2  2 ) >(x  22).
a. M ake a table of the values of g at the points x = 1.4, 1.41, 1.414, and so on through successive decimal approximations of 22. Estimate limx S 22 g(x). b. S upport your conclusion in part (a) by graphing g near c = 22 and using Zoom and Trace to estimate yvalues on the graph as x S 22.
c. Find limx S 22 g(x) algebraically.
69. Let G(x) = (x + 6)> ( x2 + 4x  12 ) . a. M ake a table of the values of G at x =  5.9, 5.99,  5.999, and so on. Then estimate limx S 6 G(x). What estimate do you arrive at if you evaluate G at x =  6.1, 6.01,  6.001, cinstead?
b. Support your conclusion in part (a) by graphing ƒ near c = 1 and using Zoom and Trace to estimate yvalues on the graph as x S 1. c. Find limx S 1 ƒ(x) algebraically.
72. Let F(x) = ( x2 + 3x + 2 ) > ( 2  0 x 0 ) .
a. Make tables of values of F at values of x that approach c = 2 from above and below. Then estimate limx S 2 F(x). b. Support your conclusion in part (a) by graphing F near c = 2 and using Zoom and Trace to estimate yvalues on the graph as x S 2. c. Find limx S 2 F(x) algebraically.
73. Let g(u) = (sin u)>u. a. Make a table of the values of g at values of u that approach u0 = 0 from above and below. Then estimate limu S 0 g(u). b. Support your conclusion in part (a) by graphing g near u0 = 0. 74. Let G(t) = (1  cos t)>t 2. a. Make tables of values of G at values of t that approach t0 = 0 from above and below. Then estimate limt S 0 G(t). b. Support your conclusion in part (a) by graphing G near t0 = 0. 75. Let ƒ(x) = x1>(1  x). a. Make tables of values of ƒ at values of x that approach c = 1 from above and below. Does ƒ appear to have a limit as x S 1? If so, what is it? If not, why not? b. Support your conclusions in part (a) by graphing ƒ near c = 1. 76. Let ƒ(x) = (3x  1)>x. a. Make tables of values of ƒ at values of x that approach c = 0 from above and below. Does ƒ appear to have a limit as x S 0? If so, what is it? If not, why not? b. Support your conclusions in part (a) by graphing ƒ near c = 0. Theory and Examples 77. If x4 … ƒ(x) … x2 for x in 31, 14 and x2 … ƒ(x) … x4 for x 6  1 and x 7 1, at what points c do you automatically know limx S c ƒ(x)? What can you say about the value of the limit at these points?
b. Support your conclusions in part (a) by graphing G and using Zoom and Trace to estimate yvalues on the graph as x S  6.
78. Suppose that g(x) … ƒ(x) … h(x) for all x ≠ 2 and suppose that
c. Find limx S 6 G(x) algebraically.
Can we conclude anything about the values of ƒ, g, and h at x = 2? Could ƒ(2) = 0? Could limx S 2 ƒ(x) = 0? Give reasons for your answers. ƒ(x)  5 79. If lim = 1, find lim ƒ(x). xS4 x  2 xS4 ƒ(x) 80. If lim 2 = 1, find x S 2 x ƒ(x) b. lim x a. lim ƒ(x) x S 2 x S 2
70. Let h(x) = ( x2  2x  3 ) > ( x2  4x + 3 ) .
a. Make a table of the values of h at x = 2.9, 2.99, 2.999, and so on. Then estimate limx S 3 h(x). What estimate do you arrive at if you evaluate h at x = 3.1, 3.01, 3.001, c instead? b. Support your conclusions in part (a) by graphing h near c = 3 and using Zoom and Trace to estimate yvalues on the graph as x S 3. c. Find limx S 3 h(x) algebraically.
71. Let ƒ(x) = ( x  1 ) > ( 0 x 0  1 ) . 2
a. M ake tables of the values of ƒ at values of x that approach c = 1 from above and below. Then estimate limx S 1 ƒ(x).
lim g(x) = lim h(x) =  5.
xS2
xS2
ƒ(x)  5 = 3, find lim ƒ(x). x  2 xS2 ƒ(x)  5 b. If lim = 4, find lim ƒ(x). xS2 x  2 xS2 81. a. If lim
xS2
2.3 The Precise Definition of a Limit
ƒ(x) = 1, find x S 0 x2
77
x4  16 xS2 x  2
85. lim
82. If lim
a. lim ƒ(x)
86. lim
xS0
x S 1
ƒ(x) b. lim x xS0
x3  x2  5x  3 (x + 1)2 3
raph g(x) = x sin (1>x) to estimate limx S 0 g(x), zooming in T 83. a. G on the origin as necessary. b. Confirm your estimate in part (a) with a proof. raph h(x) = x2 cos (1>x3) to estimate limx S 0 h(x), zooming T 84. a. G in on the origin as necessary. b. Confirm your estimate in part (a) with a proof.
87. lim
xS0
88. lim
xS3
21 + x  1
x
2
x  9 2x2 + 7  4
89. lim
1  cos x x sin x
90. lim
2x2 3  3 cos x
xS0
xS0
COMPUTER EXPLORATIONS Graphical Estimates of Limits In Exercises 85–90, use a CAS to perform the following steps:
a. Plot the function near the point c being approached. b. From your plot guess the value of the limit.
2.3 The Precise Definition of a Limit
We now turn our attention to the precise definition of a limit. We replace vague phrases like “gets arbitrarily close to” in the informal definition with specific conditions that can be applied to any particular example. With a precise definition, we can avoid misunderstandings, prove the limit properties given in the preceding section, and establish many important limits. To show that the limit of ƒ(x) as x S c equals the number L, we need to show that the gap between ƒ(x) and L can be made “as small as we choose” if x is kept “close enough” to c. Let us see what this would require if we specified the size of the gap between ƒ(x) and L.
Example 1 Consider the function y = 2x  1 near x = 4. Intuitively it appears that y is close to 7 when x is close to 4, so limx S 4 (2x  1) = 7. However, how close to x = 4 does x have to be so that y = 2x  1 differs from 7 by, say, less than 2 units?
y y = 2x − 1 Upper bound: y=9
9 To satisfy this
7 5
0
Lower bound: y=5
3 4 5
x
Restrict to this
Figure 2.15 Keeping x within 1 unit of x = 4 will keep y within 2 units of y = 7 (Example 1).
Solution We are asked: For what values of x is 0 y  7 0 6 2? To find the answer we first express 0 y  7 0 in terms of x:
0 y  7 0 = 0 (2x  1)  7 0 = 0 2x  8 0 .
The question then becomes: what values of x satisfy the inequality 0 2x  8 0 6 2? To find out, we solve the inequality:
0 2x  8 0 6 2 2 6 3 1
6 6 6 6
2x  8 6 2 2x 6 10 x 6 5 x  4 6 1.
Solve for x. Solve for x  4.
Keeping x within 1 unit of x = 4 will keep y within 2 units of y = 7 (Figure 2.15).
78
Chapter 2: Limits and Continuity
In the previous example we determined how close x must be to a particular value c to ensure that the outputs ƒ(x) of some function lie within a prescribed interval about a limit value L. To show that the limit of ƒ(x) as x S c actually equals L, we must be able to show that the gap between ƒ(x) and L can be made less than any prescribed error, no matter how small, by holding x close enough to c.
y
L+
1 10
f(x)
L L−
f (x) lies in here
Definition of Limit
1 10
for all x ≠ c in here
d 0
d x c
c−d
x
c+d
Figure 2.16 How should we define d 7 0 so that keeping x within the interval (c  d, c + d) will keep ƒ(x) within the interval aL 
1 1 ,L + b? 10 10
Suppose we are watching the values of a function ƒ(x) as x approaches c (without taking on the value of c itself). Certainly we want to be able to say that ƒ(x) stays within onetenth of a unit from L as soon as x stays within some distance d of c (Figure 2.16). But that in itself is not enough, because as x continues on its course toward c, what is to prevent ƒ(x) from jittering about within the interval from L  (1>10) to L + (1>10) without tending toward L? We can be told that the error can be no more than 1>100 or 1>1000 or 1>100,000. Each time, we find a new [email protected] about c so that keeping x within that interval satisfies the new error tolerance. And each time the possibility exists that ƒ(x) jitters away from L at some stage. The figures on the next page illustrate the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents [email protected] to prove that the limit does not exist or, more precisely, that there is room for doubt. The scholar answers every challenge with a [email protected] around c that keeps the function values within P of L. How do we stop this seemingly endless series of challenges and responses? We can do so by proving that for every error tolerance P that the challenger can produce, we can present a matching distance d that keeps x “close enough” to c to keep ƒ(x) within that [email protected] of L (Figure 2.17). This leads us to the precise definition of a limit.
y
DEFINITION Let ƒ(x) be defined on an open interval about c, except possibly at c
itself. We say that the limit of ƒ(x) as x approaches c is the number L, and write
L+P L
f(x)
lim ƒ(x) = L,
f (x) lies in here
xSc
if, for every number P 7 0, there exists a corresponding number d 7 0 such that for all x,
L−P
0 6 0x  c0 6 d
for all x ≠ c in here d x 0
c−d
d x c
c+d
Figure 2.17 The relation of d and P in the definition of limit.
1
0 ƒ(x)  L 0 6 P.
One way to think about the definition is to suppose we are machining a generator shaft to a close tolerance. We may try for diameter L, but since nothing is perfect, we must be satisfied with a diameter ƒ(x) somewhere between L  P and L + P. The d is the measure of how accurate our control setting for x must be to guarantee this degree of accuracy in the diameter of the shaft. Notice that as the tolerance for error becomes stricter, we may have to adjust d. That is, the value of d, how tight our control setting must be, depends on the value of P, the error tolerance.
Examples: Testing the Definition The formal definition of limit does not tell how to find the limit of a function, but it enables us to verify that a conjectured limit value is correct. The following examples show how the definition can be used to verify limit statements for specific functions. However, the real purpose of the definition is not to do calculations like this, but rather to prove general theorems so that the calculation of specific limits can be simplified, such as the theorems stated in the previous section.
2.3 The Precise Definition of a Limit
y
L+
y = f (x)
1 10
L−
x c c − d1/10 c + d1/10 Response: 0 x − c 0 < d1/10 (a number)
L+
y y = f (x)
y = f (x)
1 L+ 1000
L+
1 1000
L
L
1 1000
L−
1 1000
x
c
0
Response: 0 x − c 0 < d1/1000
y
y y = f (x)
L−
1 L+ 100,000
L
L
1 100,000
L−
c New challenge: 1 P = 100,000
y y = f (x)
1 100,000
0
x
c
0
New challenge: P = 1 1000
L+
0
New challenge: Make 0 f (x) − L 0 < P = 1 100
y
L−
x c c + d1/100 c − d1/100 Response: 0 x − c 0 < d1/100
x
c
0
0
The challenge: Make 0 f (x) − L 0 < P = 1 10
1 100 L 1 L− 100
1 100
L 1 L− 100
1 10
x
c
0
y = f(x)
y = f (x) L+
L
1 10
y
y = f (x)
1 L+ 10
L L−
y
y
79
x
y = f(x) L+P L L−P
1 100,000
0
c
x
c
0
Response: 0 x − c 0 < d1/100,000
x
New challenge: P = ...
Example 2 Show that lim (5x  3) = 2.
xS1
Solution Set c = 1, ƒ(x) = 5x  3, and L = 2 in the definition of limit. For any given P 7 0, we have to find a suitable d 7 0 so that if x ≠ 1 and x is within distance d of c = 1, that is, whenever 0 6 0 x  1 0 6 d,
it is true that ƒ(x) is within distance P of L = 2, so
0 ƒ(x)  2 0 6 P.
80
Chapter 2: Limits and Continuity
y
We find d by working backward from the [email protected]:
y = 5x − 3
0 (5x  3)  2 0 = 0 5x  5 0 6 P 50x  10 6 P 0 x  1 0 6 P>5.
2+P 2 2−P x
1−P 1 1+P 5 5
0
Thus, we can take d = P>5 (Figure 2.18). If 0 6 0 x  1 0 6 d = P>5, then
0 (5x  3)  2 0 = 0 5x  5 0 = 5 0 x  1 0 6 5(P>5) = P,
which proves that limx S 1(5x  3) = 2. The value of d = P>5 is not the only value that will make 0 6 0 x  1 0 6 d imply 0 5x  5 0 6 P. Any smaller positive d will do as well. The definition does not ask for a “best” positive d, just one that will work.
Example 3 Prove the following results presented graphically in Section 2.2.
−3
(a) lim x = c
NOT TO SCALE
xSc
Figure 2.18 If ƒ(x) = 5x  3, then 0 6 0 x  1 0 6 P>5 guarantees that 0 ƒ(x)  2 0 6 P (Example 2).
(b) lim k = k (k constant) xSc
Solution (a) Let P 7 0 be given. We must find d 7 0 such that for all x
y
implies
0 6 0x  c0 6 d
implies
0 x  c 0 6 P.
The implication will hold if d equals P or any smaller positive number (Figure 2.19). This proves that limx S c x = c. (b) Let P 7 0 be given. We must find d 7 0 such that for all x
y=x c+P c+d c c−d
0 k  k 0 6 P.
Since k  k = 0, we can use any positive number for d and the implication will hold (Figure 2.20). This proves that limx S c k = k.
c−P
0
c−d c c+d
x
Finding Deltas Algebraically for Given Epsilons In Examples 2 and 3, the interval of values about c for which 0 ƒ(x)  L 0 was less than P was symmetric about c and we could take d to be half the length of that interval. When such symmetry is absent, as it usually is, we can take d to be the distance from c to the interval’s nearer endpoint.
Figure 2.19 For the function ƒ(x) = x, we find that 0 6 0 x  c 0 6 d will guarantee 0 ƒ(x)  c 0 6 P whenever d … P (Example 3a).
Example 4 For the limit limx S 5 2x  1 = 2, find a d 7 0 that works for P = 1. That is, find a d 7 0 such that for all x
y
0 6 0x  50 6 d
y=k
k+P k k−P
0
0 6 0x  c0 6 d
1
Solution We organize the search into two steps.
0 2x  1  2 0 6 1.
1. Solve the inequality 0 2x  1  2 0 6 1 to find an interval containing x = 5 on which the inequality holds for all x ≠ 5. c−d
c
c+d
Figure 2.20 For the function ƒ(x) = k, we find that 0 ƒ(x)  k 0 6 P for any positive d (Example 3b).
x
0 2x  1  2 0 6 1 1 6 2x  1  2 6 1 1 6 2x  1 6 3 1 6 x  1 6 9 2 6 x 6 10
2.3 The Precise Definition of a Limit
(
3
3
2
5
)
x
10
8
Figure 2.21 An open interval of radius 3 about x = 5 will lie inside the open interval (2, 10).
y y = "x − 1
0 6 0x  50 6 3
0 2x  1  2 0 6 1.
1
How to Find Algebraically a D for a Given ƒ, L, c, and E + 0 The process of finding a d 7 0 such that for all x
2
0 6 0x  c0 6 d
1
can be accomplished in two steps.
1 3 0
The inequality holds for all x in the open interval (2, 10), so it holds for all x ≠ 5 in this interval as well. 2. Find a value of d 7 0 to place the centered interval 5  d 6 x 6 5 + d (centered at x = 5) inside the interval (2, 10). The distance from 5 to the nearer endpoint of (2, 10) is 3 (Figure 2.21). If we take d = 3 or any smaller positive number, then the inequality 0 6 0 x  5 0 6 d will automatically place x between 2 and 10 to make 0 2x  1  2 0 6 1 (Figure 2.22):
3
3 5
1 2
8
10
81
x
NOT TO SCALE
Figure 2.22 The function and intervals in Example 4.
0 ƒ(x)  L 0 6 P
1. Solve the inequality 0 ƒ(x)  L 0 6 P to find an open interval (a, b) containing c on which the inequality holds for all x ≠ c. 2. Find a value of d 7 0 that places the open interval (c  d, c + d) centered at c inside the interval (a, b). The inequality 0 ƒ(x)  L 0 6 P will hold for all x ≠ c in this [email protected]
Example 5 Prove that limx S 2 ƒ(x) = 4 if ƒ(x) = e
x2, 1,
x≠2 x = 2.
Solution Our task is to show that given P 7 0 there exists a d 7 0 such that for all x y
0 6 0x  20 6 d
y = x2
0 ƒ(x)  4 0 6 P.
1. Solve the inequality 0 ƒ(x)  4 0 6 P to find an open interval containing x = 2 on which the inequality holds for all x ≠ 2.
4+P
For x ≠ c = 2, we have ƒ(x) = x2, and the inequality to solve is 0 x2  4 0 6 P:
(2, 4)
4
0 x2  4 0 6 P
4−P (2, 1) 0
1
"4 − P
2
x "4 + P
Figure 2.23 An interval containing x = 2 so that the function in Example 5 satisfies 0 ƒ(x)  4 0 6 P.
P 6 x2  4 6 P 4  P 6 x2 6 4 + P 24  P 6 0 x 0 6 24 + P Assumes P 24  P 6 x 6 24 + P.
6 4; see below. An open interval about x = 2 that solves the inequality
The inequality 0 ƒ(x)  4 0 6 P holds for all x ≠ 2 in the open interval ( 24  P, 24 + P ) (Figure 2.23).
2. Find a value of d 7 0 that places the centered interval (2  d, 2 + d) inside the interval ( 24  P, 24 + P ) .
Take d to be the distance from x = 2 to the nearer endpoint of ( 24  P, 24 + P ) . In other words, take d = min 5 2  24  P, 24 + P  2 6 , the minimum (the
82
Chapter 2: Limits and Continuity
smaller) of the two numbers 2  24  P and 24 + P  2. If d has this or any smaller positive value, the inequality 0 6 0 x  2 0 6 d will automatically place x between 24  P and 24 + P to make 0 ƒ(x)  4 0 6 P. For all x, 0 6 0x  20 6 d
1
0 ƒ(x)  4 0 6 P.
This completes the proof for P 6 4. If P Ú 4, then we take d to be the distance from x = 2 to the nearer endpoint of the interval ( 0, 24 + P ) . In other words, take d = min 5 2, 24 + P  2 6 . (See Figure 2.23.)
Using the Definition to Prove Theorems We do not usually rely on the formal definition of limit to verify specific limits such as those in the preceding examples. Rather, we appeal to general theorems about limits, in particular the theorems of Section 2.2. The definition is used to prove these theorems (Appendix 5). As an example, we prove part 1 of Theorem 1, the Sum Rule.
Example 6 Given that limx S c ƒ(x) = L and limx S c g(x) = M, prove that lim (ƒ(x) + g(x)) = L + M.
xSc
Solution Let P 7 0 be given. We want to find a positive number d such that for all x 0 6 0x  c0 6 d
1
Regrouping terms, we get
0 ƒ(x) + g(x)  (L + M) 0 6 P.
0 ƒ(x) + g(x)  (L + M) 0 = 0 (ƒ(x)  L) + (g(x)  M) 0 Triangle Inequality: 0a + b0 … 0a0 + 0b0 … 0 ƒ(x)  L 0 + 0 g(x)  M 0 .
Since limx S c ƒ(x) = L, there exists a number d1 7 0 such that for all x 0 6 0 x  c 0 6 d1
1
0 ƒ(x)  L 0 6 P>2.
Similarly, since limx S c g(x) = M, there exists a number d2 7 0 such that for all x 0 6 0 x  c 0 6 d2
1
0 g(x)  M 0 6 P>2.
Let d = min 5d1, d2 6 , the smaller of d1 and d2. If 0 6 0 x  c 0 6 d then 0 x  c 0 6 d1, so 0 ƒ(x)  L 0 6 P>2, and 0 x  c 0 6 d2, so 0 g(x)  M 0 6 P>2. Therefore
0 ƒ(x) + g(x)  (L + M) 0 6 P + P = P. 2
2
This shows that limx S c (ƒ(x) + g(x)) = L + M. Next we prove Theorem 5 of Section 2.2.
Example 7 Given that limx S c ƒ(x) = L and limx S c g(x) = M, and that ƒ(x) … g(x) for all x in an open interval containing c (except possibly c itself), prove that L … M.
Solution We use the method of proof by contradiction. Suppose, on the contrary, that L 7 M. Then by the limit of a difference property in Theorem 1, lim (g(x)  ƒ(x)) = M  L.
xSc
83
2.3 The Precise Definition of a Limit
Therefore, for any P 7 0, there exists d 7 0 such that
0 (g(x)  ƒ(x))  (M  L) 0 6 P
whenever 0 6 0 x  c 0 6 d.
Since L  M 7 0 by hypothesis, we take P = L  M in particular and we have a number d 7 0 such that
0 (g(x)  ƒ(x))  (M  L) 0 6 L  M
whenever 0 6 0 x  c 0 6 d.
Since a … 0 a 0 for any number a, we have
whenever 0 6 0 x  c 0 6 d
(g(x)  ƒ(x))  (M  L) 6 L  M
which simplifies to
whenever 0 6 0 x  c 0 6 d.
g(x) 6 ƒ(x)
But this contradicts ƒ(x) … g(x). Thus the inequality L 7 M must be false. Therefore L … M.
Exercises 2.3 Centering Intervals About a Point In Exercises 1–6, sketch the interval (a, b) on the xaxis with the point c inside. Then find a value of d 7 0 such that for all x, 0 6 0 x  c 0 6 d 1 a 6 x 6 b.
1. a = 1, b = 7, c = 5
2. a = 1, b = 7, c = 2
3. a =  7>2, b =  1>2, c =  3 4. a =  7>2, b =  1>2, c =  3>2
9. y 5 4 1 3 4
10.
f (x) = " x c=1 L=1 P = 1 y = "x 4
f(x) = 2" x + 1 c=3 L=4 P = 0.2
0
Finding Deltas Graphically In Exercises 7–14, use the graphs to find a d 7 0 such that for all x 0 6 0 x  c 0 6 d 1 0 ƒ(x)  L 0 6 P.
7.
y y = 2x − 4 6.2 6 5.8
0
f (x) = 2x − 4 c=5 L=6 P = 0.2
4.9
5
x
25 16
11.
x
y
y = x2
y = 4 − x2
5
0
−2.9
0
NOT TO SCALE
3.25 3 2.75
3
5.1
−3
y f (x) = 4 − x 2 c = −1 L=3 P = 0.25
4
NOT TO SCALE
−3.1
x
12.
f (x) = c=2 L=4 P=1
7.65 7.5 7.35
2.61 3 3.41 NOT TO SCALE
x2
y = −3 x + 3 2
2
−1 0
y
f (x) = − 3 x + 3 2 c = −3 L = 7.5 P = 0.15
1
9 16
8.
y = 2" x + 1
4.2 4 3.8
5. a = 4>9, b = 4>7, c = 1>2 6. a = 2.7591, b = 3.2391, c = 3
y
x
"3
NOT TO SCALE
2
x "5 −
" 5 −1 " 3 − 2 2
NOT TO SCALE
0
x
84
Chapter 2: Limits and Continuity
13.
14.
y f(x) = 2 "−x c = −1 L=2 P = 0.5 y=
y f(x) = 1x c=1 2 L=2 P = 0.01
2.01
2 "−x
Using the Formal Definition Each of Exercises 31–36 gives a function ƒ(x), a point c, and a positive number P. Find L = lim ƒ(x). Then find a number d 7 0 such xSc that for all x
0 6 0x  c0 6 d
31. ƒ(x) = 3  2x,
2.5
1.99
2
y = 1x
1.5
c = 3,
32. ƒ(x) =  3x  2,
2
x2  4 , x  2
34. ƒ(x) =
x2 + 6x + 5 , x + 5
c = 2,
35. ƒ(x) = 21  5x,
16 − 9
−1
16 − 25
x
0
x
1 1 1 2 2.01 1.99
0
P = 0.02
c = 1,
33. ƒ(x) =
36. ƒ(x) = 4>x,
P = 0.05
c = 5, c =  3,
c = 2,
Finding Deltas Algebraically Each of Exercises 15–30 gives a function ƒ(x) and numbers L, c, and P 7 0. In each case, find an open interval about c on which the inequality 0 ƒ(x)  L 0 6 P holds. Then give a value for d 7 0 such that for all x satisfying 0 6 0 x  c 0 6 d the inequality 0 ƒ(x)  L 0 6 P holds.
L = 5,
c = 4,
16. ƒ(x) = 2x  2,
L =  6,
17. ƒ(x) = 2x + 1,
L = 1,
18. ƒ(x) = 2x,
L = 1>2,
19. ƒ(x) = 219  x,
20. ƒ(x) = 2x  7, 21. ƒ(x) = 1>x, 2
L = 3, L = 4,
L = 1>4,
22. ƒ(x) = x ,
L = 3,
23. ƒ(x) = x2,
L = 4,
24. ƒ(x) = 1>x, 25. ƒ(x) = x  5,
P = 0.1
c = 1>4,
P = 0.1
c = 10,
P = 1
c = 23,
c =  2,
c =  1, c = 4,
L = 5,
P = 0.02
c = 0,
c = 23,
L = 11,
26. ƒ(x) = 120>x,
c =  2,
c = 4,
L =  1,
2
P = 0.01
c = 24,
P = 1 P = 0.05
P = 0.1
m 7 0,
L = 3m,
Prove the limit statements in Exercises 37–50. 37. lim (9  x) = 5 38. lim (3x  7) = 2 xS3
39. lim 2x  5 = 2
40. lim 24  x = 2 xS0
41. lim ƒ(x) = 1 if ƒ(x) = e xS1
2
x, 2,
42. lim ƒ(x) = 4 if ƒ(x) = e x S 2
1 43. lim x = 1 x S 3
x ≠ 2 x = 2
x 2, 1,
44. lim
xS1
45. lim
x ≠ 1 x = 1
x S 23
x2  9 = 6 x + 3
46. lim
xS1
47. lim ƒ(x) = 2 if ƒ(x) = e xS1
48. lim ƒ(x) = 0 if ƒ(x) = e xS0
4  2x, 6x  4, 2x, x>2,
x 6 1 x Ú 1 x 6 0 x Ú 0
1 49. lim x sin x = 0 xS0
y
P = 1 P = 1 c = 3,
L = (m>2) + b,
30. ƒ(x) = mx + b, P = 0.05
L = m + b,
P = c 7 0
c = 1,
− 1 −p
1 2p
y = x sin 1x
1 2p 1 p
1 1 = 3 x2
x2  1 = 2 x  1
P = 0.1
29. ƒ(x) = mx + b, m 7 0, c = 1>2, P = c 7 0 m 7 0,
P = 0.5
P = 0.5
27. ƒ(x) = mx, m 7 0, L = 2m, c = 2, P = 0.03 28. ƒ(x) = mx,
P = 0.05
P = 0.4
xS9
15. ƒ(x) = x + 1,
P = 0.03
xS4
NOT TO SCALE
0 ƒ(x)  L 0 6 P.
1
x
2.3 The Precise Definition of a Limit
When Is a Number L Not the Limit of ƒ(x) as x u c? Showing L is not a limit We can prove that limx S c ƒ(x) ≠ L by providing an P 7 0 such that no possible d 7 0 satisfies the condition
1 50. lim x2 sin x = 0 xS0
85
y 1
for all x, 0 6 0 x  c 0 6 d
y = x2
0 ƒ(x)  L 0 6 P.
1
We accomplish this for our candidate P by showing that for each d 7 0 there exists a value of x such that y = x 2 sin 1x −1
0
2 −p
2 p
x
1
0 6 0x  c0 6 d
0 ƒ(x)  L 0 Ú P.
and
y y = f (x) L+P
−1
L
y = −x 2
L−P f (x)
Theory and Examples 51. Define what it means to say that lim g(x) = k.
0
xS0
52. Prove that lim ƒ(x) = L if and only if lim ƒ(h + c) = L. xSc
53. A wrong statement about limits Show by example that the following statement is wrong.
Explain why the function in your example does not have the given value of L as a limit as x S c. 54. Another wrong statement about limits Show by example that the following statement is wrong.
57. Let ƒ(x) = e
x, x 6 1 x + 1, x 7 1.
y y=x+1
The number L is the limit of ƒ(x) as x approaches c if, given any P 7 0, there exists a value of x for which 0 ƒ(x)  L 0 6 P.
2
Explain why the function in your example does not have the given value of L as a limit as x S c. T 55. Grinding engine cylinders Before contracting to grind engine cylinders to a crosssectional area of 9 in2, you need to know how much deviation from the ideal cylinder diameter of c = 3.385 in. you can allow and still have the area come within 0.01 in2 of the required 9 in2. To find out, you let A = p(x>2)2 and look for the interval in which you must hold x to make 0 A  9 0 … 0.01. What interval do you find?
56. Manufacturing electrical resistors Ohm’s law for electrical circuits like the one shown in the accompanying figure states that V = RI. In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which V will be 120 volts and I is to be 5 { 0.1 amp. In what interval does R have to lie for I to be within 0.1 amp of the value I0 = 5? − V +
I
R
x
c+ d
c
a value of x for which 0 < 0 x − c 0 < d and 0 f (x) − L 0 ≥ P
hS0
The number L is the limit of ƒ(x) as x approaches c if ƒ(x) gets closer to L as x approaches c.
c− d
y = f (x) 1 x
1 y=x
a. Let P = 1>2. Show that no possible d 7 0 satisfies the following condition: For all x, 0 6 0 x  1 0 6 d
1
0 ƒ(x)  2 0 6 1>2.
That is, for each d 7 0 show that there is a value of x such that 0 6 0x  10 6 d
and
0 ƒ(x)  2 0 Ú 1>2.
This will show that limx S 1 ƒ(x) ≠ 2. b. Show that limx S 1 ƒ(x) ≠ 1. c. Show that limx S 1 ƒ(x) ≠ 1.5.
86
Chapter 2: Limits and Continuity
60. a. For the function graphed here, show that limx S 1 g(x) ≠ 2.
x 2, x 6 2 58. Let h(x) = c 3, x = 2 2, x 7 2.
b. Does limx S 1 g(x) appear to exist? If so, what is the value of the limit? If not, why not? y
y
2 y = h(x)
4
y = g(x) 3
1 y=2
2 1
y = x2
0
2
x
0
−1 x
COMPUTER EXPLORATIONS In Exercises 61–66, you will further explore finding deltas graphically. Use a CAS to perform the following steps:
Show that a. lim h(x) ≠ 4 xS2
a. Plot the function y = ƒ(x) near the point c being approached.
b. lim h(x) ≠ 3
b. Guess the value of the limit L and then evaluate the limit symbolically to see if you guessed correctly.
xS2
c. lim h(x) ≠ 2 xS2
c. Using the value P = 0.2, graph the banding lines y1 = L  P and y2 = L + P together with the function ƒ near c.
59. For the function graphed here, explain why a. lim ƒ(x) ≠ 4
d. From your graph in part (c), estimate a d 7 0 such that for all x
xS3
b. lim ƒ(x) ≠ 4.8
0 6 0x  c0 6 d
xS3
c. lim ƒ(x) ≠ 3 xS3
y
1
0 ƒ(x)  L 0 6 P.
Test your estimate by plotting ƒ, y1, and y2 over the interval 0 6 0 x  c 0 6 d. For your viewing window use c  2d … x … c + 2d and L  2P … y … L + 2P. If any function values lie outside the interval 3L  P, L + P], your choice of d was too large. Try again with a smaller estimate.
e. Repeat parts (c) and (d) successively for P = 0.1, 0.05, and 0.001.
4.8 4
y = f (x)
3
5x3 + 9x2 , c = 0 2x5 + 3x2 x(1  cos x) 64. ƒ(x) = , c = 0 x  sin x
61. ƒ(x) =
x4  81 , c = 3 x  3
63. ƒ(x) =
sin 2x , c = 0 3x
65. ƒ(x) =
2x  1
66. ƒ(x) =
3x2  (7x + 1) 2x + 5 , c = 1 x  1
62. ƒ(x) =
3
0
3
x
x  1
, c = 1
2.4 OneSided Limits In this section we extend the limit concept to onesided limits, which are limits as x approaches the number c from the lefthand side (where x 6 c) or the righthand side (x 7 c) only.
Approaching a Limit from One Side To have a limit L as x approaches c, a function ƒ must be defined on both sides of c and its values ƒ(x) must approach L as x approaches c from either side. That is, ƒ must be defined in some open interval about c, but not necessarily at c. Because of this, ordinary limits are called twosided.
87
2.4 OneSided Limits
y y= x 0x0
1
x
0
−1
If ƒ fails to have a twosided limit at c, it may still have a onesided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a righthand limit. From the left, it is a lefthand limit. The function ƒ(x) = x> 0 x 0 (Figure 2.24) has limit 1 as x approaches 0 from the right, and limit 1 as x approaches 0 from the left. Since these onesided limit values are not the same, there is no single number that ƒ(x) approaches as x approaches 0. So ƒ(x) does not have a (twosided) limit at 0. Intuitively, if ƒ(x) is defined on an interval (c, b), where c 6 b, and approaches arbitrarily close to L as x approaches c from within that interval, then ƒ has righthand limit L at c. We write lim ƒ(x) = L.
x S c+
Figure 2.24 Different righthand and lefthand limits at the origin.
The symbol “x S c+ ” means that we consider only values of x greater than c. Similarly, if ƒ(x) is defined on an interval (a, c), where a 6 c and approaches arbitrarily close to M as x approaches c from within that interval, then ƒ has lefthand limit M at c. We write lim ƒ(x) = M.
x S c
The symbol “x S c ” means that we consider only xvalues less than c. These informal definitions of onesided limits are illustrated in Figure 2.25. For the function ƒ(x) = x> 0 x 0 in Figure 2.24 we have lim ƒ(x) = 1
lim ƒ(x) = 1.
and
x S 0+
x S 0
y
y
f (x)
L 0
c
M
f(x) x
x
0
x: c
(a) lim+ f (x) = L
x (b)
c
x
x: c
lim _ f(x) = M
Figure 2.25 (a) Righthand limit as x approaches c. (b) Lefthand limit as x approaches c. y
Example 1 The domain of ƒ(x) = 24  x2 is 3 2, 24 ; its graph is the semicircle in Figure 2.26. We have
y = "4 − x 2
−2
0
2
lim 24  x2 = 0
x S 2 +
x
Figure 2.26 The function ƒ(x) = 24  x2 has righthand limit 0 at x =  2 and lefthand limit 0 at x = 2 (Example 1).
and
lim 24  x2 = 0.
x S 2
The function does not have a lefthand limit at x = 2 or a righthand limit at x = 2. It does not have a twosided limit at either 2 or 2 because each point does not belong to an open interval over which ƒ is defined. Onesided limits have all the properties listed in Theorem 1 in Section 2.2. The righthand limit of the sum of two functions is the sum of their righthand limits, and so on. The theorems for limits of polynomials and rational functions hold with onesided limits, as do the Sandwich Theorem and Theorem 5. Onesided limits are related to limits in the following way.
THEOREM 6 A function ƒ(x) has a limit as x approaches c if and only if it has lefthand and righthand limits there and these onesided limits are equal: lim ƒ(x) = L
xSc
3
lim ƒ(x) = L
x S c
and
lim ƒ(x) = L.
x S c+
88
Chapter 2: Limits and Continuity
Example 2 For the function graphed in Figure 2.27,
y y = f (x)
2
At x = 0:
limx S 0+ ƒ(x) = 1, limx S 0 ƒ(x) and limx S 0 ƒ(x) do not exist. The function is not defined to the left of x = 0.
At x = 1:
limx S 1 ƒ(x) = 0 even though ƒ(1) = 1, limx S 1+ ƒ(x) = 1, limx S 1 ƒ(x) does not exist. The right and lefthand limits are not equal.
At x = 2:
limx S 2 ƒ(x) = 1, limx S 2+ ƒ(x) = 1, limx S 2 ƒ(x) = 1 even though ƒ(2) = 2.
At x = 3:
limx S 3 ƒ(x) = limx S 3+ ƒ(x) = limx S 3 ƒ(x) = ƒ(3) = 2.
At x = 4:
limx S 4 ƒ(x) = 1 even though ƒ(4) ≠ 1, limx S 4+ ƒ(x) and limx S 4 ƒ(x) do not exist. The function is not defined to the right of x = 4.
1
0
1
2
3
x
4
Figure 2.27 Graph of the function in Example 2.
y
At every other point c in 3 0, 44, ƒ(x) has limit ƒ(c).
L+P
f(x) f (x) lies in here
L
Precise Definitions of OneSided Limits The formal definition of the limit in Section 2.3 is readily modified for onesided limits.
L−P for all x ≠ c in here
DEFINITIONS We say that ƒ(x) has righthand limit L at c, and write
d x 0
c
c+d
x
lim ƒ(x) = L (see Figure 2.28)
x S c+
if for every number P 7 0 there exists a corresponding number d 7 0 such that for all x
Figure 2.28 Intervals associated with the definition of righthand limit.
c 6 x 6 c + d
1
0 ƒ(x)  L 0 6 P.
We say that ƒ has lefthand limit L at c, and write y
lim ƒ(x) = L (see Figure 2.29)
x S c
if for every number P 7 0 there exists a corresponding number d 7 0 such that for all x L+P L
c  d 6 x 6 c f(x) f (x) lies in here
0 ƒ(x)  L 0 6 P.
Example 3 Prove that
L−P
lim 2x = 0.
x S 0+
for all x ≠ c in here
Solution Let P 7 0 be given. Here c = 0 and L = 0, so we want to find a d 7 0 such that for all x
d x 0
1
c−d
c
Figure 2.29 Intervals associated with the definition of lefthand limit.
0 6 x 6 d
x
0 2x  0 0 6 P,
1
or 0 6 x 6 d
1
2x 6 P.
2.4 OneSided Limits
y
Squaring both sides of this last inequality gives x 6 P2
f (x) = "x
P
0 6 x 6 d.
if
If we choose d = P2 we have 0 6 x 6 d = P2
f(x)
L=0
89
x
d = P2
x
Figure 2.30 lim+ 1x = 0 in Example 3. xS0
1
2x 6 P,
or 0 6 x 6 P2
1
0 2x  0 0 6 P.
According to the definition, this shows that limx S 0+ 2x = 0 (Figure 2.30).
The functions examined so far have had some kind of limit at each point of interest. In general, that need not be the case.
Example 4 Show that y = sin (1>x) has no limit as x approaches zero from either side (Figure 2.31).
y 1
x
0
y = sin 1x −1
Figure 2.31 The function y = sin (1>x) has neither a righthand nor a lefthand limit as x approaches zero (Example 4). The graph here omits values very near the yaxis.
Solution As x approaches zero, its reciprocal, 1>x, grows without bound and the values of sin (1>x) cycle repeatedly from 1 to 1. There is no single number L that the function’s values stay increasingly close to as x approaches zero. This is true even if we restrict x to positive values or to negative values. The function has neither a righthand limit nor a lefthand limit at x = 0.
Limits Involving (sin U) , U
A central fact about (sin u)>u is that in radian measure its limit as u S 0 is 1. We can see this in Figure 2.32 and confirm it algebraically using the Sandwich Theorem. You will see the importance of this limit in Section 3.5, where instantaneous rates of change of the trigonometric functions are studied. y 1
−3p
−2p
y = sin u (radians) u
−p
p
2p
3p
u
NOT TO SCALE
Figure 2.32 The graph of ƒ(u) = (sin u)>u suggests that the rightand lefthand limits as u approaches 0 are both 1.
90
Chapter 2: Limits and Continuity
y
THEOREM 7—Limit of the Ratio sin U , U as U u 0 T
lim
uS0
1
sin u = 1 u
(u in radians)
(1)
P tan u 1 sin u u
cos u Q
O
A(1, 0)
x
1
Proof The plan is to show that the righthand and lefthand limits are both 1. Then we will know that the twosided limit is 1 as well. To show that the righthand limit is 1, we begin with positive values of u less than p>2 (Figure 2.33). Notice that Area ∆OAP 6 area sector OAP 6 area ∆OAT. We can express these areas in terms of u as follows:
FIGURE 2.33 The figure for the proof of Theorem 7. By definition, TA>OA = tan u, but OA = 1, so TA = tan u.
Area ∆OAP =
1 1 1 base * height = (1)(sin u) = sin u 2 2 2
Area sector OAP = Area ∆OAT =
u 1 2 1 r u = (1)2u = 2 2 2
(2)
1 1 1 base * height = (1)(tan u) = tan u. 2 2 2
Thus, Equation (2) is where radian measure comes in: The area of sector OAP is u>2 only if u is measured in radians.
1 1 1 sin u 6 u 6 tan u. 2 2 2 This last inequality goes the same way if we divide all three terms by the number (1>2) sin u, which is positive, since 0 6 u 6 p>2: 1 6
u 1 6 . cos u sin u
Taking reciprocals reverses the inequalities: 1 7
sin u 7 cos u. u
Since limu S 0+ cos u = 1 (Example 11b, Section 2.2), the Sandwich Theorem gives lim
u S 0+
sin u = 1. u
To consider the lefthand limit, we recall that sin u and u are both odd functions (Section 1.1). Therefore, ƒ(u) = (sin u)>u is an even function, with a graph symmetric about the yaxis (see Figure 2.32). This symmetry implies that the lefthand limit at 0 exists and has the same value as the righthand limit: lim
u S 0
sin u sin u = 1 = lim+ , u u uS0
so limu S 0 (sin u)>u = 1 by Theorem 6.
EXAMPLE 5
Show that (a) lim
hS0
cos h  1 = 0 h
and
(b) lim
xS0
sin 2x 2 = . 5x 5
2.4 OneSided Limits
91
Solution (a) Using the halfangle formula cos h = 1  2 sin2 (h>2), we calculate 2 sin2 (h>2) cos h  1 = lim h h hS0 hS0 lim
=  lim
uS0
sin u sin u u
Let u = h>2. Eq. (1) and Example 11a in Section 2.2
= (1)(0) = 0.
(b) Equation (1) does not apply to the original fraction. We need a 2x in the denominator,
not a 5x. We produce it by multiplying numerator and denominator by 2>5: (2>5) # sin 2x sin 2x = lim x S 0 5x x S 0 (2>5) # 5x lim
=
Now, Eq. (1) applies sin 2x 2 lim with u = 2x. 5 x S 0 2x
=
2 2 (1) = 5 5
Example 6 Find lim
tS0
tan t sec 2t . 3t
Solution From the definition of tan t and sec 2t, we have lim
tS0
tan t sec 2t 1 1 sin t 1 = lim # t # cos t # 3t cos 2t tS0 3 =
sin t # 1 # 1 1 lim 3 t S 0 t cos t cos 2t
=
1 1 (1)(1)(1) = . 3 3
Eq. (1) and Example 11b in Section 2.2
Exercises 2.4 Finding Limits Graphically 1. Which of the following statements about the function y = ƒ(x) graphed here are true, and which are false?
y y = f (x) 2
y y = f (x)
1
1
−1
0
1
2
x
−1
lim ƒ(x) = 0 a. lim + ƒ(x) = 1 b. x S 1
xS0
xS0
xS0
c. lim ƒ(x) = 1 d. lim ƒ(x) = lim+ ƒ(x) xS0
e. lim ƒ(x) exists. f. lim ƒ(x) = 0 xS0
xS0
g. lim ƒ(x) = 1 h. lim ƒ(x) = 1 xS0
xS1
0
1
2
3
x
lim ƒ(x) does not exist. a. lim + ƒ(x) = 1 b. x S 1
xS2
xS2
xS1
c. lim ƒ(x) = 2 d. lim ƒ(x) = 2 e. lim+ ƒ(x) = 1 f. lim ƒ(x) does not exist. xS1
xS1
g. lim+ ƒ(x) = lim ƒ(x) xS0
xS0
i. lim ƒ(x) = 0 j. lim ƒ(x) = 2
h. lim ƒ(x) exists at every c in the open interval ( 1, 1).
k. lim  ƒ(x) does not exist. l. lim+ ƒ(x) = 0
i. lim ƒ(x) exists at every c in the open interval (1, 3).
xS1
x S 1
xS2
xS2
2. Which of the following statements about the function y = ƒ(x) graphed here are true, and which are false?
xSc xSc
j.
lim ƒ(x) = 0 k. lim+ ƒ(x) does not exist.
x S 1
xS3
92
Chapter 2: Limits and Continuity
3  x,
6. Let g(x) = 2x sin(1>x).
x 6 2
3. Let ƒ(x) = c x + 1, x 7 2. 2
y 1
y = "x
y
y = "x sin 1x
y=3−x 3
1 2p
y= x+1 2 0
0
x
4
2
1 p
1
2 p
x
a. Find limx S 2+ ƒ(x) and limx S 2 ƒ(x). b. Does limx S 2 ƒ(x) exist? If so, what is it? If not, why not?
−1
c. Find limx S 4 ƒ(x) and limx S 4+ ƒ(x).
a. Does limx S 0+ g(x) exist? If so, what is it? If not, why not?
d. Does limx S 4 ƒ(x) exist? If so, what is it? If not, why not?
4. Let ƒ(x) = d
b. Does limx S 0 g(x) exist? If so, what is it? If not, why not?
3  x, x 6 2 2, x = 2 x , 2
c. Does limx S 0 g(x) exist? If so, what is it? If not, why not? x 3, x≠1 0, x = 1. b. Find limx S 1 ƒ(x) and limx S 1+ ƒ(x).
7. a. Graph ƒ(x) = e
x 7 2.
y
c. Does limx S 1 ƒ(x) exist? If so, what is it? If not, why not? 1  x 2, x≠1 2, x = 1. b. Find limx S 1+ ƒ(x) and limx S 1 ƒ(x).
y=3−x
8. a. Graph ƒ(x) = e
3 y= x 2 0
−2
y = −"x
2
c. Does limx S 1 ƒ(x) exist? If so, what is it? If not, why not?
x
Graph the functions in Exercises 9 and 10. Then answer these questions. a. Find limx S 2+ ƒ(x), limx S 2 ƒ(x), and ƒ(2).
a. What are the domain and range of ƒ?
b. Does limx S 2 ƒ(x) exist? If so, what is it? If not, why not?
b. At what points c, if any, does limx S c ƒ(x) exist?
c. Find limx S 1 ƒ(x) and limx S 1+ ƒ(x).
c. At what points does only the lefthand limit exist?
d. Does limx S 1 ƒ(x) exist? If so, what is it? If not, why not?
d. At what points does only the righthand limit exist?
0, 5. Let ƒ(x) = c
21  x2,
x … 0
9. ƒ(x) = c 1, 2,
1 sin x , x 7 0.
y 1
x
0 y=
x≤0 1 sin x , x > 0
0,
−1
x, 10. ƒ(x) = c 1, 0,
b. Does limx S 0 ƒ(x) exist? If so, what is it? If not, why not? c. Does limx S 0 ƒ(x) exist? If so, what is it? If not, why not?
1 … x 6 0, or 0 6 x … 1 x = 0 x 6  1 or x 7 1
Finding OneSided Limits Algebraically Find the limits in Exercises 11–18.
11.
x + 2 + 1
lim
x S 0.5 A x
13. lim + a x S 2
a. Does limx S 0+ ƒ(x) exist? If so, what is it? If not, why not?
0 … x 6 1 1 … x 6 2 x = 2
14. lim a xS1
15. lim+ hS0
x 2x + 5 ba b x + 1 x2 + x
12. lim+
x + 6 3  x 1 ba x ba b 7 x + 1
2h2 + 4h + 5  25
h
xS1
x  1 Ax + 2
2.5 Continuity
16. limhS0
26  25h2 + 11h + 6
17. a. lim +(x + 3) x S 2
18. a. lim+ xS1
41. lim
uS0
h
0x + 20 x + 2
22x (x  1)
0x  10
b. lim (x + 3) x S 2
0x + 20 x + 2
22x (x  1) b. lim x S 10x  10
Use the graph of the greatest integer function y = : x ; , Figure 1.10 in Section 1.1, to help you find the limits in Exercises 19 and 20. :u ; :u ; 19. a. lim+ b. lim uS3 u u S 3 u 20. a. lim+(t  : t ; ) b. lim(t  : t ; ) tS4
tS4
sin U = 1 U Find the limits in Exercises 21–42. Using lim
Uu 0
sin 22u 22u sin 3y 23. lim y S 0 4y tan 2x 2 5. lim x xS0
22. lim
21. lim
uS0
27. 29. 31. 33. 35. 37.
tS0
h h S 0 sin 3h 2t 26. lim tan t tS0 24. lim
x csc 2x lim x S 0 cos 5x x + x cos x lim x S 0 sin x cos x 1  cos u lim u S 0 sin 2u sin (1  cos t) lim 1  cos t tS0 sin u lim u S 0 sin 2u lim u cos u
28. lim 6x2(cot x)(csc 2x) xS0
xS0
x2  x + sin x 2x xS0 x  x cos x lim xS0 sin2 3x sin (sin h) lim sin h hS0 sin 5x lim x S 0 sin 4x lim sin u cot 2u
30. lim 32. 34. 36. 38.
uS0
39. lim
sin kt (k constant) t
tan 3x sin 8x
uS0
40. lim
yS0
sin 3y cot 5y y cot 4y
tan u u 2 cot 3u
42. lim
uS0
93
u cot 4u sin2 u cot2 2u
Theory and Examples 43. Once you know limx S a+ ƒ(x) and limx S a ƒ(x) at an interior point of the domain of ƒ, do you then know limx S a ƒ(x)? Give reasons for your answer.
44. If you know that limx S c ƒ(x) exists, can you find its value by calculating limx S c+ ƒ(x)? Give reasons for your answer. 45. Suppose that ƒ is an odd function of x. Does knowing that limx S 0+ ƒ(x) = 3 tell you anything about limx S 0 ƒ(x)? Give reasons for your answer. 46. Suppose that ƒ is an even function of x. Does knowing that limx S 2 ƒ(x) = 7 tell you anything about either limx S 2 ƒ(x) or limx S 2+ ƒ(x)? Give reasons for your answer. Formal Definitions of OneSided Limits 47. Given P 7 0, find an interval I = (5, 5 + d), d 7 0, such that if x lies in I, then 2x  5 6 P. What limit is being verified and what is its value?
48. Given P 7 0, find an interval I = (4  d, 4), d 7 0, such that if x lies in I, then 24  x 6 P. What limit is being verified and what is its value? Use the definitions of righthand and lefthand limits to prove the limit statements in Exercises 49 and 50. x x  2 = 1 50. lim+ = 1 49. limxS0 0 x 0 xS2 0 x  2 0 51. Greatest integer function Find (a) limx S 400+ : x ; and (b) limx S 400 : x ; ; then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can you say anything about limx S 400 : x ; ? Give reasons for your answer.
52. Onesided limits Let ƒ(x) = e
x2 sin (1>x), x 6 0 2x, x 7 0.
Find (a) limx S 0+ ƒ(x) and (b) limx S 0 ƒ(x); then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can you say anything about limx S 0 ƒ(x)? Give reasons for your answer.
2.5 Continuity y
Distance fallen (m)
500
Q4 Q3
375 Q2
250 125 0
When we plot function values generated in a laboratory or collected in the field, we often connect the plotted points with an unbroken curve to show what the function’s values are likely to have been at the points we did not measure (Figure 2.34). In doing so, we are assuming that we are working with a continuous function, so its outputs vary regularly and consistently with the inputs, and do not jump abruptly from one value to another without taking on the values in between. Intuitively, any function y = ƒ(x) whose graph can be sketched over its domain in one unbroken motion is an example of a continuous function. Such functions play an important role in the study of calculus and its applications.
Q1
5 Elapsed time (sec)
10
t
Figure 2.34 Connecting plotted points by an unbroken curve from experimental data Q1, Q2, Q3, c for a falling object.
Continuity at a Point To understand continuity, it helps to consider a function like that in Figure 2.35, whose limits we investigated in Example 2 in the last section.
94
Chapter 2: Limits and Continuity
Example 1 At which numbers does the function ƒ in Figure 2.35 appear to be not continuous? Explain why. What occurs at other numbers in the domain? Solution First we observe that the domain of the function is the closed interval 3 0, 44, so we will be considering the numbers x within that interval. From the figure, we notice right away that there are breaks in the graph at the numbers x = 1, x = 2, and x = 4. The breaks appear as jumps, which we identify later as “jump discontinuities.” These are numbers for which the function is not continuous, and we discuss each in turn.
y y = f (x)
2 1
1
0
2
3
Numbers at which the graph of ƒ has breaks:
x
4
Figure 2.35 The function is not continuous at x = 1, x = 2, and x = 4 (Example 1).
At x = 1, the function fails to have a limit. It does have both a lefthand limit, limx S 1 ƒ(x) = 0, as well as a righthand limit, limx S 1+ ƒ(x) = 1, but the limit values are different, resulting in a jump in the graph. The function is not continuous at x = 1. At x = 2, the function does have a limit, limx S 2 ƒ(x) = 1, but the value of the function is ƒ(2) = 2. The limit and function values are not the same, so there is a break in the graph and ƒ is not continuous at x = 2. At x = 4, the function does have a lefthand limit at this right endpoint, limx S 4 ƒ(x) = 1, but again the value of the function ƒ(4) = 12 differs from the value of the limit. We see again a break in the graph of the function at this endpoint and the function is not continuous from the left. Numbers at which the graph of ƒ has no breaks: At x = 0, the function has a righthand limit at this left endpoint, limx S 0+ ƒ(x) = 1, and the value of the function is the same, ƒ(0) = 1. So no break occurs in the graph of the function at this endpoint, and the function is continuous from the right at x = 0. At x = 3, the function has a limit, limx S 3 ƒ(x) = 2. Moreover, the limit is the same value as the function there, ƒ(3) = 2. No break occurs in the graph and the function is continuous at x = 3. At all other numbers x = c in the domain, which we have not considered, the function has a limit equal to the value of the function at the point, so limx S c ƒ(x) = ƒ(c). For example, limx S 5>2 ƒ(x) = ƒ1 52 2 = 32 . No breaks appear in the graph of the function at any of these remaining numbers and the function is continuous at each of them. The following definitions capture the continuity ideas we observed in Example 1.
DEFINITIONS Let c be a real number on the xaxis. The function ƒ is continuous at c if lim ƒ(x) = ƒ(c).
xSc
The function ƒ is rightcontinuous at c (or continuous from the right) if lim ƒ(x) = ƒ(c).
x S c+
Continuity from the right
Twosided continuity
The function ƒ is leftcontinuous at c (or continuous from the left) if Continuity from the left
lim ƒ(x) = ƒ(c).
x S c
y = f (x) a
c
b
x
Figure 2.36 Continuity at points a, b, and c.
From Theorem 6, it follows immediately that a function ƒ is continuous at an interior point c of its domain if and only if it is both rightcontinuous and leftcontinuous at c (Figure 2.36). We say that a function is continuous over a closed interval 3 a, b4 if it is rightcontinuous at a, leftcontinuous at b, and continuous at all interior points of the interval.
2.5 Continuity
This definition applies to the infinite closed intervals 3 a, q) and ( q, b4 as well, but only one endpoint is involved. If a function is not continuous at an interior point c of its domain, we say that ƒ is discontinuous at c, and that c is a point of discontinuity of ƒ. Note that a function ƒ can be continuous, rightcontinuous, or leftcontinuous only at a point c for which ƒ(c) is defined.
y y = "4 −
2
x2
x
2
0
−2
95
Example 2 The function ƒ(x) = 24  x2 is continuous over its domain 3 2, 24
Figure 2.37 A function that is continuous over its domain (Example 2).
(Figure 2.37). It is rightcontinuous at x = 2, and leftcontinuous at x = 2.
Example 3 The unit step function U(x), graphed in Figure 2.38, is rightcontinuous at x = 0, but is neither leftcontinuous nor continuous there. It has a jump discontinuity at x = 0.
y
We summarize continuity at an interior point in the form of a test. y = U(x)
1
Continuity Test
x
0
A function ƒ(x) is continuous at a point x = c if and only if it meets the following three conditions. 1. ƒ(c) exists (c lies in the domain of ƒ). 2. limx S c ƒ(x) exists (ƒ has a limit as x S c). 3. limx S c ƒ(x) = ƒ(c) (the limit equals the function value).
Figure 2.38 A function that has a jump discontinuity at the origin (Example 3).
For onesided continuity and continuity at an endpoint of an interval, the limits in parts 2 and 3 of the test should be replaced by the appropriate onesided limits.
Example 4 The function y = : x ; introduced in Section 1.1 is graphed in Figure 2.39. It is discontinuous at every integer because the lefthand and righthand limits are not equal as x S n:
y 4
lim : x ; = n  1 and lim+ : x ; = n .
x S n
3 2 1 1
−1
2
xSn
Since : n ; = n, the greatest integer function is rightcontinuous at every integer n (but not leftcontinuous). The greatest integer function is continuous at every real number other than the integers. For example,
y = :x;
3
4
lim : x ; = 1 = : 1.5 ; .
x
x S 1.5
In general, if n  1 6 c 6 n, n an integer, then −2
Figure 2.39 The greatest integer function is continuous at every noninteger point. It is rightcontinuous, but not leftcontinuous, at every integer point (Example 4).
lim : x ; = n  1 = : c ; .
xSc
Figure 2.40 displays several common types of discontinuities. The function in Figure 2.40a is continuous at x = 0. The function in Figure 2.40b would be continuous if it had ƒ(0) = 1. The function in Figure 2.40c would be continuous if ƒ(0) were 1 instead of 2. The discontinuity in Figure 2.40c is removable. The function has a limit as x S 0, and we can remove the discontinuity by setting ƒ(0) equal to this limit. The discontinuities in Figure 2.40d through f are more serious: limx S 0 ƒ(x) does not exist, and there is no way to improve the situation by changing ƒ at 0. The step function in Figure 2.40d has a jump discontinuity: The onesided limits exist but have different values. The function ƒ(x) = 1>x2 in Figure 2.40e has an infinite discontinuity. The function in Figure 2.40f has an oscillating discontinuity: It oscillates too much to have a limit as x S 0.
96
Chapter 2: Limits and Continuity
y = f (x)
0
2
y = f (x)
1
1 x
x
1 x
0
(b)
(c) y
y y = f(x) = 12 x
1
−1 0
y = f (x)
1
0
(a)
y
y
y
y
y = f(x)
0
x
(d) y = sin 1x
0
x
x −1 (f)
(e)
Figure 2.40 The function in (a) is continuous at x = 0; the functions in (b) through (f ) are not.
Continuous Functions Generally, we want to describe the continuity behavior of a function throughout its entire domain, not only at a single point. We know how to do that if the domain is a closed interval. In the same way, we define a continuous function as one that is continuous at every point in its domain. This is a property of the function. A function always has a specified domain, so if we change the domain, we change the function, and this may change its continuity property as well. If a function is discontinuous at one or more points of its domain, we say it is a discontinuous function.
Example 5 (a) The function y = 1>x (Figure 2.41) is a continuous function because it is continuous
at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = 0. (b) The identity function ƒ(x) = x and constant functions are continuous everywhere by Example 3, Section 2.3.
y y = 1x
Algebraic combinations of continuous functions are continuous wherever they are defined. 0
x
Figure 2.41 The function y = 1>x is continuous over its natural domain. It has a point of discontinuity at the origin, so it is discontinuous on any interval containing x = 0 (Example 5).
THEOREM 8—Properties of Continuous Functions If the functions ƒ and g are continuous at x = c, then the following algebraic combinations are continuous at x = c. 1. Sums: 2. Differences: 3. Constant multiples: 4. Products: 5. Quotients: 6. Powers:
ƒ + g ƒ  g k # ƒ, for any number k ƒ#g ƒ>g, provided g(c) ≠ 0 ƒn, n a positive integer n
7. Roots: 2ƒ, provided it is defined on an open interval containing c, where n is a positive integer
2.5 Continuity
97
Most of the results in Theorem 8 follow from the limit rules in Theorem 1, Section 2.2. For instance, to prove the sum property we have lim (ƒ + g)(x) = lim (ƒ(x) + g(x))
xSc
xSc
= lim ƒ(x) + lim g(x)
Sum Rule, Theorem 1
= ƒ(c) + g(c) = (ƒ + g)(c).
Continuity of ƒ, g at c
xSc
xSc
This shows that ƒ + g is continuous.
Example 6
(a) Every polynomial P(x) = an xn + an  1xn  1 + g + a0 is continuous because
lim P(x) = P(c) by Theorem 2, Section 2.2.
xSc
(b) If P(x) and Q(x) are polynomials, then the rational function P(x)>Q(x) is continuous
wherever it is defined (Q(c) ≠ 0) by Theorem 3, Section 2.2.
Example 7 The function ƒ(x) = 0 x 0 is continuous. If x 7 0, we have ƒ(x) = x,
a polynomial. If x 6 0, we have ƒ(x) = x, another polynomial. Finally, at the origin, limx S 0 0 x 0 = 0 = 0 0 0 .
The functions y = sin x and y = cos x are continuous at x = 0 by Example 11 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 70). It follows from Theorem 8 that all six trigonometric functions are then continuous wherever they are defined. For example, y = tan x is continuous on g ∪ (p>2, p>2) ∪ (p>2, 3p>2) ∪ g.
Inverse Functions and Continuity The inverse function of any function continuous on an interval is continuous over its domain. This result is suggested by the observation that the graph of ƒ 1 , being the reflection of the graph of ƒ across the line y = x, cannot have any breaks in it when the graph of ƒ has no breaks. A rigorous proof that ƒ 1 is continuous whenever ƒ is continuous on an interval is given in more advanced texts. It follows that the inverse trigonometric functions are all continuous over their domains. We defined the exponential function y = ax in Section 1.5 informally by its graph. Recall that the graph was obtained from the graph of y = ax for x a rational number by “filling in the holes” at the irrational points x, so the function y = ax was defined to be continuous over the entire real line. The inverse function y = loga x is also continuous. In particular, the natural exponential function y = ex and the natural logarithm function y = ln x are both continuous over their domains.
Composites All composites of continuous functions are continuous. The idea is that if ƒ(x) is continuous at x = c and g(x) is continuous at x = ƒ(c), then g ∘ ƒ is continuous at x = c (Figure 2.42). In this case, the limit as x S c is g(ƒ(c)). g f ˚ Continuous at c
c
f
g
Continuous at c
Continuous at f(c) f (c)
Figure 2.42 Composites of continuous functions are continuous.
g( f(c))
98
Chapter 2: Limits and Continuity
THEOREM 9—Composite of Continuous Functions If ƒ is continuous at c and g is continuous at ƒ(c), then the composite g ∘ ƒ is continuous at c. Intuitively, Theorem 9 is reasonable because if x is close to c, then ƒ(x) is close to ƒ(c), and since g is continuous at ƒ(c), it follows that g(ƒ(x)) is close to g(ƒ(c)). The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of a proof of Theorem 9, see Exercise 6 in Appendix 4.
EXAMPLE 8
Show that the following functions are continuous on their natural
domains. (a) y = 2x2  2x  5 (c) y = ` y
x  2 ` x2  2
(b) y =
x2>3 1 + x4
(d) y = `
x sin x ` x2 + 2
Solution
(a) The square root function is continuous on 3 0, q) because it is a root of the continu0.4 0.3 0.2 0.1 −2p
−p
0
p
2p
FIGURE 2.43 The graph suggests that y = 0 (x sin x)>(x2 + 2) 0 is continuous (Example 8d).
x
ous identity function ƒ(x) = x (Part 7, Theorem 8). The given function is then the composite of the polynomial ƒ(x) = x2  2x  5 with the square root function g(t) = 2t , and is continuous on its natural domain. (b) The numerator is the cube root of the identity function squared; the denominator is an everywherepositive polynomial. Therefore, the quotient is continuous. (c) The quotient (x  2)>(x2  2) is continuous for all x ≠ { 22, and the function is the composition of this quotient with the continuous absolute value function (Example 7). (d) Because the sine function is everywherecontinuous (Exercise 70), the numerator term x sin x is the product of continuous functions, and the denominator term x2 + 2 is an everywherepositive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.43). Theorem 9 is actually a consequence of a more general result, which we now state and prove. THEOREM 10—Limits of Continuous Functions and limx S c ƒ(x) = b, then
If g is continuous at the point b
limx S c g(ƒ(x)) = g(b) = g(limx S c ƒ(x)).
Proof Let P 7 0 be given. Since g is continuous at b, there exists a number d1 7 0 such that
0 g(y)  g(b) 0 6 P whenever 0 6 0 y  b 0 6 d1. Since limx S c ƒ(x) = b, there exists a d 7 0 such that
0 ƒ(x)  b 0 6 d1 whenever 0 6 0 x  c 0 6 d. If we let y = ƒ(x), we then have that
0 y  b 0 6 d1 whenever 0 6 0 x  c 0 6 d, which implies from the first statement that 0 g(y)  g(b) 0 = 0 g(ƒ(x))  g(b) 0 6 P whenever 0 6 0 x  c 0 6 d. From the definition of limit, this proves that limx S c g(ƒ(x)) = g(b).
2.5 Continuity
99
Example 9 As an application of Theorem 10, we have the following calculations. (a)
lim cos a2x + sin a
x S p/2
(b) lim sin1 a xS1
3p 3p + xb b = cos a lim 2x + lim sin a + xb b 2 2 x S p/2 x S p/2 = cos (p + sin 2p) = cos p = 1.
1  x 1  x b = sin1 a lim b 2 S x 1 1  x 1  x2 = sin1 a lim
xS1
= sin1 We sometimes denote eu by exp u when u is a complicated mathematical expression.
(c)
Arcsine is continuous.
1 b 1 + x
Cancel common factor (1  x).
1 p = 2 6
lim 2x + 1 etan x = lim 2x + 1 # exp a lim tan xb
xS0
xS0
= 1 # e0 = 1
xS0
Exponential is continuous.
Intermediate Value Theorem for Continuous Functions Functions that are continuous on intervals have properties that make them particularly useful in mathematics and its applications. One of these is the Intermediate Value Property. A function is said to have the Intermediate Value Property if whenever it takes on two values, it also takes on all the values in between. THEOREM 11—The Intermediate Value Theorem for Continuous Functions If ƒ is a continuous function on a closed interval 3 a, b4, and if y0 is any value between ƒ(a) and ƒ(b), then y0 = ƒ(c) for some c in 3 a, b4. y y = f (x) f (b)
y0 f (a)
y
0
a
c
b
x
3 2 1
0
1
2
3
4
x
Figure 2.44 The function 2x  2, 1 … x 6 2 ƒ(x) = e 3, 2 … x … 4 does not take on all values between ƒ(1) = 0 and ƒ(4) = 3; it misses all the values between 2 and 3.
Theorem 11 says that continuous functions over finite closed intervals have the Intermediate Value Property. Geometrically, the Intermediate Value Theorem says that any horizontal line y = y0 crossing the yaxis between the numbers ƒ(a) and ƒ(b) will cross the curve y = ƒ(x) at least once over the interval 3 a, b4 . The proof of the Intermediate Value Theorem depends on the completeness property of the real number system (Appendix 7) and can be found in more advanced texts. The continuity of ƒ on the interval is essential to Theorem 11. If ƒ is discontinuous at even one point of the interval, the theorem’s conclusion may fail, as it does for the function graphed in Figure 2.44 (choose y0 as any number between 2 and 3). A Consequence for Graphing: Connectedness Theorem 11 implies that the graph of a function continuous on an interval cannot have any breaks over the interval. It will be connected—a single, unbroken curve. It will not have jumps like the graph of the greatest integer function (Figure 2.39), or separate branches like the graph of 1>x (Figure 2.41).
100
Chapter 2: Limits and Continuity
A Consequence for Root Finding We call a solution of the equation ƒ(x) = 0 a root of the equation or zero of the function ƒ. The Intermediate Value Theorem tells us that if ƒ is continuous, then any interval on which ƒ changes sign contains a zero of the function. In practical terms, when we see the graph of a continuous function cross the horizontal axis on a computer screen, we know it is not stepping across. There really is a point where the function’s value is zero.
Example 10 Show that there is a root of the equation x3  x  1 = 0 between 1 and 2. Solution Let ƒ(x) = x3  x  1. Since ƒ(1) = 1  1  1 = 1 6 0 and ƒ(2) = 23  2  1 = 5 7 0, we see that y0 = 0 is a value between ƒ(1) and ƒ(2). Since ƒ is continuous, the Intermediate Value Theorem says there is a zero of ƒ between 1 and 2. Figure 2.45 shows the result of zooming in to locate the root near x = 1.32. 5
1
1
1.6
2
−1 −2
−1 (a)
(b)
0.02
0.003
1.320
1.330
−0.02
1.3240
1.3248
−0.003 (c)
(d)
Figure 2.45 Zooming in on a zero of the function ƒ(x) = x3  x  1. The zero is near x = 1.3247 (Example 10).
Example 11 Use the Intermediate Value Theorem to prove that the equation y 4
has a solution (Figure 2.46).
y = 4 − x2
Solution We rewrite the equation as
3
22x + 5 + x2 = 4,
2 y = " 2x + 5 1 0
22x + 5 = 4  x2
c
2
Figure 2.46 The curves y = 22x + 5 and y = 4  x2 have the same value at x = c where 22x + 5 = 4  x2 (Example 11).
x
and set ƒ(x) = 22x + 5 + x2 . Now g(x) = 22x + 5 is continuous on the interval 3 5>2, q) since it is the composite of the square root function with the nonnegative linear function y = 2x + 5. Then ƒ is the sum of the function g and the quadratic function y = x2, and the quadratic function is continuous for all values of x. It follows that ƒ(x) = 22x + 5 + x2 is continuous on the interval 3 5>2, q). By trial and error, we find the function values ƒ(0) = 25 ≈ 2.24 and ƒ(2) = 29 + 4 = 7, and note that ƒ is also continuous on the finite closed interval 3 0, 24 ⊂ 3 5>2, q). Since the value y0 = 4 is between the numbers 2.24 and 7, by the Intermediate Value Theorem there is a number c∊[0, 24 such that ƒ(c) = 4. That is, the number c solves the original equation.
101
2.5 Continuity
Continuous Extension to a Point Sometimes the formula that describes a function ƒ does not make sense at a point x = c. It might nevertheless be possible to extend the domain of ƒ, to include x = c, creating a new function that is continuous at x = c. For example, the function y = ƒ(x) = (sin x)>x is continuous at every point except x = 0, since the origin is not in its domain. Since y = (sin x)>x has a finite limit as x S 0 (Theorem 7), we can extend the function’s domain to include the point x = 0 in such a way that the extended function is continuous at x = 0. We define the new function sin x , F(x) = • x 1,
x ≠ 0 x = 0.
The function F(x) is continuous at x = 0 because sin x lim x = F(0),
xS0
so it meets the requirements for continuity (Figure 2.47). y
y (0, 1)
−
p 2
p 2 a− , b 2 p
(0, 1)
f (x) p 2 a , b 2 p
0
x
p 2
−
p 2
p 2 a− , b 2 p
(a)
0
F(x) p 2 a , b 2 p
p 2
x
(b)
Figure 2.47 The graph (a) of ƒ(x) = (sin x)>x for  p>2 … x … p>2 does not include the point (0, 1) because the function is not defined at x = 0. (b) We can remove the discontinuity from the graph by defining the new function F(x) with F(0) = 1 and F(x) = ƒ(x) everywhere else. Note that F(0) = limx S 0 ƒ(x).
More generally, a function (such as a rational function) may have a limit at a point where it is not defined. If ƒ(c) is not defined, but limx S c ƒ(x) = L exists, we can define a new function F(x) by the rule y 2
y=
F(x) = e
2
x +x−6 x2 − 4
1 −1
0
1
−1
2
3
4
x
The function F is continuous at x = c. It is called the continuous extension of ƒ to x = c. For rational functions ƒ, continuous extensions are often found by canceling common factors in the numerator and denominator.
ƒ(x) =
x+3 y= x+2
x2 + x  6 , x ≠ 2 x2  4
has a continuous extension to x = 2, and find that extension.
1 0
if x is in the domain of ƒ if x = c.
Example 12 Show that
(a)
y 5 4
2
ƒ(x), L,
1
2
3
4
x
Solution Although ƒ(2) is not defined, if x ≠ 2 we have
(b)
Figure 2.48 (a) The graph of ƒ(x) and (b) the graph of its continuous extension F(x) (Example 12).
ƒ(x) =
x2 + x  6 (x  2)(x + 3) x + 3 = = . (x  2)(x + 2) x + 2 x2  4
The new function F(x) =
x + 3 x + 2
102
Chapter 2: Limits and Continuity
is equal to ƒ(x) for x ≠ 2, but is continuous at x = 2, having there the value of 5>4. Thus F is the continuous extension of ƒ to x = 2, and x2 + x  6 5 = lim ƒ(x) = . 4 xS2 xS2 x2  4 lim
The graph of ƒ is shown in Figure 2.48. The continuous extension F has the same graph except with no hole at (2, 5>4). Effectively, F is the function ƒ with its point of discontinuity at x = 2 removed.
Exercises 2.5 Continuity from Graphs In Exercises 1–4, say whether the function graphed is continuous on 3 1, 34. If not, where does it fail to be continuous and why?
1.
2.
y
y
−1
1
2
x
3
d. Is ƒ continuous at x = 1?
0
−1
3.
1
3
x
4.
2
2
1
1
0
1
2
x
3
b. Is ƒ continuous at x = 2?
10. To what new value should ƒ(1) be changed to remove the discontinuity?
y = k(x)
1
0
−1
2
Exercises 5–10 refer to the function
3
x
Applying the Continuity Test At which points do the functions in Exercises 11 and 12 fail to be continuous? At which points, if any, are the discontinuities removable? Not removable? Give reasons for your answers.
11. Exercise 1, Section 2.4
2
x  1, 2x, ƒ(x) = e 1,  2x + 4, 0,
1 0 x 1 2
… x 6 x = 1 6 x 6 x
6 0 6 1 6 2 6 3
y
13. y =
1  3x x  2
14. y =
1 + 4 (x + 2)2
15. y =
x + 1 x2  4x + 3
16. y =
x + 3 x2  3x  10
y = −2x + 4
18. y =
21. y = csc 2x
22. y = tan
(1, 1) −1 y = x2 − 1
0
1
2
−1
The graph for Exercises 5–10.
3
x
x2 1 0x0 + 1 2
17. y = 0 x  1 0 + sin x
19. y =
(1, 2)
y = 2x
12. Exercise 2, Section 2.4
At what points are the functions in Exercises 13–30 continuous?
graphed in the accompanying figure.
y = f (x) 2
7. a. Is ƒ defined at x = 2? (Look at the definition of ƒ.)
9. What value should be assigned to ƒ(2) to make the extended function continuous at x = 2?
y
y = h(x)
−1
2
8. At what values of x is ƒ continuous?
y
d. Is ƒ continuous at x = 1?
c. Does limx S 1 ƒ(x) = ƒ(1)?
1
0
c. Does limx S 1+ ƒ(x) = ƒ( 1)?
b. Does limx S 1 ƒ(x) exist?
y = g(x) 2
1
b. Does limx S 1+ ƒ(x) exist?
6. a. Does ƒ(1) exist?
y = f (x) 2
5. a. Does ƒ(1) exist?
23. y =
cos x x
x tan x x2 + 1
25. y = 22x + 3
27. y = (2x  1)1>3
x + 2 20. y = cos x
24. y =
px 2
2x4 + 1
1 + sin2 x
4 26. y = 2 3x  1
28. y = (2  x)1>5
2.5 Continuity
47. For what values of a and b is
x2  x  6 , x ≠ 3 29. g(x) = c x  3 5, x = 3 x3  8 , x ≠ 2, x ≠  2 x2  4 30. ƒ(x) = d 3, x = 2 4, x = 2 Limits Involving Trigonometric Functions Find the limits in Exercises 31–38. Are the functions continuous at the point being approached? p 32. lim sina cos (tan t)b 31. lim sin (x  sin x) 2 xSp tS0
33. lim sec (y sec2 y  tan2 y  1) yS1
34. lim tana xS0
p cos ( sin x1>3 ) b 4
35. lim cos a
p b 36. lim 2csc2 x + 5 13 tan x x S p/6 219  3 sec 2t
xS0
xS1
tS0
37. lim+ sin a
p 2x e b 2
38. lim cos1 1 ln 2x 2
Continuous Extensions 39. Define g(3) in a way that extends g(x) = (x2  9)>(x  3) to be continuous at x = 3.
40. Define h(2) in a way that extends h(t) = (t 2 + 3t  10)>(t  2) to be continuous at t = 2. 41. Define ƒ(1) in a way that extends ƒ(s) = (s3  1)>(s2  1) to be continuous at s = 1. 42. Define g(4) in a way that extends g(x) = (x2  16)> (x2  3x  4) to be continuous at x = 4. 43. For what value of a is x2  1, x 6 3 ƒ(x) = e 2ax, x Ú 3
continuous at every x? 44. For what value of b is
x, x 6 2 bx2, x Ú  2
continuous at every x?
continuous at every x? 48. For what values of a and b is ax + 2b, x … 0 g(x) = c x2 + 3a  b, 0 6 x … 2 3x  5, x 7 2 continuous at every x? T In Exercises 49–52, graph the function ƒ to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at x = 0. If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function’s value(s) should be? 49. ƒ(x) = 51. ƒ(x) =
10 x  1 x sin x 0x0
50. ƒ(x) =
10 0 x 0  1 x
52. ƒ(x) = (1 + 2x)1>x
Theory and Examples 53. A continuous function y = ƒ(x) is known to be negative at x = 0 and positive at x = 1. Why does the equation ƒ(x) = 0 have at least one solution between x = 0 and x = 1? Illustrate with a sketch.
54. Explain why the equation cos x = x has at least one solution. 55. Roots of a cubic Show that the equation x3  15x + 1 = 0 has three solutions in the interval 3 4, 4]. 56. A function value Show that the function F(x) = (x  a)2 # (x  b)2 + x takes on the value (a + b)>2 for some value of x.
57. Solving an equation If ƒ(x) = x3  8x + 10, show that there are values c for which ƒ(c) equals (a) p; (b)  23; (c) 5,000,000. 58. Explain why the following five statements ask for the same information. b. Find the xcoordinates of the points where the curve y = x3 crosses the line y = 3x + 1. c. Find all the values of x for which x3  3x = 1. d. Find the xcoordinates of the points where the cubic curve y = x3  3x crosses the line y = 1.
45. For what values of a is a2x  2a, x Ú 2 12, x 6 2
continuous at every x? 46. For what value of b is x  b , x 6 0 g(x) = c b + 1 x2 + b, x 7 0 continuous at every x?
2, x … 1 ƒ(x) = c ax  b,  1 6 x 6 1 3, x Ú 1
a. Find the roots of ƒ(x) = x3  3x  1.
g(x) = e
ƒ(x) = b
103
e. Solve the equation x3  3x  1 = 0. 59. Removable discontinuity Give an example of a function ƒ(x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how you know that ƒ is discontinuous at x = 2, and how you know the discontinuity is removable. 60. Nonremovable discontinuity Give an example of a function g(x) that is continuous for all values of x except x =  1, where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.
104
Chapter 2: Limits and Continuity
61. A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function ƒ(x) = e
1, if x is rational 0, if x is irrational
68. The signpreserving property of continuous functions Let ƒ be defined on an interval (a, b) and suppose that ƒ(c) ≠ 0 at some c where ƒ is continuous. Show that there is an interval (c  d, c + d) about c where ƒ has the same sign as ƒ(c). 69. Prove that ƒ is continuous at c if and only if lim ƒ(c + h) = ƒ(c).
hS0
70. Use Exercise 69 together with the identities
is discontinuous at every point. b. Is ƒ rightcontinuous or leftcontinuous at any point? 62. If functions ƒ(x) and g(x) are continuous for 0 … x … 1, could ƒ(x)>g(x) possibly be discontinuous at a point of 30, 14? Give reasons for your answer.
63. If the product function h(x) = ƒ(x) # g(x) is continuous at x = 0, must ƒ(x) and g(x) be continuous at x = 0? Give reasons for your answer.
64. Discontinuous composite of continuous functions Give an example of functions ƒ and g, both continuous at x = 0, for which the composite ƒ ∘ g is discontinuous at x = 0. Does this contradict Theorem 9? Give reasons for your answer. 65. Neverzero continuous functions Is it true that a continuous function that is never zero on an interval never changes sign on that interval? Give reasons for your answer. 66. Stretching a rubber band Is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for your answer. 67. A fixed point theorem Suppose that a function ƒ is continuous on the closed interval 30, 14 and that 0 … ƒ(x) … 1 for every x in 30, 14 . Show that there must exist a number c in 30, 14 such that ƒ(c) = c (c is called a fixed point of ƒ).
sin (h + c) = sin h cos c + cos h sin c, cos (h + c) = cos h cos c  sin h sin c to prove that both ƒ(x) = sin x and g(x) = cos x are continuous at every point x = c. Solving Equations Graphically T Use the Intermediate Value Theorem in Exercises 71–78 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations.
71. x3  3x  1 = 0 72. 2x3  2x2  2x + 1 = 0 73. x(x  1)2 = 1 (one root) 74. xx = 2 75. 2x + 21 + x = 4
76. x3  15x + 1 = 0 (three roots) 77. cos x = x (one root). Make sure you are using radian mode. 78. 2 sin x = x (three roots). Make sure you are using radian mode.
2.6 Limits Involving Infinity; Asymptotes of Graphs In this section we investigate the behavior of a function when the magnitude of the independent variable x becomes increasingly large, or x S { q . We further extend the concept of limit to infinite limits, which are not limits as before, but rather a new use of the term limit. Infinite limits provide useful symbols and language for describing the behavior of functions whose values become arbitrarily large in magnitude. We use these limit ideas to analyze the graphs of functions having horizontal or vertical asymptotes.
y 4 3 y = 1x
2 1 −1 0 −1
1
2
3
4
Figure 2.49 The graph of y = 1>x approaches 0 as x S q or x S  q .
x
Finite Limits as x u t H The symbol for infinity (q) does not represent a real number. We use q to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For example, the function ƒ(x) = 1>x is defined for all x ≠ 0 (Figure 2.49). When x is positive and becomes increasingly large, 1>x becomes increasingly small. When x is negative and its magnitude becomes increasingly large, 1>x again becomes small. We summarize these observations by saying that ƒ(x) = 1>x has limit 0 as x S q or x S  q, or that 0 is a limit of ƒ(x) = 1>x at infinity and negative infinity. Here are precise definitions.
2.6 Limits Involving Infinity; Asymptotes of Graphs
105
DEFINITIONS 1. We say that ƒ(x) has the limit L as x approaches infinity and write lim ƒ(x) = L
xS q
if, for every number P 7 0, there exists a corresponding number M such that for all x 1
x 7 M
0 ƒ(x)  L 0 6 P.
2. We say that ƒ(x) has the limit L as x approaches minus infinity and write lim ƒ(x) = L
xS q
if, for every number P 7 0, there exists a corresponding number N such that for all x x 6 N
y y = 1x
1 N = −P
0
y = –P
We prove the second result in Example 1, and leave the first to Exercises 87 and 88.
Example 1 Show that
y=P
M = 1P
0 ƒ(x)  L 0 6 P.
Intuitively, limx Sq ƒ(x) = L if, as x moves increasingly far from the origin in the positive direction, ƒ(x) gets arbitrarily close to L. Similarly, limx S q ƒ(x) = L if, as x moves increasingly far from the origin in the negative direction, ƒ(x) gets arbitrarily close to L. The strategy for calculating limits of functions as x S { q is similar to the one for finite limits in Section 2.2. There we first found the limits of the constant and identity functions y = k and y = x. We then extended these results to other functions by applying Theorem 1 on limits of algebraic combinations. Here we do the same thing, except that the starting functions are y = k and y = 1>x instead of y = k and y = x. The basic facts to be verified by applying the formal definition are 1 and lim lim k = k = 0. (1) x S {q x S {q x
No matter what positive number P is, the graph enters 1 this band at x = P and stays.
P
1
1
x
–P
No matter what positive number P is, the graph enters 1 this band at x = − P and stays.
Figure 2.50 The geometry behind the argument in Example 1.
1
(a) lim x = 0 x Sq
(b) lim x = 0. x S q
Solution (a) Let P 7 0 be given. We must find a number M such that for all x x 7 M
1
1 1 ` x  0 ` = ` x ` 6 P.
The implication will hold if M = 1>P or any larger positive number (Figure 2.50). This proves limx Sq (1>x) = 0. (b) Let P 7 0 be given. We must find a number N such that for all x x 6 N
1
1 1 ` x  0 ` = ` x ` 6 P.
The implication will hold if N = 1>P or any number less than 1>P (Figure 2.50). This proves limx S  q (1>x) = 0. Limits at infinity have properties similar to those of finite limits. THEOREM 12 All the Limit Laws in Theorem 1 are true when we replace limx S c by limx S q or limx S  q . That is, the variable x may approach a finite number c or { q.
106
Chapter 2: Limits and Continuity
Example 2 The properties in Theorem 12 are used to calculate limits in the same way as when x approaches a finite number c. 1
1
(a) lim a5 + x b = lim 5 + lim x xS q xS q xS q
= 5 + 0 = 5
(b)
lim
xS  q
Sum Rule Known limits
p 23 1 1 = lim p 23 # x # x xS  q x2
1 1 = lim p 23 # lim x # lim x xS  q xS  q xS  q
= p 23 # 0 # 0 = 0
2
−5
1
Line y = 5 3
0
5
10
Known limits
Limits at Infinity of Rational Functions
2 y = 5x +2 8x − 3 3x + 2
y
Product Rule
To determine the limit of a rational function as x S { q, we first divide the numerator and denominator by the highest power of x in the denominator. The result then depends on the degrees of the polynomials involved. x
Example 3 These examples illustrate what happens when the degree of the numerator is less than or equal to the degree of the denominator.
−1
5 + (8>x)  (3>x2) Divide numerator and 5x2 + 8x  3 = lim denominator by x2 . xS q xS q 3x2 + 2 3 + (2>x2)
(a) lim −2
NOT TO SCALE
Figure 2.51 The graph of the function in Example 3a. The graph approaches the line y = 5>3 as 0 x 0 increases. y 8 y= 6
0
(11>x2) + (2>x3) 11x + 2 = lim x S  q 2x 3  1 xS  q 2  (1>x3)
Divide numerator and denominator by x3 .
lim
=
0 + 0 = 0 2  0
See Fig. 2.52.
Horizontal Asymptotes
2 −2
(b)
5 + 0  0 5 = See Fig. 2.51. 3 3 + 0
Cases for which the degree of the numerator is greater than the degree of the denominator are illustrated in Examples 10 and 14.
11x + 2 2x 3 − 1
4
−4
=
2
4
6
x
If the distance between the graph of a function and some fixed line approaches zero as a point on the graph moves increasingly far from the origin, we say that the graph approaches the line asymptotically and that the line is an asymptote of the graph. Looking at ƒ(x) = 1>x (see Figure 2.49), we observe that the xaxis is an asymptote of the curve on the right because
−2
1 lim x = 0
xS q
−4 −6
and on the left because 1 lim x = 0.
−8
Figure 2.52 The graph of the function in Example 3b. The graph approaches the xaxis as 0 x 0 increases.
xS  q
We say that the xaxis is a horizontal asymptote of the graph of ƒ(x) = 1>x. DEFINITION A line y = b is a horizontal asymptote of the graph of a function y = ƒ(x) if either lim ƒ(x) = b
xS q
or
lim ƒ(x) = b.
xS  q
2.6 Limits Involving Infinity; Asymptotes of Graphs
107
The graph of the function ƒ(x) =
5x2 + 8x  3 3x2 + 2
sketched in Figure 2.51 (Example 3a) has the line y = 5>3 as a horizontal asymptote on both the right and the left because lim ƒ(x) =
xS q
5 3
xS  q
5 . 3
Example 4 Find the horizontal asymptotes of the graph of
y 2
ƒ(x) =
y=1
y = −1
lim ƒ(x) =
and
x
0 f(x) = −2
x3  2 . 0x03 + 1
Solution We calculate the limits as x S { q.
x3 − 2 0x03 + 1
For x Ú 0:
Figure 2.53 The graph of the function in Example 4 has two horizontal asymptotes.
For x 6 0:
1  (2>x3) x3  2 x3  2 = lim = lim = 1. 3 xS q 0 x 0 3 + 1 xS q x + 1 x S q 1 + (1>x 3) lim
1  ( 2>x3 ) x3  2 x3  2 = lim = lim = 1. x S  q ( x ) 3 + 1 x S  q 1 + ( 1>x 3 ) xS  q 0 x 0 3 + 1 lim
The horizontal asymptotes are y = 1 and y = 1. The graph is displayed in Figure 2.53. Notice that the graph crosses the horizontal asymptote y = 1 for a positive value of x.
Example 5 The xaxis (the line y = 0) is a horizontal asymptote of the graph of
y
y = ex because
lim ex = 0.
xS  q
y = ex
To see this, we use the definition of a limit as x approaches  q . So let P 7 0 be given, but arbitrary. We must find a constant N such that for all x,
1
x 6 N P N = ln P
x
1
0 ex  0 0 6 P.
Now 0 ex  0 0 = ex , so the condition that needs to be satisfied whenever x 6 N is ex 6 P.
Figure 2.54 The graph of y = e approaches the xaxis as x S  q (Example 5).
x
Let x = N be the number where ex = P. Since ex is an increasing function, if x 6 N , then ex 6 P. We find N by taking the natural logarithm of both sides of the equation eN = P, so N = ln P (see Figure 2.54). With this value of N the condition is satisfied, and we conclude that limx S  q ex = 0.
Example 6 Find (a) limq sin (1>x) and (b) limq x sin (1>x). xS{
xS
Solution (a) We introduce the new variable t = 1>x. From Example 1, we know that t S 0+ as x S q (see Figure 2.49). Therefore, 1 lim sin x = lim+ sin t = 0. tS0
xS q
108
Chapter 2: Limits and Continuity
(b) We calculate the limits as x S q and x S  q:
y
sin t 1 lim x sin x = lim+ t = 1
1
xS q
x
1
Example 7 Find lime1>x . xS0
Solution We let t = 1>x. From Figure 2.49, we can see that t S  q as x S 0 . (We make this idea more precise further on.) Therefore,
y
−3
−2
lim e1>x = lim et = 0 Example 5
1 0.8 0.6 0.4 0.2
tS  q
x S 0
(Figure 2.56). The Sandwich Theorem also holds for limits as x S { q. You must be sure, though, that the function whose limit you are trying to find stays between the bounding functions at very large values of x in magnitude consistent with whether x S q or x S  q.
x
0
−1
tS0
Likewise, we can investigate the behavior of y = ƒ(1>x) as x S 0 by investigating y = ƒ(t) as t S { q , where t = 1>x.
Figure 2.55 The line y = 1 is a horizontal asymptote of the function graphed here (Example 6b).
y = e1x
sin t 1 lim x sin x = lim t = 1.
xS  q
The graph is shown in Figure 2.55, and we see that the line y = 1 is a horizontal asymptote.
1 y = x sin x −1
and
tS0
Figure 2.56 The graph of y = e1>x for x 6 0 shows limx S 0 e1>x = 0 (Example 7).
Example 8 Using the Sandwich Theorem, find the horizontal asymptote of the curve y = 2 +
sin x x .
Solution We are interested in the behavior as x S { q. Since y
sin x 1 0 … ` x ` … `x`
y = 2 + sinx x
and limx S {q 0 1>x 0 = 0, we have limx S {q (sin x)>x = 0 by the Sandwich Theorem. Hence,
2 1 −3p −2p −p
0
p
2p
3p
lim a2 +
x S {q
x
Figure 2.57 A curve may cross one of its asymptotes infinitely often (Example 8).
sin x x b = 2 + 0 = 2,
and the line y = 2 is a horizontal asymptote of the curve on both left and right (Figure 2.57). This example illustrates that a curve may cross one of its horizontal asymptotes many times.
Example 9 Find limq 1 x  2x2 + 16 2. xS
Solution Both of the terms x and 2x2 + 16 approach infinity as x S q, so what happens to the difference in the limit is unclear (we cannot subtract q from q because the symbol does not represent a real number). In this situation we can multiply the numerator and the denominator by the conjugate radical expression to obtain an equivalent algebraic result: lim 1 x  2x2 + 16 2 = lim 1 x  2x2 + 16 2
xS q
xS q
= lim
x2  (x2 + 16)
xS q x
2
+ 2x + 16
x + 2x2 + 16 x + 2x2 + 16
= lim
xS q x
16 . + 2x2 + 16
2.6 Limits Involving Infinity; Asymptotes of Graphs
109
As x S q, the denominator in this last expression becomes arbitrarily large, so we see that the limit is 0. We can also obtain this result by a direct calculation using the Limit Laws:
16 lim = lim x S q x + 2x 2 + 16 xS q
Oblique Asymptotes 2 y= x −3=x+1+ 1 2x − 4 2 2x − 4
y
The vertical distance between curve and line goes to zero as x : ∞
6
3
Oblique asymptote
x=2 y= x+1 2
2
1
2
3
4
ƒ(x) =
x2  3 2x  4
in Figure 2.58.
1 −1 0 −1
If the degree of the numerator of a rational function is 1 greater than the degree of the denominator, the graph has an oblique or slant line asymptote. We find an equation for the asymptote by dividing numerator by denominator to express ƒ as a linear function plus a remainder that goes to zero as x S { q.
Example 10 Find the oblique asymptote of the graph of
5 4
16  x 0 = = 0. 2 1 + 21 + 0 16 x 1 + + x2 A x2
x
x
Solution We are interested in the behavior as x S { q . We divide (2x  4) into (x2  3):
−2
x + 1 2 2x  4) x2  3 x2  2x 2x  3 2x  4 1
−3
Figure 2.58 The graph of the function in Example 10 has an oblique asymptote.
This tells us that ƒ(x) =
x2  3 x 1 ≤. = ¢ + 1≤ + ¢ 2x  4 2()* 2x  4 (1)1*
linear g(x) remainder y
B
You can get as high as you want by taking x close enough to 0. No matter how high B is, the graph goes higher. y = 1x
x 0
You can get as low as you want by taking x close enough to 0.
x
x No matter how low −B is, the graph goes lower.
−B
Figure 2.59 Onesided infinite limits: 1 1 and lim =  q. lim = q x S 0+ x x S 0 x
As x S { q , the remainder, whose magnitude gives the vertical distance between the graphs of ƒ and g, goes to zero, making the slanted line g(x) =
x + 1 2
an asymptote of the graph of ƒ (Figure 2.58). The line y = g(x) is an asymptote both to the right and to the left. The next subsection will confirm that the function ƒ(x) grows arbitrarily large in absolute value as x S 2 (where the denominator is zero), as shown in the graph. Notice in Example 10 that if the degree of the numerator in a rational function is greater than the degree of the denominator, then the limit as 0 x 0 becomes large is + q or  q, depending on the signs assumed by the numerator and denominator.
Infinite Limits Let us look again at the function ƒ(x) = 1>x. As x S 0+, the values of ƒ grow without bound, eventually reaching and surpassing every positive real number. That is, given any positive real number B, however large, the values of ƒ become larger still (Figure 2.59).
110
Chapter 2: Limits and Continuity
Thus, ƒ has no limit as x S 0+. It is nevertheless convenient to describe the behavior of ƒ by saying that ƒ(x) approaches q as x S 0+. We write 1 lim ƒ(x) = lim+ x = q.
x S 0+
xS0
In writing this equation, we are not saying that the limit exists. Nor are we saying that there is a real number q, for there is no such number. Rather, we are saying that limx S 0+ (1>x) does not exist because 1>x becomes arbitrarily large and positive as x S 0+. As x S 0 , the values of ƒ(x) = 1>x become arbitrarily large and negative. Given any negative real number B, the values of ƒ eventually lie below B. (See Figure 2.59.) We write 1 lim ƒ(x) = lim x =  q. xS0
x S 0
Again, we are not saying that the limit exists and equals the number  q. There is no real number  q. We are describing the behavior of a function whose limit as x S 0 does not exist because its values become arbitrarily large and negative.
y
y=
1 x−1
EXAMPLE 11
1 −1
0
1
2
3
x
Find lim+ xS1
lim
lim
x S 1
1 . x  1
1 = q x  1
and
lim
x S 1
1 =  q. x  1
Analytic Solution Think about the number x  1 and its reciprocal. As x S 1+, we have (x  1) S 0+ and 1>(x  1) S q. As x S 1, we have (x  1) S 0  and 1>(x  1) S  q.
EXAMPLE 12
Discuss the behavior of ƒ(x) =
1 x2
as
x S 0.
Solution As x approaches zero from either side, the values of 1>x2 are positive and become arbitrarily large (Figure 2.61). This means that
y
lim ƒ(x) = lim
No matter how high B is, the graph goes higher.
B
xS0
x
xS0
1 = q. x2
The function y = 1>x shows no consistent behavior as x S 0. We have 1>x S q if x S 0+, but 1>x S  q if x S 0. All we can say about limx S 0 (1>x) is that it does not exist. The function y = 1>x2 is different. Its values approach infinity as x approaches zero from either side, so we can say that limx S 0 (1>x2) = q.
f (x) = 12 x
x 0
and
Geometric Solution The graph of y = 1>(x  1) is the graph of y = 1>x shifted 1 unit to the right (Figure 2.60). Therefore, y = 1>(x  1) behaves near 1 exactly the way y = 1>x behaves near 0: x S 1+
FIGURE 2.60 Near x = 1, the function y = 1>(x  1) behaves the way the function y = 1>x behaves near x = 0. Its graph is the graph of y = 1>x shifted 1 unit to the right (Example 11).
1 x  1
x
FIGURE 2.61 The graph of ƒ(x) in Example 12 approaches infinity as x S 0.
EXAMPLE 13
These examples illustrate that rational functions can behave in various ways near zeros of the denominator. (x  2)2 (x  2)2 x  2 = lim = lim = 0 x S 2 x2  4 x S 2 (x  2)(x + 2) xS2 x + 2
(a) lim (b) lim
xS2
x  2 x  2 1 1 = lim = lim = x2  4 x S 2 (x  2)(x + 2) x S 2 x + 2 4
2.6 Limits Involving Infinity; Asymptotes of Graphs
(c) lim+
x  3 x  3 = lim =  q x2  4 x S 2+ (x  2)(x + 2)
The values are negative for x 7 2, x near 2.
(d) lim
x  3 x  3 = lim = q x2  4 x S 2 (x  2)(x + 2)
The values are positive for x 6 2, x near 2.
xS2
xS2
(e) lim
x  3 x  3 = lim does not exist. x2  4 x S 2 (x  2)(x + 2)
(f ) lim
(x  2) 2  x 1 = q 3 = xlim 3 = xlim S S 2 2 (x  2) (x  2) (x  2)2
xS2
xS2
111
See parts (c) and (d).
In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f ), where cancellation still leaves a zero factor in the denominator. y
2x5  6x4 + 1 . x S  q 3x 2 + x  7
Example 14 Find lim
Solution We are asked to find the limit of a rational function as x S  q , so we divide the numerator and denominator by x2 , the highest power of x in the denominator:
y = f (x)
lim
B
xS  q
2x5  6x4 + 1 = 3x2 + x  7
lim
xS  q
2x3  6x2 + x2 3 + x1  7x2
2x2 (x  3) + x2 x S  q 3 + x 1  7x2 =  q, xn S 0, x  3 S  q
= lim 0
c−d
c
x c+d
because the numerator tends to  q while the denominator approaches 3 as x S  q .
Figure 2.62 For c  d 6 x 6 c + d, the graph of ƒ(x) lies above the line y = B.
y c−d
c
c+d x
0
Precise Definitions of Infinite Limits Instead of requiring ƒ(x) to lie arbitrarily close to a finite number L for all x sufficiently close to c, the definitions of infinite limits require ƒ(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seen before. Figures 2.62 and 2.63 accompany these definitions.
DEFINITIONs 1. We say that ƒ(x) approaches infinity as x approaches c, and write lim ƒ(x) = q,
xSc
if for every positive real number B there exists a corresponding d 7 0 such that for all x
−B
y = f (x)
0 6 0x  c0 6 d
1
ƒ(x) 7 B.
2. We say that ƒ(x) approaches minus infinity as x approaches c, and write lim ƒ(x) =  q,
xSc
Figure 2.63 For c  d 6 x 6 c + d, the graph of ƒ(x) lies below the line y = B.
if for every negative real number B there exists a corresponding d 7 0 such that for all x 0 6 0x  c0 6 d
1
ƒ(x) 6 B.
The precise definitions of onesided infinite limits at c are similar and are stated in the exercises.
112
Chapter 2: Limits and Continuity
Example 15 Prove that lim 12 = q. xS0
x
Solution Given B 7 0, we want to find d 7 0 such that 0 6 x  0 6 d implies
1 7 B. x2
1 7 B x2
x2 6
Now, if and only if
1 B
or, equivalently,
0x0 6 1 . 2B
Thus, choosing d = 1> 2B (or any smaller positive number), we see that
0 x 0 6 d implies
Therefore, by definition,
lim
xS0
1 1 7 2 Ú B. x2 d
1 = q. x2
Vertical Asymptotes Notice that the distance between a point on the graph of ƒ(x) = 1>x and the yaxis approaches zero as the point moves vertically along the graph and away from the origin (Figure 2.64). The function ƒ(x) = 1>x is unbounded as x approaches 0 because
y Vertical asymptote
Horizontal asymptote
1 0
lim
1
Horizontal asymptote, y=0
1
x S 0+ x
y = 1x x
Vertical asymptote, x=0
and
lim
1
x S 0 x
=  q.
We say that the line x = 0 (the yaxis) is a vertical asymptote of the graph of ƒ(x) = 1>x. Observe that the denominator is zero at x = 0 and the function is undefined there. DEFINITION A line x = a is a vertical asymptote of the graph of a function y = ƒ(x) if either lim ƒ(x) = { q
x S a+
Figure 2.64 The coordinate axes are asymptotes of both branches of the hyperbola y = 1>x.
= q
or
lim ƒ(x) = { q.
x S a
Example 16 Find the horizontal and vertical asymptotes of the curve y =
x + 3 . x + 2
Solution We are interested in the behavior as x S { q and the behavior as x S 2, where the denominator is zero. The asymptotes are quickly revealed if we recast the rational function as a polynomial with a remainder, by dividing (x + 2) into (x + 3): 1 x + 2) x + 3 x + 2 1
2.6 Limits Involving Infinity; Asymptotes of Graphs
This result enables us to rewrite y as:
y Vertical asymptote, x = −2
6 5
y=
4
y = 1 +
x+3 x+2
=1+
3
Horizontal asymptote, y=1
1 x+2
2 1
−5 −4 −3 −2 −1 0 −1
113
1
2
3
x
−2
1 . x + 2
As x S { q , the curve approaches the horizontal asymptote y = 1; as x S 2, the curve approaches the vertical asymptote x = 2. We see that the curve in question is the graph of ƒ(x) = 1>x shifted 1 unit up and 2 units left (Figure 2.65). The asymptotes, instead of being the coordinate axes, are now the lines y = 1 and x = 2.
Example 17 Find the horizontal and vertical asymptotes of the graph of
−3 −4
ƒ(x) = 
Figure 2.65 The lines y = 1 and x = 2 are asymptotes of the curve in Example 16.
8 . x2  4
Solution We are interested in the behavior as x S { q and as x S {2, where the denominator is zero. Notice that ƒ is an even function of x, so its graph is symmetric with respect to the yaxis. (a) The behavior as x S { q. Since lim x Sq ƒ(x) = 0, the line y = 0 is a horizontal
asymptote of the graph to the right. By symmetry it is an asymptote to the left as well (Figure 2.66). Notice that the curve approaches the xaxis from only the negative side (or from below). Also, ƒ(0) = 2. (b) The behavior as x S {2. Since
y 8 7 6 5 4 3 2 1
Vertical asymptote, x = −2
y=−
8 x2 − 4
Vertical asymptote, x = 2
lim ƒ(x) =  q
Horizontal asymptote, y = 0
−4−3−2−1 0
1 2 3 4
and
x S 2+
x
lim ƒ(x) = q,
x S 2
the line x = 2 is a vertical asymptote both from the right and from the left. By symmetry, the line x = 2 is also a vertical asymptote. There are no other asymptotes because ƒ has a finite limit at all other points.
Example 18 The graph of the natural logarithm function has the yaxis (the line Figure 2.66 Graph of the function in Example 17. Notice that the curve approaches the xaxis from only one side. Asymptotes do not have to be twosided.
y
x = 0) as a vertical asymptote. We see this from the graph sketched in Figure 2.67 (which is the reflection of the graph of the natural exponential function across the line y = x) and the fact that the xaxis is a horizontal asymptote of y = ex (Example 5). Thus, lim ln x =  q.
x S 0+
The same result is true for y = loga x whenever a 7 1.
Example 19 The curves y = ex
1 y = sec x = cos x
4 3 2
and
sin x y = tan x = cos x
both have vertical asymptotes at oddinteger multiples of p>2, where cos x = 0 (Figure 2.68).
y = ln x
1 −1 −1
1
2
3
4
x
Figure 2.67 The line x = 0 is a vertical asymptote of the natural logarithm function (Example 18).
Dominant Terms In Example 10 we saw that by long division we could rewrite the function ƒ(x) =
x2  3 2x  4
114
Chapter 2: Limits and Continuity y
y
y = sec x
1
1 −
3p −p p − 2 2
0
y = tan x
p 2
p
3p 2
x −
0 p 3p −p p − −1 2 2 2
p
3p 2
x
Figure 2.68 The graphs of sec x and tan x have infinitely many vertical asymptotes (Example 19).
as a linear function plus a remainder term:
y
y
This tells us immediately that
20
500,000
ƒ(x) ≈
15
300,000
ƒ(x) ≈
10 f (x)
5 −2
−1
g(x) = 3x 4
0
1
2
x
−5 (a) y
2
100,000 x
−20
−10
0
For 0 x 0 large,
1 is near 0. 2x  4
1 For x near 2, this term is very large in absolute value. 2x  4
100,000 If we want to know how ƒ behaves, this is the way to find out. It behaves like y = (x>2) + 1 when x is large xand the contribution of 1>(2x  4) to the total value of ƒ −20 −10 0 It behaves 10 20 1>(2x  4) when x is so close to 2 that 1>(2x  4) makes is insignificant. like −100,000 the dominant contribution. We say that (x>2) + 1 dominates when x is numerically large, and we say that (b) 1>(2x  4) dominates when x is near 2. Dominant terms like these help us predict a function’s behavior.
although ƒ and g are quite different for numerically small values of x, they are virtually identical for 0 x 0 very large, in the sense that their ratios approach 1 as x S q or x S  q .
300,000
1
x + 1 2
x 1 b. + 1b + a 2 2x  4
Example 20 Let ƒ(x) = 3x4  2x3 + 3x2  5x + 6 and g(x) = 3x4. Show that
500,000
g(x) = 3x 4
ƒ(x) = a
10
20
x
Solution The graphs of ƒ and g behave quite differently near the origin (Figure 2.69a), but appear as virtually identical on a larger scale (Figure 2.69b). We can test that the term 3x4 in ƒ, represented graphically by g, dominates the polynomial ƒ for numerically large values of x by examining the ratio of the two functions as x S { q. We find that
−100,000
lim
x S {q
(b)
Figure 2.69 The graphs of ƒ and g are (a) distinct for 0 x 0 small, and (b) nearly identical for 0 x 0 large (Example 20).
ƒ(x) 3x4  2x3 + 3x2  5x + 6 = lim g(x) x S {q 3x4 = lim a1 x S {q
= 1,
5 2 1 2 + + b 3x x2 3x3 x4
which means that ƒ and g appear nearly identical when 0 x 0 is large.
Summary
In this chapter we presented several important calculus ideas that are made meaningful and precise by the concept of the limit. These include the three ideas of the exact rate of change of a function, the slope of the graph of a function at a point, and the continuity of a function. The primary methods used for calculating limits of many functions are captured in the algebraic
2.6 Limits Involving Infinity; Asymptotes of Graphs
115
Limit Laws of Theorem 1 and in the Sandwich Theorem, all of which are proved from the precise definition of the limit. We saw that these computational rules also apply to onesided limits and to limits at infinity. Moreover, we can sometimes apply these rules when calculating limits of simple transcendental functions, as illustrated by our examples or in cases like the following: lim
xS0
ex  1 ex  1 1 1 1 = lim x = lim = = . 2x e  1 x S 0 (e  1)(ex + 1) x S 0 ex + 1 1 + 1 2
However, calculating more complicated limits involving transcendental functions such as lim
xS0
ln x 1 x lim x , and lim a1 + x b xS0 xS0
x , e  1 2x
requires more than simple algebraic techniques. The derivative is exactly the tool we need to calculate limits such as these (see Section 4.5), and this notion is the main subject of our next chapter.
Exercises 2.6 Finding Limits 1. For the function ƒ whose graph is given, determine the following limits.
a. lim ƒ(x) xS2
b. lim + ƒ(x) c. lim  ƒ(x) x S 3
2 3. ƒ(x) = x  3
x S 3
d. lim ƒ(x) e. lim+ ƒ(x) f. lim ƒ(x) x S 3
xS0
In Exercises 3–8, find the limit of each function (a) as x S q and (b) as x S  q. (You may wish to visualize your answer with a graphing calculator or computer.)
xS0
g. lim ƒ(x) h. lim ƒ(x) i. lim ƒ(x)
5. g(x) =
7. h(x) =
xS0
x S q
x Sq
y 3 2
−6 −5 −4 −3 −2 −1 −1
1
2
3
4 5
6
x
11. lim
t S q
−2
2. For the function ƒ whose graph is given, determine the following limits. a. lim ƒ(x) b. lim+ ƒ(x) c. lim ƒ(x) xS2
xS2
d. lim ƒ(x) e. lim + ƒ(x) f. lim  ƒ(x) xS2
x S 3
x S 3
g. lim ƒ(x) h. lim+ ƒ(x) i. lim ƒ(x) x S 3
xS0
x S q
x Sq
3  (1>x )
2x + 3 5x + 7 x + 1 15. ƒ(x) = 2 x + 3 13. ƒ(x) =
8. h(x) =
3
f
2 1
12. lim
r Sq
4 + ( 22>x2)
r + sin r 2r + 7  5 sin r
1
2
3
4 5
6
x
2x3 + 7 x3  x2 + x + 7 3x + 7 16. ƒ(x) = 2 x  2 14. ƒ(x) =
7x3 x  3x2 + 6x
18. h(x) =
9x4 + x 2x + 5x2  x + 6
19. g(x) =
10x5 + x4 + 31 x6
20. g(x) =
x3 + 7x2  2 x2  x + 1
21. f(x) =
3x7 + 5x2  1 6x3  7x + 3
22. h(x) =
5x8  2x3 + 9 3 + x  4x5
3
4
Limits as x u H or x u − H The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36.
−2 −3
3  (2>x)
17. h(x) =
y
−6 −5 −4 −3 −2 −1 −1
2
2  t + sin t t + cos t
xS0
j. lim ƒ(x) k. lim ƒ(x) l. lim ƒ(x) xS0
5 + (7>x)
Limits of Rational Functions In Exercises 13–22, find the limit of each rational function (a) as x S q and (b) as x S  q.
−3
xS4
1 2 + (1>x)
Find the limits in Exercises 9–12. cos u sin 2x 10. lim 9. lim x x Sq u S q 3u
f
1
2 x2 1 6. g(x) = 8  (5>x2)
4. ƒ(x) = p 
8x2  3 x Sq A 2x 2 + x
23. lim
24. lim ¢ x S q
x2 + x  1 1>3 ≤ 8x2  3
116
Chapter 2: Limits and Continuity
25. lim ¢ xS  q
1  x3 5 ≤ x2 + 7x
27. lim
2 2x + x1 3x  7
29. lim
2x  2x
xS q
xS  q
3
5
3
5
2x + 2x
28. lim
xS q
2x5>3  x1>3 + 7 x S q x 8>5 + 3x + 2x
31. lim 33. lim
xS q
2x2 + 1
x + 1
x  3 x S q 24x 2 + 25
35. lim
x2  5x x S q A x3 + x  2
26. lim
2 + 2x 2  2x
x1 + x4 30. lim 2 xS q x  x3 3
32. lim
xS  q
34. lim
xS  q
2x  5x + 3
2x + x2>3  4 2x2 + 1
x + 1
4  3x3 q x S  2x 6 + 9
36. lim
Infinite Limits Find the limits in Exercises 37–48. 1 37. lim+ 38. x S 0 3x 3 39. lim 40. xS2 x  2 2x 41. lim + 42. x S 8 x + 8 4 43. lim 44. x S 7 (x  7)2
lim
x S 0
5 2x
1 x  3 3x lim x S 5 2x + 10 1 lim x S 0 x 2(x + 1) lim
x S 3+
51. lim (1 + csc u) uS0
50.
lim
x S (p>2) +
sec x
52. lim (2  cot u) uS0
Find the limits in Exercises 53–58. 1 53. lim 2 as x  4 a. x S 2+ b. x S 2+ c. x S  2 d. x S  2x 54. lim 2 as x  1 a. x S 1+ b. x S 1+ c. x S  1 d. x S  1x2 1 55. lim a  x b as 2 a. x S 0+ b. x S 03 c. x S 2 2 d. x S 1
x2  1 as 2x + 4 a. x S 2+ b. x S 2 56. lim
c. x S 1 d. xS0 +
e. What, if anything, can be said about the limit as x S 0? Find the limits in Exercises 59–62. 3 b as t 1>3 a. t S 0+ b. t S 0
59. lima2 

1 2 + b as x2>3 (x  1)2>3 a. x S 0+ b. x S 0+ c. x S 1 d. x S 1 61. lima
1 1 b as x1>3 (x  1)4>3 a. x S 0+ b. x S 0 c. x S 1+ d. x S 1 62. lima
Graphing Simple Rational Functions Graph the rational functions in Exercises 63–68. Include the graphs and equations of the asymptotes and dominant terms.
Find the limits in Exercises 49–52. lim tan x
x2  3x + 2 as x3  4x + a. x S 2 b. x S 2+ c. x S 0 d. x S 1+
58. lim
1 + 7b as t 3>5 a. t S 0+ b. t S 0
2 2 b. lim x S 0 3x 1>3 x S 0 3x 1>3 2 2 46. a. lim+ 1>5 b. lim xS0 x x S 0 x 1>5 4 1 47. lim 2>5 48. lim 2>3 xS0 x xS0 x
x S (p>2)
e. What, if anything, can be said about the limit as x S 0?
60. lima
45. a. lim+
49.
x2  3x + 2 as x3  2x2 a. x S 0+ b. x S 2+ c. x S 2 d. xS2
57. lim
1 x  1 1 65. y = 2x + 4 x + 3 6 7. y = x + 2
1 x + 1 3 66. y = x  3 2x 68. y = x + 1
63. y =
64. y =
Inventing Graphs and Functions In Exercises 69–72, sketch the graph of a function y = ƒ(x) that satisfies the given conditions. No formulas are required—just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.)
69. ƒ(0) = 0, ƒ(1) = 2, ƒ( 1) = 2, lim ƒ(x) = 1, and x S q
lim ƒ(x) = 1 x Sq
70. ƒ(0) = 0, lim ƒ(x) = 0, lim+ ƒ(x) = 2, and x S {q
lim ƒ(x) =  2
xS0
xS0
71. ƒ(0) = 0, lim ƒ(x) = 0, lim ƒ(x) = lim + ƒ(x) = q, x S {q xS1 x S 1 lim ƒ(x) =  q, and lim ƒ(x) =  q xS1 +
x S 1
72. ƒ(2) = 1, ƒ(1) = 0, lim ƒ(x) = 0, lim+ ƒ(x) = q, x Sq xS0 lim ƒ(x) =  q, and lim ƒ(x) = 1 x S 0
x S q
2.6 Limits Involving Infinity; Asymptotes of Graphs
In Exercises 73–76, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) 73. lim ƒ(x) = 0, lim ƒ(x) = q, and lim ƒ(x) = q x S {q
xS2
x S {q
xS3
xS2

+
74. lim g(x) = 0, lim g(x) =  q, and lim+ g(x) = q xS3
75. lim h(x) =  1, lim h(x) = 1, lim h(x) =  1, and xS  q
lim+ h(x) = 1
xS q
xS0
xS0
76. lim k(x) = 1, lim k(x) = q, and lim+ k(x) =  q x S {q
xS1
xS1
77. Suppose that ƒ(x) and g(x) are polynomials in x and that limx S q (ƒ(x)>g(x)) = 2. Can you conclude anything about limx S  q (ƒ(x)>g(x))? Give reasons for your answer. 78. Suppose that ƒ(x) and g(x) are polynomials in x. Can the graph of ƒ(x)>g(x) have an asymptote if g(x) is never zero? Give reasons for your answer. 79. How many horizontal asymptotes can the graph of a given rational function have? Give reasons for your answer. Finding Limits of Differences When x u t H Find the limits in Exercises 80–86. xS q
81. lim ( 2x2 + 25  2x2  1 ) xS 
( 2x2 + 3 + x ) q ( 2x + q
xS 
x S c
b. lim+ ƒ(x) =  q xSc
c. lim ƒ(x) =  q xSc
Use the formal definitions from Exercise 93 to prove the limit statements in Exercises 94–98. 1 94. lim+ x = q xS0 1 95. lim x =  q xS0 1 96. lim= q xS2 x  2 1 97. lim+ = q xS2 x  2 1 98. lim= q x S 1 1  x2 Oblique Asymptotes Graph the rational functions in Exercises 99–104. Include the graphs and equations of the asymptotes.
100. y =
xS q
82. lim 83. lim
Modify the definition to cover the following cases. a. lim ƒ(x) = q
99. y =
80. lim ( 2x + 9  2x + 4 )
101. y =
24x2 + 3x  2 )
102. y =
84. lim ( 29x2  x  3x ) xS q
85. lim ( 2x2 + 3x  2x2  2x )
103. y =
xS q
86. lim ( 2x2 + x  2x2  x )
104. y =
xS q
Using the Formal Definitions Use the formal definitions of limits as x S { q to establish the limits in Exercises 87 and 88.
87. If ƒ has the constant value ƒ(x) = k, then lim ƒ(x) = k. x Sq
88. If ƒ has the constant value ƒ(x) = k, then lim ƒ(x) = k. x S q
Use formal definitions to prove the limit statements in Exercises 89–92. 1 =  q x2 2 =  q 91. lim x S 3 (x  3)2 89. lim
90. lim
xS0
xS0
92. lim
x S 5
1
0x0
= q
1 = q (x + 5)2
93. Here is the definition of infinite righthand limit. We say that ƒ(x) approaches infinity as x approaches c from the right, and write lim+ ƒ(x) = q,
c 6 x 6 c + d
1
x2 x  1 x2 + 1 x  1 x2  4 x  1 x2  1 2x + 4 x2  1 x x3 + 1 x2
Additional Graphing Exercises T Graph the curves in Exercises 105–108. Explain the relationship between the curve’s formula and what you see. x 105. y = 24  x2 1 106. y = 24  x2 1 107. y = x2>3 + 1>3 x p 108. y = sin a 2 b x + 1
T Graph the functions in Exercises 109 and 110. Then answer the following questions. a. How does the graph behave as x S 0+? b. How does the graph behave as x S { q? c. How does the graph behave near x = 1 and x =  1?
xSc
if, for every positive real number B, there exists a corresponding number d 7 0 such that for all x ƒ(x) 7 B.
117
Give reasons for your answers. 109. y =
3 1 2>3 ax  x b 2
110. y =
2>3 x 3 a b 2 x  1
118
Chapter 2: Limits and Continuity
Chapter 2
Questions to Guide Your Review
1. What is the average rate of change of the function g(t) over the interval from t = a to t = b? How is it related to a secant line?
11. What conditions must be satisfied by a function if it is to be continuous at an interior point of its domain? At an endpoint?
2. What limit must be calculated to find the rate of change of a function g(t) at t = t0?
12. How can looking at the graph of a function help you tell where the function is continuous?
3. Give an informal or intuitive definition of the limit
13. What does it mean for a function to be rightcontinuous at a point? Leftcontinuous? How are continuity and onesided continuity related?
lim ƒ(x) = L.
xSc
Why is the definition “informal”? Give examples. 4. Does the existence and value of the limit of a function ƒ(x) as x approaches c ever depend on what happens at x = c? Explain and give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples. 6. What theorems are available for calculating limits? Give examples of how the theorems are used. 7. How are onesided limits related to limits? How can this relationship sometimes be used to calculate a limit or prove it does not exist? Give examples. 8. What is the value of limu S 0 ((sin u)>u)? Does it matter whether u is measured in degrees or radians? Explain. 9. What exactly does limx S c ƒ(x) = L mean? Give an example in which you find a d 7 0 for a given ƒ, L, c, and P 7 0 in the precise definition of limit. 10. Give precise definitions of the following statements. a. limx S 2 ƒ(x) = 5 c. limx S 2 ƒ(x) = q
Chapter 2
b. limx S 2+ ƒ(x) = 5 d. limx S 2 ƒ(x) =  q
14. What does it mean for a function to be continuous on an interval? Give examples to illustrate the fact that a function that is not continuous on its entire domain may still be continuous on selected intervals within the domain. 15. What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example. 16. What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graphing and solving the equation ƒ(x) = 0? 17. Under what circumstances can you extend a function ƒ(x) to be continuous at a point x = c? Give an example. 18. What exactly do limx Sq ƒ(x) = L and limx S q ƒ(x) = L mean? Give examples. 19. What are limx S {q k (k a constant) and limx S {q (1>x)? How do you extend these results to other functions? Give examples. 20. How do you find the limit of a rational function as x S { q? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.
Practice Exercises
Limits and Continuity 1. Graph the function
1,  x, ƒ(x) = e 1,  x, 1,
x 1 x 0 x
… 6 = 6 Ú
1 x 6 0 0 x 6 1 1.
Then discuss, in detail, limits, onesided limits, continuity, and onesided continuity of ƒ at x =  1, 0, and 1. Are any of the discontinuities removable? Explain. 2. Repeat the instructions of Exercise 1 for 0, 1>x, ƒ(x) = d 0, 1,
x 0 x x
… 6 = 7
1 0x0 6 1 1 1.
3. Suppose that ƒ(t) and ƒ(t) are defined for all t and that limt S t0 ƒ(t) =  7 and limt S t0 g(t) = 0. Find the limit as t S t0 of the following functions. (ƒ(t))2 a. 3ƒ(t) b. ƒ(t) c. ƒ(t) # g(t) d. g(t)  7
0 ƒ(t) 0 e. cos (g(t)) f. g. ƒ(t) + g(t) h. 1>ƒ(t)
4. Suppose the functions ƒ(x) and g(x) are defined for all x and that limx S 0 ƒ(x) = 1>2 and limx S 0 g(x) = 22. Find the limits as x S 0 of the following functions.
g(x) # ƒ(x) a. g(x) b. c. ƒ(x) + g(x) d. 1>ƒ(x)
ƒ(x) # cos x e. x + ƒ(x) f. x  1
Chapter 2 Practice Exercises
In Exercises 5 and 6, find the value that limx S 0 g(x) must have if the given limit statements hold. 4  g(x) 5. lim a b = 1 6. lim ax lim g(x)b = 2 x xS0 x S 4 xS0
7. On what intervals are the following functions continuous? g(x) = x3>4 a. ƒ(x) = x1>3 b. c. h(x) = x2>3 d. k(x) = x1>6
119
c. It can be shown that the exact value of the solution in part (b) is a
269 1>3 269 1>3 1 1 + b + a b . 2 18 2 18
Evaluate this exact answer and compare it with the value you found in part (b). T 34. Let ƒ(u) = u 3  2u + 2.
8. On what intervals are the following functions continuous?
a. Use the Intermediate Value Theorem to show that ƒ has a zero between 2 and 0.
g(x) = csc x a. ƒ(x) = tan x b. cos x sin x c. h(x) = x  p d. k(x) = x
b. Solve the equation ƒ(u) = 0 graphically with an error of magnitude at most 104. c. It can be shown that the exact value of the solution in part (b) is
Finding Limits In Exercises 9–28, find the limit or explain why it does not exist.
x2  4x + 4 x3 + 5x2  14x a. as x S 0 b. as x S 2 9. lim
x2 + x 10. lim 5 x + 2x4 + x3 a. as x S 0 b. as x S 1 1  2x xS1 1  x
11. lim
(x + h)2  x2 h hS0 1 1 2 + x 2 15. lim x xS0 13. lim
x1>3  1 x S 1 2x  1 tan (2x) 19. lim x S 0 tan (px) 17. lim
x 21. lim sin a + sin xb 2 xSp
x 2  a2 x S a x 4  a4 (x + h)2  x2 14. lim h xS0 12. lim
(2 + x)3  8 x xS0
16. lim
x2>3  16 x S 64 2x  8
18. lim
20. lim csc x xSp
2
22. lim cos (x  tan x) xSp
cos 2x  1 sin x
8x 23. lim x S 0 3 sin x  x
24. lim
25. lim+ ln (t  3)
26. lim t ln ( 2  2t )
27. lim+ 2uecos (p>u)
28. lim+
tS3
uS0
xS0
2
tS1
zS0
2e1>z e + 1 1>z
In Exercises 29–32, find the limit of g(x) as x approaches the indicated value. 1 = 2 30. lim 29. lim+ (4g(x))1>3 = 2 xS0 x S 25 x + g(x)
3x2 + 1 q = 31. lim g(x) xS1
5  x2 32. lim = 0 x S 2 2g(x)
T Roots 33. Let ƒ(x) = x3  x  1. a. Use the Intermediate Value Theorem to show that ƒ has a zero between  1 and 2. b. Solve the equation ƒ(x) = 0 graphically with an error of magnitude at most 108.
a
1>3 1>3 19 19  1b  a + 1b . A 27 A 27
Evaluate this exact answer and compare it with the value you found in part (b). Continuous Extension 35. Can ƒ(x) = x(x2  1)> 0 x2  1 0 be extended to be continuous at x = 1 or 1? Give reasons for your answers. (Graph the function—you will find the graph interesting.)
36. Explain why the function ƒ(x) = sin (1>x) has no continuous extension to x = 0. T In Exercises 37–40, graph the function to see whether it appears to have a continuous extension to the given point a. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at a. If the function does not appear to have a continuous extension, can it be extended to be continuous from the right or left? If so, what do you think the extended function’s value should be? x  1 37. ƒ(x) = , a = 1 4 x  2 x 5 cos u , a = p>2 38. g(u) = 4u  2p 39. h(t) = ( 1 + 0 t 0 ) 1>t, a = 0 x 40. k(x) = , a = 0 1  20 x 0
Limits at Infinity Find the limits in Exercises 41–54. 2x + 3 2x2 + 3 41. lim 42. lim x S q 5x + 7 x S  q 5x 2 + 7
x2  4x + 8 3x3 x2  7x 45. lim xS  q x + 1 43. lim
xS  q
47. lim
xS q
sin x :x ;
1 x2  7x + 1 x4 + x3 46. lim x S q 12x 3 + 128 (If you have a grapher, try graphing the function for  5 … x … 5.) 44. lim
xS q
(If you have a grapher, try graphing ƒ(x) = x (cos (1>x)  1) near the origin to “see” the limit at infinity.) x + sin x + 2 2x x2>3 + x1 49. lim 50. lim 2>3 x + sin x xS q xS q x + cos2 x cos u  1 u uS q
48. lim
1 51. lim e1>x cos x xS q
53. lim tan1 x xS  q
1 52. lim ln a1 + t b tS q
1 54. lim e3t sin1 t tS  q
120
Chapter 2: Limits and Continuity
Horizontal and Vertical Asymptotes 55. Use limits to determine the equations for all vertical asymptotes.
56. Use limits to determine the equations for all horizontal asymptotes.
x2 + 4 x2  x  2 a. y = b. ƒ(x) = 2 x  3 x  2x + 1 x2 + x  6 c. y = 2 x + 2x  8
a. y =
Chapter 2
2x + 4 1  x2 b. ƒ(x) = x2 + 1 2x + 4
c. g(x) =
2x2 + 4
x
d. y =
x2 + 9 B 9x2 + 1
Additional and Advanced Exercises
0 T 1. Assigning a value to 0 The rules of exponents tell us that 0 a = 1 if a is any number different from zero. They also tell us that 0n = 0 if n is any positive number. If we tried to extend these rules to include the case 00, we would get conflicting results. The first rule would say 00 = 1, whereas the second would say 00 = 0. We are not dealing with a question of right or wrong here. Neither rule applies as it stands, so there is no contradiction. We could, in fact, define 00 to have any value we wanted as long as we could persuade others to agree. What value would you like 00 to have? Here is an example that might help you to decide. (See Exercise 2 below for another example.)
a. C alculate xx for x = 0.1, 0.01, 0.001, and so on as far as your calculator can go. Record the values you get. What pattern do you see? b. G raph the function y = xx for 0 6 x … 1. Even though the function is not defined for x … 0, the graph will approach the yaxis from the right. Toward what yvalue does it seem to be headed? Zoom in to further support your idea. T 2. A reason you might want 00 to be something other than 0 or 1 As the number x increases through positive values, the numbers 1>x and 1 > (ln x) both approach zero. What happens to the number 1 1>(ln x) ƒ(x) = a x b
as x increases? Here are two ways to find out. a. E valuate ƒ for x = 10, 100, 1000, and so on as far as your calculator can reasonably go. What pattern do you see? b. G raph ƒ in a variety of graphing windows, including windows that contain the origin. What do you see? Trace the yvalues along the graph. What do you find? 3. Lorentz contraction In relativity theory, the length of an object, say a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket’s length as L 0 at rest, then at speed y the length will appear to be L = L0
B
1 
y2 . c2
This equation is the Lorentz contraction formula. Here, c is the speed of light in a vacuum, about 3 * 108 m>sec. What happens to L as y increases? Find limy S c L. Why was the lefthand limit needed?
4. Controlling the flow from a draining tank Torricelli’s law says that if you drain a tank like the one in the figure shown, the rate y at which water runs out is a constant times the square root of the water’s depth x. The constant depends on the size and shape of the exit valve.
x Exit rate y ft3min
Suppose that y = 2x>2 for a certain tank. You are trying to maintain a fairly constant exit rate by adding water to the tank with a hose from time to time. How deep must you keep the water if you want to maintain the exit rate a. within 0.2 ft3 >min of the rate y0 = 1 ft3 >min?
b. within 0.1 ft3 >min of the rate y0 = 1 ft3 >min?
5. Thermal expansion in precise equipment As you may know, most metals expand when heated and contract when cooled. The dimensions of a piece of laboratory equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laboratory where the equipment is to be used. A typical aluminum bar that is 10 cm wide at 70°F will be y = 10 + (t  70) * 104 centimeters wide at a nearby temperature t. Suppose that you are using a bar like this in a gravity wave detector, where its width must stay within 0.0005 cm of the ideal 10 cm. How close to t0 = 70°F must you maintain the temperature to ensure that this tolerance is not exceeded? 6. Stripes on a measuring cup The interior of a typical 1L measuring cup is a right circular cylinder of radius 6 cm (see accompanying figure). The volume of water we put in the cup is therefore a function of the level h to which the cup is filled, the formula being V = p62h = 36ph. How closely must we measure h to measure out 1 L of water (1000 cm3) with an error of no more than 1% (10 cm3)?
Chapter 2 Additional and Advanced Exercises
17. A function continuous at only one point Let ƒ(x) = e
b. U se the fact that every nonempty open interval of real numbers contains both rational and irrational numbers to show that ƒ is not continuous at any nonzero value of x. 18. The Dirichlet ruler function If x is a rational number, then x can be written in a unique way as a quotient of integers m>n where n 7 0 and m and n have no common factors greater than 1. (We say that such a fraction is in lowest terms. For example, 6>4 written in lowest terms is 3>2.) Let ƒ(x) be defined for all x in the interval 30, 14 by
(a) r = 6 cm
Liquid volume V = 36ph
h
A 1L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6 cm Precise Definition of Limit In Exercises 7–10, use the formal definition of limit to prove that the function is continuous at c.
7. ƒ(x) = x2  7, c = 1
8. g(x) = 1>(2x), c = 1>4
9. h(x) = 22x  3, c = 2 10. F(x) = 29  x, c = 5
11. Uniqueness of limits Show that a function cannot have two different limits at the same point. That is, if limx S c ƒ(x) = L 1 and limx S c ƒ(x) = L 2, then L 1 = L 2.
12. Prove the limit Constant Multiple Rule: lim kƒ(x) = k lim ƒ(x) for any constant k. xSc
13. Onesided limits If limx S 0+ ƒ(x) = A and limx S 0 ƒ(x) = B, find limx S 0 ƒ(x3  x) a. limx S 0+ ƒ(x3  x) b. limx S 0 ƒ(x2  x4) c. limx S 0+ ƒ(x2  x4) d. 14. Limits and continuity Which of the following statements are true, and which are false? If true, say why; if false, give a counterexample (that is, an example confirming the falsehood). a. If limx S c ƒ(x) exists but limx S c g(x) does not exist, then limx S c(ƒ(x) + g(x)) does not exist. b. If neither limx S c ƒ(x) nor limx S c g(x) exists, then limx S c (ƒ(x) + g(x)) does not exist. c. If ƒ is continuous at x, then so is 0 ƒ 0 .
d. If 0 ƒ 0 is continuous at c, then so is ƒ.
In Exercises 15 and 16, use the formal definition of limit to prove that the function has a continuous extension to the given value of x. x2  1 , x =  1 x + 1
ƒ(x) = e
1>n, if x = m>n is a rational number in lowest terms 0, if x is irrational.
For instance, ƒ(0) = ƒ(1) = 1, ƒ(1>2) = 1>2, ƒ(1>3) = ƒ(2>3) = 1>3, ƒ(1>4) = ƒ(3>4) = 1>4, and so on. a. Show that ƒ is discontinuous at every rational number in 30, 14 .
(b)
15. ƒ(x) =
x, if x is rational 0, if x is irrational.
a. Show that ƒ is continuous at x = 0.
Stripes about 1 mm wide
xSc
121
16. g(x) =
x2  2x  3 , x = 3 2x  6
b. S how that ƒ is continuous at every irrational number in 30, 14 . (Hint: If P is a given positive number, show that there are only finitely many rational numbers r in 30, 14 such that ƒ(r) Ú P.)
c. Sketch the graph of ƒ. Why do you think ƒ is called the “ruler function”?
19. Antipodal points Is there any reason to believe that there is always a pair of antipodal (diametrically opposite) points on Earth’s equator where the temperatures are the same? Explain. 20. If limx S c (ƒ(x) + g(x)) = 3 and limx S c (ƒ(x)  g(x)) = 1, find limx S c ƒ(x)g(x). 21. Roots of a quadratic equation that is almost linear The equation ax2 + 2x  1 = 0, where a is a constant, has two roots if a 7 1 and a ≠ 0, one positive and one negative: r+(a) =
1 + 21 + a , a
r(a) =
 1  21 + a , a
a. What happens to r+(a) as a S 0? As a S  1+? b. What happens to r(a) as a S 0? As a S  1+? c. S upport your conclusions by graphing r+(a) and r(a) as functions of a. Describe what you see. d. For added support, graph ƒ(x) = ax2 + 2x  1 simultaneously for a = 1, 0.5, 0.2, 0.1, and 0.05. 22. Root of an equation Show that the equation x + 2 cos x = 0 has at least one solution. 23. Bounded functions A realvalued function ƒ is bounded from above on a set D if there exists a number N such that ƒ(x) … N for all x in D. We call N, when it exists, an upper bound for ƒ on D and say that ƒ is bounded from above by N. In a similar manner, we say that ƒ is bounded from below on D if there exists a number M such that ƒ(x) Ú M for all x in D. We call M, when it exists, a lower bound for ƒ on D and say that ƒ is bounded from below by M. We say that ƒ is bounded on D if it is bounded from both above and below. a. S how that ƒ is bounded on D if and only if there exists a number B such that 0 ƒ(x) 0 … B for all x in D.
122
Chapter 2: Limits and Continuity
b. Suppose that ƒ is bounded from above by N. Show that if limx S c ƒ(x) = L, then L … N. c. Suppose that ƒ is bounded from below by M. Show that if limx S c ƒ(x) = L, then L Ú M. 24. Max 5a, b6 and min 5a, b6 a. Show that the expression
max 5a, b6 =
0a  b0 a + b + 2 2
equals a if a Ú b and equals b if b Ú a. In other words, max 5a, b6 gives the larger of the two numbers a and b.
b. Find a similar expression for min 5a, b6, the smaller of a and b.
sin U U The formula limu S 0 (sin u)>u = 1 can be generalized. If limx S c ƒ(x) = 0 and ƒ(x) is never zero in an open interval containing the point x = c, except possibly c itself, then Generalized Limits Involving
lim
xSc
sin ƒ(x) = 1. ƒ(x)
Here are several examples.
sin (x2  x  2) sin (x2  x  2) # = lim x + 1 x S 1 x S 1 (x 2  x  2)
c. lim lim
x S 1
(x2  x  2) (x + 1)(x  2) = 1 # lim = 3 x + 1 x + 1 x S 1
sin 1 1  2x 2 1  2x sin 1 1  2x 2 = = lim x  1 x  1 xS1 xS1 1  2x
d. lim
1 # lim
xS1
11
 2x 21 1 + 2x 2
(x  1)1 1 + 2x 2
28. lim
sin ( x2  4 ) x  2 xS2
30. lim
29. lim
xS9
sin 1 2x  3 2 x  9
Oblique Asymptotes Find all possible oblique asymptotes in Exercises 31–34.
2x3>2 + 2x  3 2x + 1
sin x2 sin x2 x2 = lim 2 lim x = 1 # 0 = 0 x xS0 xS0 x xS0
33. y = 2x2 + 1
Chapter 2
1  x 1 = 2  1)1 1 + 2x 2
sin ( x2 + x ) x xS0
sin (sin x) x xS0
27. lim
31. y =
b. lim
x S 1 (x
Find the limits in Exercises 25–30. sin (1  cos x) sin x 25. lim 26. lim+ x xS0 x S 0 sin 2x
sin x2 = 1 x S 0 x2
a. lim
= lim
1 32. y = x + x sin x 34. y = 2x2 + 2x
Technology Application Projects
Mathematica/Maple Modules: Take It to the Limit Part I Part II (Zero Raised to the Power Zero: What Does It Mean?) Part III (OneSided Limits) Visualize and interpret the limit concept through graphical and numerical explorations. Part IV (What a Difference a Power Makes) See how sensitive limits can be with various powers of x. Going to Infinity Part I (Exploring Function Behavior as x u H or x u − H ) This module provides four examples to explore the behavior of a function as x S q or x S  q . Part II (Rates of Growth) Observe graphs that appear to be continuous, yet the function is not continuous. Several issues of continuity are explored to obtain results that you may find surprising.
3
Derivatives
OVERVIEW In the beginning of Chapter 2, we discussed how to determine the slope of a
curve at a point and how to measure the rate at which a function changes. Now that we have studied limits, we can define these ideas precisely and see that both are interpretations of the derivative of a function at a point. We then extend this concept from a single point to the derivative function, and we develop rules for finding this derivative function easily, without having to calculate any limits directly. These rules are used to find derivatives of most of the common functions reviewed in Chapter 1, as well as various combinations of them. The derivative is one of the key ideas in calculus, and is used to study a wide range of problems in mathematics, science, economics, and medicine. These problems include finding points where a continuous function is zero, calculating the velocity and acceleration of a moving object, determining how the rate of flow of a liquid into a container changes the level of the liquid within it, describing the path followed by a light ray going from a point in air to a point in water, finding the number of items a manufacturing company should produce in order to maximize its profits, studying the spread of an infectious disease within a given population, or calculating the amount of blood the heart pumps in a minute based on how well the lungs are functioning.
3.1 Tangents and the Derivative at a Point In this section we define the slope and tangent to a curve at a point, and the derivative of a function at a point. The derivative gives a way to find both the slope of a graph and the instantaneous rate of change of a function.
Finding a Tangent to the Graph of a Function y y = f (x) Q(x 0 + h, f(x 0 + h)) f (x 0 + h) − f (x 0)
P(x 0, f(x 0))
DEFINITIONS The slope of the curve y = ƒ(x) at the point P(x0, ƒ(x0)) is the number
h 0
x0
FIGURE 3.1
To find a tangent to an arbitrary curve y = ƒ(x) at a point P(x0, ƒ(x0)), we use the procedure introduced in Section 2.1. We calculate the slope of the secant through P and a nearby point Q(x0 + h, ƒ(x0 + h)). We then investigate the limit of the slope as h S 0 (Figure 3.1). If the limit exists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope.
x0 + h
The slope of the tangent ƒ(x0 + h)  ƒ(x0) . line at P is lim h hS0
x
ƒ(x0 + h)  ƒ(x0) h hS0
m = lim
(provided the limit exists).
The tangent line to the curve at P is the line through P with this slope.
123
124
Chapter 3: Derivatives
In Section 2.1, Example 3, we applied these definitions to find the slope of the parabola ƒ(x) = x2 at the point P(2, 4) and the tangent line to the parabola at P. Let’s look at another example.
y y = 1x slope is −
1 a2
x
a
0
EXAMPLE 1
slope is −1 at x = −1
(a) Find the slope of the curve y = 1>x at any point x = a ≠ 0. What is the slope at the point x = 1? (b) Where does the slope equal 1>4? (c) What happens to the tangent to the curve at the point (a, 1>a) as a changes? Solution (a) Here ƒ(x) = 1>x. The slope at (a, 1>a) is 1 1 a + h a ƒ(a + h)  ƒ(a) 1 a  (a + h) = lim lim = lim h hS0 hS0 h S 0 h a(a + h) h
FIGURE 3.2 The tangent slopes, steep near the origin, become more gradual as the point of tangency moves away (Example 1).
= lim
hS0
y y = 1x
slope is −
1 4
1 a2, b 2 x
1 a−2, − 2b
1 slope is − 4
FIGURE 3.3 The two tangent lines to y = 1>x having slope  1>4 (Example 1).
h 1 1 = lim =  2. ha(a + h) h S 0 a(a + h) a
Notice how we had to keep writing “limh S 0” before each fraction until the stage at which we could evaluate the limit by substituting h = 0. The number a may be positive or negative, but not 0. When a = 1, the slope is 1>(1)2 = 1 (Figure 3.2). (b) The slope of y = 1>x at the point where x = a is 1>a2. It will be 1>4 provided that 
1 1 =  . 4 a2
This equation is equivalent to a2 = 4, so a = 2 or a = 2. The curve has slope 1>4 at the two points (2, 1>2) and (2, 1>2) (Figure 3.3). (c) The slope 1>a2 is always negative if a ≠ 0. As a S 0+, the slope approaches  q and the tangent becomes increasingly steep (Figure 3.2). We see this situation again as a S 0. As a moves away from the origin in either direction, the slope approaches 0 and the tangent levels off becoming more and more horizontal.
Rates of Change: Derivative at a Point The expression ƒ(x0 + h)  ƒ(x0) , h ≠ 0 h is called the difference quotient of ƒ at x0 with increment h. If the difference quotient has a limit as h approaches zero, that limit is given a special name and notation.
DEFINITION The derivative of a function ƒ at a point x0 , denoted ƒ′(x0), is The notation ƒ′(x0) is read “ƒ prime of x0.”
ƒ(x0 + h)  ƒ(x0) h hS0
ƒ′(x0) = lim provided this limit exists.
3.1 Tangents and the Derivative at a Point
125
If we interpret the difference quotient as the slope of a secant line, then the derivative gives the slope of the curve y = ƒ(x) at the point P(x0, ƒ(x0)). Exercise 33 shows that the derivative of the linear function ƒ(x) = mx + b at any point x0 is simply the slope of the line, so ƒ′(x0) = m, which is consistent with our definition of slope. If we interpret the difference quotient as an average rate of change (Section 2.1), the derivative gives the function’s instantaneous rate of change with respect to x at the point x = x0. We study this interpretation in Section 3.4.
EXAMPLE 2 In Examples 1 and 2 in Section 2.1, we studied the speed of a rock falling freely from rest near the surface of the earth. We knew that the rock fell y = 16t 2 feet during the first t sec, and we used a sequence of average rates over increasingly short intervals to estimate the rock’s speed at the instant t = 1. What was the rock’s exact speed at this time? Solution We let ƒ(t) = 16t 2. The average speed of the rock over the interval between t = 1 and t = 1 + h seconds, for h 7 0, was found to be ƒ(1 + h)  ƒ(1) 16(1 + h)2  16(1)2 16(h2 + 2h) = = = 16(h + 2). h h h The rock’s speed at the instant t = 1 is then ƒ′(1) = lim 16(h + 2) = 16(0 + 2) = 32 ft>sec. hS0
Our original estimate of 32 ft>sec in Section 2.1 was right.
Summary We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, and the derivative of a function at a point. All of these ideas refer to the same limit.
The following are all interpretations for the limit of the difference quotient, ƒ(x0 + h)  ƒ(x0) . h hS0 lim
1. The slope of the graph of y = ƒ(x) at x = x0 2. The slope of the tangent to the curve y = ƒ(x) at x = x0 3. The rate of change of ƒ(x) with respect to x at x = x0 4. The derivative ƒ′(x0) at a point
In the next sections, we allow the point x0 to vary across the domain of the function ƒ.
126
Chapter 3: Derivatives
Exercises
3.1
Slopes and Tangent Lines In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in yunits per xunit) at the points P1 and P2.
1.
2. y
y
2
P2
2
P2
p
1 1
−2
−1
0
1
x
2
0
x
1
3.
c. The quadratic curve capturing the trend of the data points (see Section 1.4) is given by P(t) = 6.10t 2  9.28t + 16.43. Find the instantaneous rate of growth when t = 5 hours.
y 4
24. Effectiveness of a drug On a scale from 0 to 1, the effectiveness E of a painkilling drug t hours after entering the bloodstream is displayed in the accompanying figure.
2 3 2
P1
P2
E
1 0
1
2
x
−2
−1
0
1
2
x
In Exercises 5–10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. 5. y = 4  x2, (1, 3)
6. y = (x  1)2 + 1, (1, 1)
7. y = 2 2x, (1, 2)
1 8. y = 2 , (1, 1) x
9. y = x , ( 2,  8) 3
1 10. y = 3 , x
1 a 2,  b 8
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there. 11. ƒ(x) = x2 + 1, (2, 5)
12. ƒ(x) = x  2x2, (1,  1)
x 13. g(x) = , (3, 3) x  2
8 14. g(x) = 2 , (2, 2) x
15. h(t) = t 3, (2, 8)
16. h(t) = t 3 + 3t, (1, 4)
17. ƒ(x) = 2x, (4, 2)
18. ƒ(x) = 2x + 1, (8, 3)
In Exercises 19–22, find the slope of the curve at the point indicated. 19. y = 5x  3x2, x = 1 21. y =
1 , x = 3 x  1
t
b. Which is larger, P′(2) or P′(3)? Give a reason for your answer.
4.
P2
1 2 3 4 5 6 7
a. Explain what is meant by the derivative P′(5). What are its units?
−2
y
P1
250 200 150 100 50 0
P1 −1
P1
1
Interpreting Derivative Values 23. Growth of yeast cells In a controlled laboratory experiment, yeast cells are grown in an automated cell culture system that counts the number P of cells present at hourly intervals. The number after t hours is shown in the accompanying figure.
20. y = x3  2x + 7, x =  2 22. y =
x  1 , x = 0 x + 1
1.0 0.8 0.6 0.4 0.2 0
1
2
3
4
5
t
a. At what times does the effectiveness appear to be increasing? What is true about the derivative at those times? b. At what time would you estimate that the drug reaches its maximum effectiveness? What is true about the derivative at that time? What is true about the derivative as time increases in the 1 hour before your estimated time? At what points do the graphs of the functions in Exercises 25 and 26 have horizontal tangents? 25. ƒ(x) = x2 + 4x  1
26. g(x) = x3  3x
27. Find equations of all lines having slope 1 that are tangent to the curve y = 1>(x  1). 28. Find an equation of the straight line having slope 1>4 that is tangent to the curve y = 2x. Rates of Change 29. Object dropped from a tower An object is dropped from the top of a 100mhigh tower. Its height above ground after t sec is 100  4.9t 2 m. How fast is it falling 2 sec after it is dropped?
3.1 Tangents and the Derivative at a Point
30. Speed of a rocket At t sec after liftoff, the height of a rocket is 3t 2 ft. How fast is the rocket climbing 10 sec after liftoff?
127
does not exist, because the limit is q from the right and  q from the left.
31. Circle’s changing area What is the rate of change of the area of a circle ( A = pr 2 ) with respect to the radius when the radius is r = 3?
y y = g(x) = x 23
32. Ball’s changing volume What is the rate of change of the volume of a ball ( V = (4>3)pr 3 ) with respect to the radius when the radius is r = 2? 33. Show that the line y = mx + b is its own tangent line at any point (x0, mx0 + b). 34. Find the slope of the tangent to the curve y = 1> 2x at the point where x = 4.
NO VERTICAL TANGENT AT ORIGIN
37. Does the graph of 1, x 6 0 ƒ(x) = c 0, x = 0 1, x 7 0
Testing for Tangents 35. Does the graph of
ƒ(x) = e
x2 sin (1>x), x ≠ 0 0, x = 0
have a vertical tangent at the origin? Give reasons for your answer. 38. Does the graph of
have a tangent at the origin? Give reasons for your answer.
U(x) = e
36. Does the graph of g(x) = e
x sin (1>x), x ≠ 0 0, x = 0
have a tangent at the origin? Give reasons for your answer. Vertical Tangents We say that a continuous curve y = ƒ(x) has a vertical tangent at the point where x = x0 if the limit of the difference quotient is q or  q. For example, y = x1>3 has a vertical tangent at x = 0 (see accompanying figure):
ƒ(0 + h)  ƒ(0) h1>3  0 lim = lim h h hS0 hS0 1 = lim 2>3 = q. hS0 h
x
0
0, x 6 0 1, x Ú 0
have a vertical tangent at the point (0, 1)? Give reasons for your answer. T Graph the curves in Exercises 39–48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38. 39. y = x2>5
40. y = x4>5
41. y = x1>5
42. y = x3>5
43. y = 4x
2>5
 2x
44. y = x5>3  5x2>3
45. y = x2>3  (x  1)1>3
46. y = x1>3 + (x  1)1>3
47. y = e
48. y = 2 0 4  x 0
 20x0, x … 0 2x, x 7 0
y y = f (x) = x 13
COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exercises 49–52:
a. Plot y = ƒ(x) over the interval (x0  1>2) … x … (x0 + 3). x
0
b. Holding x0 fixed, the difference quotient q(h) =
VERTICAL TANGENT AT ORIGIN
However, y = x2>3 has no vertical tangent at x = 0 (see next figure): lim
hS0
g(0 + h)  g(0) h2>3  0 = lim h h hS0 1 = lim 1>3 hS0 h
ƒ(x0 + h)  ƒ(x0) h
at x0 becomes a function of the step size h. Enter this function into your CAS workspace. c. Find the limit of q as h S 0. d. Define the secant lines y = ƒ(x0) + q # (x  x0) for h = 3, 2, and 1. Graph them together with ƒ and the tangent line over the interval in part (a).
49. ƒ(x) = x3 + 2x, x0 = 0
5 50. ƒ(x) = x + x , x0 = 1
51. ƒ(x) = x + sin (2x), x0 = p>2 52. ƒ(x) = cos x + 4 sin (2x), x0 = p
128
Chapter 3: Derivatives
3.2 The Derivative as a Function In the last section we defined the derivative of y = ƒ(x) at the point x = x0 to be the limit HISTORICAL ESSAY
ƒ′(x0) = lim
The Derivative
hS0
ƒ(x0 + h)  ƒ(x0) . h
We now investigate the derivative as a function derived from ƒ by considering the limit at each point x in the domain of ƒ.
DEFINITION The derivative of the function ƒ(x) with respect to the variable x is the function ƒ′ whose value at x is ƒ′(x) = lim
hS0
ƒ(x + h)  ƒ(x) , h
provided the limit exists. y = f (x) Secant slope is f (z) − f (x) z−x
Q(z, f(z))
f(z) − f (x)
P(x, f(x)) h=z−x
z=x+h
x
We use the notation ƒ(x) in the definition to emphasize the independent variable x with respect to which the derivative function ƒ′(x) is being defined. The domain of ƒ′ is the set of points in the domain of ƒ for which the limit exists, which means that the domain may be the same as or smaller than the domain of ƒ. If ƒ′ exists at a particular x, we say that ƒ is differentiable (has a derivative) at x. If ƒ′ exists at every point in the domain of ƒ, we call ƒ differentiable. If we write z = x + h, then h = z  x and h approaches 0 if and only if z approaches x. Therefore, an equivalent definition of the derivative is as follows (see Figure 3.4). This formula is sometimes more convenient to use when finding a derivative function, and focuses on the point z that approaches x.
Derivative of f at x is f(x + h) − f (x) f '(x) = lim h h:0 = lim
z :x
f(z) − f(x) z−x
FIGURE 3.4 Two forms for the difference quotient.
Alternative Formula for the Derivative ƒ′(x) = lim
zSx
ƒ(z)  ƒ(x) z  x
Calculating Derivatives from the Definition The process of calculating a derivative is called differentiation. To emphasize the idea that differentiation is an operation performed on a function y = ƒ(x), we use the notation d ƒ(x) dx
Derivative of the Reciprocal Function d 1 1 a b =  2, x ≠ 0 dx x x
as another way to denote the derivative ƒ′(x). Example 1 of Section 3.1 illustrated the differentiation process for the function y = 1>x when x = a. For x representing any point in the domain, we get the formula d 1 1 a b =  2. dx x x Here are two more examples in which we allow x to be any point in the domain of ƒ.
3.2 The Derivative as a Function
EXAMPLE 1
Differentiate ƒ(x) =
129
x . x  1
Solution We use the definition of derivative, which requires us to calculate ƒ(x + h) and then subtract ƒ(x) to obtain the numerator in the difference quotient. We have (x + h) x , so and ƒ(x + h) = x  1 (x + h)  1 ƒ(x + h)  ƒ(x) ƒ′(x) = lim Definition h hS0 x x + h x + h  1 x  1 = lim hS0 h (x + h) (x  1)  x(x + h  1) 1 a c ad  cb = lim #  = b d bd (x + h  1) (x  1) hS0 h ƒ(x) =
= lim
1#
hS0 h
= lim
h S 0 (x
h (x + h  1) (x  1)
Simplify.
1 1 = . + h  1) (x  1) (x  1)2
Cancel h ≠ 0.
EXAMPLE 2 (a) Find the derivative of ƒ(x) = 2x for x 7 0. (b) Find the tangent line to the curve y = 2x at x = 4. Solution (a) We use the alternative formula to calculate ƒ′:
Derivative of the Square Root Function
ƒ(z)  ƒ(x) z  x zSx
d 1 2x = , x 7 0 dx 2 2x
ƒ′(x) = lim = lim
1z  1x
z  x
zSx
= lim
zSx
y = 1x + 1 4
(b) The slope of the curve at x = 4 is ƒ′(4) =
y = "x
1 0
4
FIGURE 3.5 The curve y = 2x and its tangent at (4, 2). The tangent’s slope is found by evaluating the derivative at x = 4 (Example 2).
 1x 21 1z + 1x 2
1 1 = lim = . z S x 1z + 1x 2 1x
y
(4, 2)
1 1z
1z  1x
x
1 2 24
=
1 . 4
The tangent is the line through the point (4, 2) with slope 1>4 (Figure 3.5): y = 2 + y =
1 (x  4) 4
1 x + 1. 4
Notations There are many ways to denote the derivative of a function y = ƒ(x), where the independent variable is x and the dependent variable is y. Some common alternative notations for the derivative are ƒ′(x) = y′ =
dy dƒ d = = ƒ(x) = D(ƒ)(x) = Dx ƒ(x). dx dx dx
130
Chapter 3: Derivatives y
The symbols d>dx and D indicate the operation of differentiation. We read dy>dx as “the derivative of y with respect to x,” and dƒ>dx and (d>dx) ƒ(x) as “the derivative of ƒ with respect to x.” The “prime” notations y′ and ƒ′ come from notations that Newton used for derivatives. The d>dx notations are similar to those used by Leibniz. The symbol dy>dx should not be regarded as a ratio (until we introduce the idea of “differentials” in Section 3.11). To indicate the value of a derivative at a specified number x = a, we use the notation
Slope 0 A 10
Slope −1 B C
y = f (x)
Slope − 4 3 E
≈8
D Slope 0
5
0
≈ 4 xunits 10 15
5
ƒ′(a) = x
For instance, in Example 2
(a)
ƒ′(4) =
Slope
4
A'
E′
2 1 −1 −2
D′ 10
5 C′
d 1 1 1 2x ` = ` = = . 4 dx 2 1 x 2 24 x=4 x=4
Graphing the Derivative
y = f '(x)
3
dy df d ` = ` = ƒ(x) ` . dx x = a dx x = a dx x=a
15
x
B′ Vertical coordinate −1
We can often make a reasonable plot of the derivative of y = ƒ(x) by estimating the slopes on the graph of ƒ. That is, we plot the points (x, ƒ′(x)) in the xyplane and connect them with a smooth curve, which represents y = ƒ′(x).
EXAMPLE 3
Graph the derivative of the function y = ƒ(x) in Figure 3.6a.
Solution We sketch the tangents to the graph of ƒ at frequent intervals and use their slopes to estimate the values of ƒ′(x) at these points. We plot the corresponding (x, ƒ′(x)) pairs and connect them with a smooth curve as sketched in Figure 3.6b.
(b)
FIGURE 3.6 We made the graph of y = ƒ′(x) in (b) by plotting slopes from the graph of y = ƒ(x) in (a). The vertical coordinate of B′ is the slope at B and so on. The slope at E is approximately 8>4 = 2. In (b) we see that the rate of change of ƒ is negative for x between A′ and D′; the rate of change is positive for x to the right of D′.
What can we learn from the graph of y = ƒ′(x)? At a glance we can see 1. 2. 3.
where the rate of change of ƒ is positive, negative, or zero; the rough size of the growth rate at any x and its size in relation to the size of ƒ(x); where the rate of change itself is increasing or decreasing.
Differentiable on an Interval; OneSided Derivatives A function y = ƒ(x) is differentiable on an open interval (finite or infinite) if it has a derivative at each point of the interval. It is differentiable on a closed interval 3 a, b4 if it is differentiable on the interior (a, b) and if the limits lim
ƒ(a + h)  ƒ(a) h
Righthand derivative at a
lim
ƒ(b + h)  ƒ(b) h
Lefthand derivative at b
h S 0+
Slope = f(a + h) − f(a) lim + h h:0
Slope = f (b + h) − f (b) lim h h:0−
h S 0
exist at the endpoints (Figure 3.7). Righthand and lefthand derivatives may be defined at any point of a function’s domain. Because of Theorem 6, Section 2.4, a function has a derivative at a point if and only if it has lefthand and righthand derivatives there, and these onesided derivatives are equal.
y = f (x)
Show that the function y = x is differentiable on ( q, 0) and (0, q) but has no derivative at x = 0.
EXAMPLE 4
a
a+h h>0
b+h hx) near the origin, where it is discontinuous (see Figure 2.31).
Differentiable Functions Are Continuous A function is continuous at every point where it has a derivative.
THEOREM 1—Differentiability Implies Continuity x = c, then ƒ is continuous at x = c.
If ƒ has a derivative at
Proof Given that ƒ′(c) exists, we must show that limx S c ƒ(x) = ƒ(c), or equivalently, that limh S 0 ƒ(c + h) = ƒ(c). If h ≠ 0, then ƒ(c + h) = ƒ(c) + (ƒ(c + h)  ƒ(c)) ƒ(c + h)  ƒ(c) # = ƒ(c) + h. h Now take limits as h S 0. By Theorem 1 of Section 2.2, lim ƒ(c + h) = lim ƒ(c) + lim
hS0
hS0
hS0
= ƒ(c) + ƒ′(c) # 0 = ƒ(c) + 0 = ƒ(c).
ƒ(c + h)  ƒ(c) # lim h h hS0
Similar arguments with onesided limits show that if ƒ has a derivative from one side (right or left) at x = c, then ƒ is continuous from that side at x = c. Theorem 1 says that if a function has a discontinuity at a point (for instance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function y = : x ; fails to be differentiable at every integer x = n (Example 4, Section 2.5). Caution The converse of Theorem 1 is false. A function need not have a derivative at a point where it is continuous, as we saw with the absolute value function in Example 4.
133
3.2 The Derivative as a Function
Exercises
3.2
Finding Derivative Functions and Values Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.
Graphs Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d).
1. ƒ(x) = 4  x2; ƒ′( 3), ƒ′(0), ƒ′(1)
y′
y′
2. F(x) = (x  1)2 + 1; F′( 1), F′(0), F′(2) 3. g(t) =
1 ; g′( 1), g′(2), g′1 23 2 t2
4. k(z) =
1  z ; k′( 1), k′(1), k′1 22 2 2z
x
0 x
0
5. p(u) = 23u ; p′(1), p′(3), p′(2>3)
(a)
(b)
y′
y′
6. r (s) = 22s + 1 ; r′(0), r′(1), r′(1>2) In Exercises 7–12, find the indicated derivatives. 7.
dy if y = 2x3 dx
8.
ds t 9. if s = 2t + 1 dt 11.
dp if p = q3>2 dq
dr if r = s3  2s2 + 3 ds
x
0
12.
dz 1 if z = 2 dw 2w  1
9 13. ƒ(x) = x + x , x = 3
14. k(x) =
15. s = t 3  t 2, t = 1
x + 3 , x = 2 16. y = 1  x
27.
28.
In Exercises 19–22, find the values of the derivatives. dy ds 1 ` ` if s = 1  3t 2 20. if y = 1  x 19. dt t = 1 dx x = 23 dw 0 22. if w = z + 1z dz z = 4
y = f 2 (x)
y = f1(x) x
0
x
0
29.
30. y
8 , (x, y) = (6, 4) 17. y = ƒ(x) = 2x  2 18. w = g(z) = 1 + 24  z, (z, w) = (3, 2)
y
y
1 , x = 2 2 + x
In Exercises 17–18, differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function.
(d)
(c)
Slopes and Tangent Lines In Exercises 13–16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable.
dr 2 21. ` if r = du u = 0 24  u
x
0
dy 1 10. if y = t  t dt
y y = f3(x)
0
y = f4(x) x
x
0
31. a. The graph in the accompanying figure is made of line segments joined end to end. At which points of the interval 34, 64 is ƒ′ not defined? Give reasons for your answer. y
Using the Alternative Formula for Derivatives Use the formula
ƒ′(x) = lim
zSx
1 x + 2
x 25. g(x) = x  1
y = f (x)
ƒ(z)  ƒ(x) z  x
to find the derivative of the functions in Exercises 23–26. 23. ƒ(x) =
(6, 2)
(0, 2)
24. ƒ(x) = x2  3x + 4 26. g(x) = 1 + 1x
(−4, 0)
0
1
(1, −2)
6
(4, −2)
b. Graph the derivative of ƒ. The graph should show a step function.
x
134
Chapter 3: Derivatives
32. Recovering a function from its derivative a. Use the following information to graph the function ƒ over the closed interval 32, 54.
35. Temperature The given graph shows the temperature T in °F at Davis, CA, on April 18, 2008, between 6 a.m. and 6 p.m. T
i) The graph of ƒ is made of closed line segments joined end to end. Temperature (°F)
80
ii) The graph starts at the point ( 2, 3). iii) The derivative of ƒ is the step function in the figure shown here. y′ y′ = f ′(x) 1 −2
0
1
3
5
x
70 60 50 40 0
3
6
9
12
6 A.M.
9 A.M.
12 NOON Time (hr)
3 P.M.
6 P.M.
t
a. Estimate the rate of temperature change at the times −2
i) 7 a.m.
b. Repeat part (a), assuming that the graph starts at ( 2, 0) instead of ( 2, 3). 33. Growth in the economy The graph in the accompanying figure shows the average annual percentage change y = ƒ(t) in the U.S. gross national product (GNP) for the years 2005–2011. Graph dy>dt (where defined). 7% 6 5 4 3 2 1 0 2005 2006 2007 2008 2009 2010 2011
iv) 4 p.m.
36. Weight loss Jared Fogle, also known as the “Subway Sandwich Guy,” weighed 425 lb in 1997 before losing more than 240 lb in 12 months (http://en.wikipedia.org/wiki/Jared_Fogle). A chart showing his possible dramatic weight loss is given in the accompanying figure. W 500 Weight (lb)
425 300 200 100 0
1 2 3 4 5 6 7 8 9 10 11 12 Time (months)
t
a. Estimate Jared’s rate of weight loss when i) t = 1
a. Use the graphical technique of Example 3 to graph the derivative of the fruit fly population. The graph of the population is reproduced here.
ii) t = 4
iii) t = 11
b. When does Jared lose weight most rapidly and what is this rate of weight loss? c. Use the graphical technique of Example 3 to graph the derivative of weight W.
p
Number of flies
iii) 2 p.m.
c. Use the graphical technique of Example 3 to graph the derivative of temperature T versus time t.
34. Fruit flies (Continuation of Example 4, Section 2.1.) Populations starting out in closed environments grow slowly at first, when there are relatively few members, then more rapidly as the number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment.
350 300 250 200 150 100 50 0
ii) 9 a.m.
b. At what time does the temperature increase most rapidly? Decrease most rapidly? What is the rate for each of those times?
OneSided Derivatives Compute the righthand and lefthand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.
37.
38.
y
y y = f (x)
10
20 30 Time (days)
40
50
t
y = x2
y = 2x
y = f (x)
y=2 2 P(1, 2)
y=x
b. During what days does the population seem to be increasing fastest? Slowest? P(0, 0)
1 x
0
1
2
x
3.2 The Derivative as a Function
39.
40. y
Theory and Examples In Exercises 49–52,
y
P(1, 1) y = "x
1 0
a. Find the derivative ƒ′(x) of the given function y = ƒ(x).
y = f (x)
y = f (x)
P(1, 1)
y = 2x − 1
b. Graph y = ƒ(x) and y = ƒ′(x) side by side using separate sets of coordinate axes, and answer the following questions.
y = 1x
1
x
1
y=x x
1
In Exercises 41 and 42, determine if the piecewisedefined function is differentiable at the origin. 41. ƒ(x) = e
2x  1, x2 + 2x + 7,
42. g(x) = e
x2>3, x1>3,
x Ú 0 x 6 0
x Ú 0 x 6 0
a. differentiable? b. continuous but not differentiable? Give reasons for your answers. 44.
y
y = f (x) D: −3 ≤ x ≤ 2 2
2
1
1
−3 −2 −1
0 1
2
x
d. Over what intervals of xvalues, if any, does the function y = ƒ(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.) 50. y =  1>x
49. y = x2
51. y = x3 >3
52. y = x4 >4
−2 −1
y = f (x) D: −2 ≤ x ≤ 3
0
−1
−1
−2
−2
54. Tangent to y = 2x Does any tangent to the curve y = 2x cross the xaxis at x =  1? If so, find an equation for the line and the point of tangency. If not, why not? 55. Derivative of −ƒ Does knowing that a function ƒ(x) is differentiable at x = x0 tell you anything about the differentiability of the function ƒ at x = x0? Give reasons for your answer. 56. Derivative of multiples Does knowing that a function g(t) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7? Give reasons for your answer.
c. neither continuous nor differentiable?
y
c. For what values of x, if any, is ƒ′ positive? Zero? Negative?
53. Tangent to a parabola Does the parabola y = 2x2  13x + 5 have a tangent whose slope is  1? If so, find an equation for the line and the point of tangency. If not, why not?
Differentiability and Continuity on an Interval Each figure in Exercises 43–48 shows the graph of a function over a closed interval D. At what domain points does the function appear to be
43.
1
2
3
x
57. Limit of a quotient Suppose that functions g(t) and h(t) are defined for all values of t and g(0) = h(0) = 0. Can limt S 0 (g(t))>(h(t)) exist? If it does exist, must it equal zero? Give reasons for your answers.
58. a. Let ƒ(x) be a function satisfying 0 ƒ(x) 0 … x2 for 1 … x … 1. Show that ƒ is differentiable at x = 0 and find ƒ′(0). b. Show that ƒ(x) = c
1 x2 sin x , x ≠ 0 x = 0
0, 45.
46.
y
is differentiable at x = 0 and find ƒ′(0).
y
y = f (x) D: −3 ≤ x ≤ 3
−3 −2 −1 0 −1
3 1
2
3
y =
2
x
1
−2
−2 −1
47.
1
0
2
3
x
y y = f (x) D: −1 ≤ x ≤ 2
4
0
y = f (x) D: −3 ≤ x ≤ 3
1
2
x
−3 −2 −1 0
h
for h = 1, 0.5, 0.1. Then try h = 1, 0.5,  0.1. Explain what is going on.
y =
(x + h)3  x3 h
for h = 2, 1, 0.2. Then try h = 2,  1, 0.2. Explain what is going on.
2
1
2x + h  2x
T 60. Graph y = 3x2 in a window that has 2 … x … 2, 0 … y … 3. Then, on the same screen, graph
48. y
−1
T 59. Graph y = 1> 1 2 2x 2 in a window that has 0 … x … 2. Then, on the same screen, graph
y = f (x) D: −2 ≤ x ≤ 3
1
135
1 2
3
x
61. Derivative of y = ∣ x ∣ Graph the derivative of ƒ(x) = 0 x 0 . Then graph y = ( 0 x 0  0 ) >(x  0) = 0 x 0 >x. What can you conclude?
136
Chapter 3: Derivatives
T 62. Weierstrass’s nowhere differentiable continuous function The sum of the first eight terms of the Weierstrass function q ƒ(x) = g n = 0 (2>3)n cos (9npx) is
e. Substitute various values for x larger and smaller than x0 into the formula obtained in part (c). Do the numbers make sense with your picture? f. Graph the formula obtained in part (c). What does it mean when its values are negative? Zero? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer.
g(x) = cos (px) + (2>3)1 cos (9px) + (2>3)2 cos (92px) + (2>3) cos (9 px) + g + (2>3) cos (9 px). 3
3
7
7
Graph this sum. Zoom in several times. How wiggly and bumpy is this graph? Specify a viewing window in which the displayed portion of the graph is smooth. COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exercises 63–68.
a. Plot y = ƒ(x) to see that function’s global behavior. b. Define the difference quotient q at a general point x, with general step size h. c. Take the limit as h S 0. What formula does this give?
63. ƒ(x) = x3 + x2  x, x0 = 1 64. ƒ(x) = x1>3 + x2>3, x0 = 1 65. ƒ(x) =
4x , x0 = 2 x2 + 1
66. ƒ(x) =
x  1 , x0 = 1 3x2 + 1
67. ƒ(x) = sin 2x, x0 = p>2 68. ƒ(x) = x2 cos x, x0 = p>4
d. Substitute the value x = x0 and plot the function y = ƒ(x) together with its tangent line at that point.
3.3 Differentiation Rules This section introduces several rules that allow us to differentiate constant functions, power functions, polynomials, exponential functions, rational functions, and certain combinations of them, simply and directly, without having to take limits each time.
Powers, Multiples, Sums, and Differences A simple rule of differentiation is that the derivative of every constant function is zero. y c
(x + h, c)
(x, c)
y=c
Derivative of a Constant Function If ƒ has the constant value ƒ(x) = c, then dƒ d = (c) = 0. dx dx
h 0
x
x+h
FIGURE 3.9 The rule (d>dx)(c) = 0 is another way to say that the values of constant functions never change and that the slope of a horizontal line is zero at every point.
x
Proof We apply the definition of the derivative to ƒ(x) = c, the function whose outputs have the constant value c (Figure 3.9). At every value of x, we find that ƒ(x + h)  ƒ(x) c  c = lim = lim 0 = 0. h h hS0 hS0 hS0
ƒ′(x) = lim
From Section 3.1, we know that d 1 1 a b =  2, dx x x
or
d 1 ( x ) = x  2. dx
From Example 2 of the last section we also know that d 1 2x 2 = 1 , dx 2 2x
or
d 1>2 ( x ) = 12 x  1>2 . dx
These two examples illustrate a general rule for differentiating a power xn . We first prove the rule when n is a positive integer.
3.3 Differentiation Rules
137
Derivative of a Positive Integer Power If n is a positive integer, then d n x = nxn  1. dx
HISTORICAL BIOGRAPHY Richard Courant (1888–1972)
Proof of the Positive Integer Power Rule z  x = (z  x)(z n
n
n1
+ z
n2
The formula x + g + zxn  2 + xn  1)
can be verified by multiplying out the righthand side. Then from the alternative formula for the definition of the derivative, ƒ(z)  ƒ(x) zn  x n = lim z  x z x zSx zSx
ƒ′(x) = lim
= lim (zn  1 + zn  2x + g + zxn  2 + xn  1) S z
x
n terms
= nxn  1. The Power Rule is actually valid for all real numbers n. We have seen examples for a negative integer and fractional power, but n could be an irrational number as well. To apply the Power Rule, we subtract 1 from the original exponent n and multiply the result by n. Here we state the general version of the rule, but postpone its proof until Section 3.8.
Power Rule (General Version)
If n is any real number, then d n x = nxn  1, dx for all x where the powers xn and xn  1 are defined.
EXAMPLE 1 (a) x3
Applying the Power Rule Subtract 1 from the exponent and multiply the result by the original exponent.
Differentiate the following powers of x.
(b) x2/3
(c) x 22
(d)
1 x4
(e) x4>3
(f) 2x2 + p
Solution d 3 ( x ) = 3x3  1 = 3x2 (a) dx (b)
d 2>3 ( x ) = 23 x(2>3)  1 = 23 x1>3 dx
(c)
d 22 1 x 2 = 22x 22  1 dx
(d)
d 1 d 4 ( x ) = 4x4  1 = 4x5 =  45 a b = dx x4 dx x
(e)
d 4>3 ( x ) =  43 x(4>3)  1 =  43 x7>3 dx
(f)
d d 1 + (p>2) 1 2x2 + p 2 = dx 1x 2 = a1 + p2 b x1 + (p>2)  1 = 21 (2 + p) 2xp dx
The next rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.
138
Chapter 3: Derivatives
Derivative Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then d du (cu) = c . dx dx
In particular, if n is any real number, then
y y=
3x 2
d (cxn) = cnxn  1. dx
Proof 3
Slope = 3(2x) = 6x Slope = 6(1) = 6
(1, 3)
cu(x + h)  cu(x) d cu = lim dx h S h 0
2
1
0
u(x + h)  u(x) h hS0
Constant Multiple Limit Property
du dx
u is differentiable.
= clim
y = x2
= c Slope = 2x Slope = 2(1) = 2 (1, 1)
1
2
x
FIGURE 3.10 The graphs of y = x2 and y = 3x2. Tripling the ycoordinate triples the slope (Example 2).
Derivative definition with ƒ(x) = cu(x)
EXAMPLE 2 (a) The derivative formula d (3x2) = 3 # 2x = 6x dx says that if we rescale the graph of y = x2 by multiplying each ycoordinate by 3, then we multiply the slope at each point by 3 (Figure 3.10). (b) Negative of a function The derivative of the negative of a differentiable function u is the negative of the function’s derivative. The Constant Multiple Rule with c = 1 gives d d d du (u) = (1 # u) = 1 # (u) =  . dx dx dx dx
Denoting Functions by u and Y The functions we are working with when we need a differentiation formula are likely to be denoted by letters like ƒ and g. We do not want to use these same letters when stating general differentiation rules, so instead we use letters like u and y that are not likely to be already in use.
The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives.
Derivative Sum Rule If u and y are differentiable functions of x, then their sum u + y is differentiable at every point where u and y are both differentiable. At such points, d du dy (u + y) = + . dx dx dx
For example, if y = x4 + 12x, then y is the sum of u(x) = x4 and y(x) = 12x. We then have dy d 4 ( x ) + d (12x) = 4x3 + 12. = dx dx dx
3.3 Differentiation Rules
Proof
139
We apply the definition of the derivative to ƒ(x) = u(x) + y(x):
3 u(x + h) + y(x + h) 4  3 u(x) + y(x)4 d 3 u(x) + y(x) 4 = lim dx h hS0 = lim c hS0
= lim
hS0
u(x + h)  u(x) y(x + h)  y(x) + d h h
u(x + h)  u(x) y(x + h)  y(x) du dy + lim = + . h h dx dx hS0
Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives: d du dy du dy d (u  y) = 3 u + (1)y 4 = + (1) = . dx dx dx dx dx dx The Sum Rule also extends to finite sums of more than two functions. If u1, u2, c, un are differentiable at x, then so is u1 + u2 + g + un , and dun du1 du2 d (u + u2 + g + un) = + + + . dx 1 dx dx g dx For instance, to see that the rule holds for three functions we compute du3 du1 du2 du3 d d d (u1 + u2 + u3) = ((u1 + u2) + u3) = (u1 + u2) + = + + . dx dx dx dx dx dx dx A proof by mathematical induction for any finite number of terms is given in Appendix 2.
EXAMPLE 3 Solution
Find the derivative of the polynomial y = x3 +
dy d 3 d d 4 2 d = (5x) + (1) x + a x b dx dx dx 3 dx dx = 3x2 +
4 2 x  5x + 1. 3
Sum and Difference Rules
8 4# 2x  5 + 0 = 3x2 + x  5 3 3
We can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 3. All polynomials are differentiable at all values of x.
EXAMPLE 4
Does the curve y = x4  2x2 + 2 have any horizontal tangents? If so,
where? Solution The horizontal tangents, if any, occur where the slope dy>dx is zero. We have y
y = x 4 − 2x 2 + 2
dy d 4 = (x  2x2 + 2) = 4x3  4x. dx dx Now solve the equation
(0, 2)
(−1, 1) −1
1
0
(1, 1)
1
x
FIGURE 3.11 The curve in Example 4 and its horizontal tangents.
dy = 0 for x: dx 4x3  4x = 0 4x(x2  1) = 0 x = 0, 1, 1.
The curve y = x4  2x2 + 2 has horizontal tangents at x = 0, 1, and 1. The corresponding points on the curve are (0, 2), (1, 1), and (1, 1). See Figure 3.11. We will see in Chapter 4 that finding the values of x where the derivative of a function is equal to zero is an important and useful procedure.
140
Chapter 3: Derivatives
Derivatives of Exponential Functions We briefly reviewed exponential functions in Section 1.5. When we apply the definition of the derivative to ƒ(x) = ax , we get d x ax + h  ax (a ) = lim dx h hS0
ax # ah  ax h hS0
= lim
= lim ax # hS0
ah  1 h
= ax # lim
ah  1 h hS0
Derivative definition ax + h = ax # ah Factoring out ax ax is constant as h S 0.
ah  1 # x b a. = a lim h hS0 (1+)1+*
(1)
a fixed number L
Thus we see that the derivative of ax is a constant multiple L of ax . The constant L is a limit unlike any we have encountered before. Note, however, that it equals the derivative of ƒ(x) = ax at x = 0:
y
a = 3 a = e a = 2.5
a=2
L = 1.0 1.1 0.92 0.69
0
h y = a − 1, a > 0 h
h
FIGURE 3.12 The position of the curve y = (ah  1)>h, a 7 0, varies continuously with a. The limit L of y as h S 0 changes with different values of a. The number for which L = 1 as h S 0 is the number e between a = 2 and a = 3.
ah  a0 ah  1 = lim = L. h h hS0 hS0
ƒ′(0) = lim
The limit L is therefore the slope of the graph of ƒ(x) = ax where it crosses the yaxis. In Chapter 7, where we carefully develop the logarithmic and exponential functions, we prove that the limit L exists and has the value ln a. For now we investigate values of L by graphing the function y = (ah  1)>h and studying its behavior as h approaches 0. Figure 3.12 shows the graphs of y = (ah  1)>h for four different values of a. The limit L is approximately 0.69 if a = 2, about 0.92 if a = 2.5, and about 1.1 if a = 3. It appears that the value of L is 1 at some number a chosen between 2.5 and 3. That number is given by a = e ≈ 2.718281828. With this choice of base we obtain the natural exponential function ƒ(x) = ex as in Section 1.5, and see that it satisfies the property eh  1 = 1 h hS0
ƒ′(0) = lim
(2)
because it is the exponential function whose graph has slope 1 when it crosses the yaxis. That the limit is 1 implies an important relationship between the natural exponential function ex and its derivative: d x eh  1 # x (e ) = lim ¢ ≤ e dx h hS0 = 1 # ex = ex.
Eq. (1) with a = e Eq. (2)
Therefore the natural exponential function is its own derivative. Derivative of the Natural Exponential Function d x (e ) = ex dx
EXAMPLE 5
Find an equation for a line that is tangent to the graph of y = ex and goes through the origin. Solution Since the line passes through the origin, its equation is of the form y = mx, where m is the slope. If it is tangent to the graph at the point (a, ea), the slope is m = (ea  0)>(a  0). The slope of the natural exponential at x = a is ea . Because these
3.3 Differentiation Rules
slopes are the same, we then have that ea = ea >a. It follows that a = 1 and m = e, so the equation of the tangent line is y = ex. See Figure 3.13.
y 6
y=
ex
4 (a, e a )
2 −1
a
141
x
We might ask if there are functions other than the natural exponential function that are their own derivatives. The answer is that the only functions that satisfy the property that ƒ′(x) = ƒ(x) are functions that are constant multiples of the natural exponential function, ƒ(x) = c # ex , c any constant. We prove this fact in Section 7.2. Note from the Constant Multiple Rule that indeed
FIGURE 3.13 The line through the origin is tangent to the graph of y = ex when a = 1 (Example 5).
d d # x (c e ) = c # (ex) = c # ex. dx dx
Products and Quotients While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d 2 d # (x x) = (x ) = 2x, dx dx
while
d d (x) # (x) = 1 # 1 = 1. dx dx
The derivative of a product of two functions is the sum of two products, as we now explain.
Derivative Product Rule If u and y are differentiable at x, then so is their product uy, and d dy du (uy) = u + y . dx dx dx
Equation (3) is equivalent to saying that (ƒg)′ = ƒ′g + ƒg′. This form of the Product Rule is useful and applies to dot products and cross products of vectorvalued functions, studied in Chapter 13.
The derivative of the product uy is u times the derivative of y plus y times the derivative of u. In prime notation, (uy)′ = uy′ + yu′. In function notation, d 3 ƒ(x)g(x) 4 = ƒ(x)g′(x) + g(x)ƒ′(x). dx
EXAMPLE 6
1 Find the derivative of (a) y = x 1 x2 + ex 2 ,
(3) (b) y = e2x.
Solution (a) We apply the Product Rule with u = 1>x and y = x2 + ex: d 1 2 1 1 c ( x + ex ) d = x ( 2x + ex ) + ( x2 + ex ) a 2 b dx x x
d dy du (uy) = u + y , and dx dx dx d
1 1 a b =  2 x
dx x ex ex = 2 + x  1  2 x ex = 1 + (x  1) 2. x d 2x d x# x d d (b) (e ) = (e e ) = ex # (ex) + ex # (ex) = 2ex # ex = 2e2x dx dx dx dx
EXAMPLE 7
Find the derivative of y = (x2 + 1)(x3 + 3).
Solution (a) From the Product Rule with u = x2 + 1 and y = x3 + 3, we find d 3 ( x2 + 1 ) ( x3 + 3 ) 4 = ( x2 + 1 ) ( 3x2 ) + ( x3 + 3 ) (2x) dx = 3x4 + 3x2 + 2x4 + 6x = 5x4 + 3x2 + 6x.
d dy du (uy) = u + y dx dx dx
142
Chapter 3: Derivatives
(b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial: y = ( x2 + 1 ) ( x3 + 3 ) = x5 + x3 + 3x2 + 3 dy = 5x4 + 3x2 + 6x. dx This is in agreement with our first calculation. Picturing the Product Rule Suppose u(x) and y(x) are positive and increase when x increases, and h 7 0.
Proof of the Derivative Product Rule u(x + h)y(x + h)  u(x)y(x) d (uy) = lim dx h S h 0
y(x + h) Δy
To change this fraction into an equivalent one that contains difference quotients for the derivatives of u and y, we subtract and add u(x + h)y(x) in the numerator:
u(x + h) Δy
y(x) u(x)y(x)
0
y(x) Δu
u(x)
u(x + h)y(x + h)  u(x + h)y(x) + u(x + h)y(x)  u(x)y(x) d (uy) = lim dx h hS0 = lim c u(x + h)
Δu u(x + h)
Then the change in the product uy is the difference in areas of the larger and smaller “squares,” which is the sum of the upper and righthand reddishshaded rectangles. That is,
hS0
= lim u(x + h) # lim hS0
∆(uy) ∆u ∆y = u(x + h) + y(x) . h h h The limit as h S 0 + gives the Product Rule.
hS0
y(x + h)  y(x) u(x + h)  u(x) + y(x) # lim . h h hS0
As h approaches zero, u(x + h) approaches u(x) because u, being differentiable at x, is continuous at x. The two fractions approach the values of dy>dx at x and du>dx at x. Therefore, dy du d (uy) = u + y . dx dx dx
∆(uy) = u(x + h)y(x + h)  u(x)y(x) = u(x + h)∆y + y(x)∆u. Division by h gives
u(x + h)  u(x) y(x + h)  y(x) + y(x) d h h
The derivative of the quotient of two functions is given by the Quotient Rule.
Derivative Quotient Rule If u and y are differentiable at x and if y(x) ≠ 0, then the quotient u>y is differentiable at x, and d u a b = dx y
y
du dy  u dx dx y2
.
In function notation, g(x)ƒ′(x)  ƒ(x)g′(x) d ƒ(x) . c d = dx g(x) g2(x)
EXAMPLE 8
Find the derivative of (a) y =
t2  1 , t3 + 1
(b) y = ex.
Solution (a) We apply the Quotient Rule with u = t 2  1 and y = t 3 + 1: dy (t 3 + 1) # 2t  (t 2  1) # 3t 2 = dt (t 3 + 1)2 4 2t + 2t  3t 4 + 3t 2 = (t 3 + 1)2 t 4 + 3t 2 + 2t = . (t 3 + 1)2
y(du>dt)  u(dy>dt) d u a b = dt y y2
3.3 Differentiation Rules
(b)
143
d x d 1 ex # 0  1 # ex 1 (e ) = a xb = = x = e  x e e dx dx (ex)2
Proof of the Derivative Quotient Rule d u a b = lim dx y hS0
u(x) u(x + h) y(x + h) y(x) h
y(x)u(x + h)  u(x)y(x + h) hy(x + h)y(x) hS0
= lim
To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and y, we subtract and add y(x)u(x) in the numerator. We then get y(x)u(x + h)  y(x)u(x) + y(x)u(x)  u(x)y(x + h) d u a b = lim dx y hy(x + h)y(x) S h 0 y(x) = lim
u(x + h)  u(x) y(x + h)  y(x)  u(x) h h y(x + h)y(x)
hS0
.
Taking the limits in the numerator and denominator now gives the Quotient Rule. Exercise 74 outlines another proof. The choice of which rules to use in solving a differentiation problem can make a difference in how much work you have to do. Here is an example.
EXAMPLE 9
Find the derivative of
(x  1)(x2  2x) . x4 Solution Using the Quotient Rule here will result in a complicated expression with many terms. Instead, use some algebra to simplify the expression. First expand the numerator and divide by x4: y =
y =
(x  1)(x2  2x) x3  3x2 + 2x = = x1  3x2 + 2x3 . x4 x4
Then use the Sum and Power Rules: dy = x2  3(2)x3 + 2(3)x4 dx 6 6 1 =  2 + 3  4. x x x
Second and HigherOrder Derivatives If y = ƒ(x) is a differentiable function, then its derivative ƒ′(x) is also a function. If ƒ′ is also differentiable, then we can differentiate ƒ′ to get a new function of x denoted by ƒ″. So ƒ″ = (ƒ′)′. The function ƒ″ is called the second derivative of ƒ because it is the derivative of the first derivative. It is written in several ways: ƒ″(x) =
d 2y dy′ d dy = y″ = D2(ƒ)(x) = Dx2 ƒ(x). = a b = 2 dx dx dx dx
The symbol D2 means the operation of differentiation is performed twice. If y = x6, then y′ = 6x5 and we have y″ = Thus D2 ( x6 ) = 30x4.
dy′ d = (6x5) = 30x4. dx dx
144
Chapter 3: Derivatives
If y″ is differentiable, its derivative, y‴ = dy″>dx = d 3y>dx3, is the third derivative of y with respect to x. The names continue as you imagine, with
How to Read the Symbols for Derivatives y′ “y prime” y″
“y double prime”
d 2y dx2
“d squared y dx squared”
n d (n  1) d y = n = Dny y dx dx
y(n) =
y‴
“y triple prime”
y(n)
“y super n”
d ny dxn
“d to the n of y by dx to the n”
denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y = ƒ(x) at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.
Dn
“D to the n”
EXAMPLE 10
The first four derivatives of y = x3  3x2 + 2 are First derivative: Second derivative: Third derivative: Fourth derivative:
y′ = 3x2  6x y″ = 6x  6 y‴ = 6 y(4) = 0.
All polynomial functions have derivatives of all orders. In this example, the fifth and later derivatives are all zero.
Exercises
3.3
Derivative Calculations In Exercises 1–12, find the first and second derivatives.
1. y =  x + 3
2. y = x + x + 8
3. s = 5t  3t
4. w = 3z7  7z3 + 21z2
2
3
5. y =
5
4x3  x + 2ex 3
1 7. w = 3z2  z 9. y = 6x2  10x  5x2 11. r =
5 1 2s 3s2
25. y =
1 + x  4 2x x
2
6. y =
x2 x3 + + ex 3 2
8. s = 2t 1 +
4 t2
10. y = 4  2x  x3 12. r =
4 12 1  3 + 4 u u u
In Exercises 13–16, find y′ (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate.
26. r = 2a
1 2u
+ 2ub
28. y =
(x + 1) (x + 2) (x  1) (x  2)
29. y = 2ex + e3x
30. y =
x2 + 3ex 2ex  x
31. y = x3ex
32. w = rer
27. y =
1
( x2  1 ) ( x2 + x + 1 )
33. y = x
9>4
2x
+ e
34. y = x3>5 + p3>2 p 1 + z1.4 2z
35. s = 2t 3>2 + 3e2
36. w =
7 2 37. y = 2 x  xe
3 9.6 38. y = 2 x + 2e1.3
es 39. r = s
40. r = eu a
1 + u p>2 b u2
13. y = ( 3  x2 ) ( x3  x + 1 ) 14. y = (2x + 3) ( 5x2  4x )
Find the derivatives of all orders of the functions in Exercises 41–44.
1 15. y = ( x2 + 1 ) ax + 5 + x b 16. y = ( 1 + x2 )( x3>4  x3 )
41. y =
Find the derivatives of the functions in Exercises 17–40. 2x + 5 4  3x 17. y = 18. z = 2 3x  2 3x + x 19. g(x) =
x2  4 x + 0.5
21. y = (1  t) ( 1 + t 2 ) 1 23. ƒ(s) =
1s  1 1s + 1
20. ƒ(t) =
t2  1 t2 + t  2
22. w = (2x  7)1(x + 5) 24. u =
5x + 1 2 1x
3 x4  x2  x 2 2
42. y =
x5 120
43. y = (x  1) (x + 2)(x + 3) 44. y = (4x2 + 3)(2  x) x Find the first and second derivatives of the functions in Exercises 45–52. x3 + 7 t 2 + 5t  1 45. y = 46. s = x t2 47. r =
(u  1)(u 2 + u + 1) u3
48. u =
(x2 + x)(x2  x + 1) x4
1 + 3z b(3  z) 3z
50. p =
q2 + 3 (q  1)3 + (q + 1)3
49. w = a
3.3 Differentiation Rules
51. w = 3z2e2z
52. w = ez(z  1)(z2 + 1)
53. Suppose u and y are functions of x that are differentiable at x = 0 and that u(0) = 5, u′(0) =  3, y(0) = 1, y′(0) = 2. Find the values of the following derivatives at x = 0. a.
d (uy) dx
b.
d u a b dx y
c.
d y a b dx u
d.
d (7y  2u) dx
61. Find all points (x, y) on the graph of ƒ(x) = 3x2  4x with tangent lines parallel to the line y = 8x + 5. 62. Find all points (x, y) on the graph of g(x) = 13 x3  32 x2 + 1 with tangent lines parallel to the line 8x  2y = 1. 63. Find all points (x, y) on the graph of y = x>(x  2) with tangent lines perpendicular to the line y = 2x + 3. 64. Find all points (x, y) on the graph of ƒ(x) = x2 with tangent lines passing through the point (3, 8).
54. Suppose u and y are differentiable functions of x and that
y 10
u(1) = 2, u′(1) = 0, y(1) = 5, y′(1) = 1. d (uy) dx
b.
d u a b dx y
c.
d y a b dx u
d.
f (x) = x 2 (3, 8)
Find the values of the following derivatives at x = 1. a.
145
6
d (7y  2u) dx
(x, y) 2
Slopes and Tangents 55. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x3  4x + 1 at the point (2, 1).
2
−2
4
x
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8.
65. a. Find an equation for the line that is tangent to the curve y = x3  x at the point ( 1, 0).
56. a. Horizontal tangents Find equations for the horizontal tangents to the curve y = x3  3x  2. Also find equations for the lines that are perpendicular to these tangents at the points of tangency.
T b. Graph the curve and tangent line together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates.
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve’s tangent at this point. 57. Find the tangents to Newton’s serpentine (graphed here) at the origin and the point (1, 2). y y=
4x x2 + 1
66. a. Find an equation for the line that is tangent to the curve y = x3  6x2 + 5x at the origin. T b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates. T c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
(1, 2) 2 1 0
T c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
x
1 2 3 4
Theory and Examples For Exercises 67 and 68 evaluate each limit by first converting each to a derivative at a particular xvalue.
67. lim 58. Find the tangent to the Witch of Agnesi (graphed here) at the point (2, 1). y y= 2 1 0
8 x2 + 4
x50  1 x  1
68. lim
x S 1
x
59. Quadratic tangent to identity function The curve y = ax2 + bx + c passes through the point (1, 2) and is tangent to the line y = x at the origin. Find a, b, and c. 60. Quadratics having a common tangent The curves y = x2 + ax + b and y = cx  x2 have a common tangent line at the point (1, 0). Find a, b, and c.
x2>9  1 x + 1
69. Find the value of a that makes the following function differentiable for all xvalues. g(x) = e
(2, 1) 1 2 3
xS1
ax, x2  3x,
if x 6 0 if x Ú 0
70. Find the values of a and b that make the following function differentiable for all xvalues. ƒ(x) = e
ax + b, bx2  3,
x 7 1 x … 1
71. The general polynomial of degree n has the form P(x) = an xn + an  1 xn  1 + g + a2 x2 + a1 x + a0 where an ≠ 0. Find P′(x).
146
Chapter 3: Derivatives
72. The body’s reaction to medicine The reaction of the body to a dose of medicine can sometimes be represented by an equation of the form C M R = M a  b, 2 3
76. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d m (x ) = mxm  1 dx
2
where m is a positive integer.
where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dR>dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see how to find the amount of medicine to which the body is most sensitive. 73. Suppose that the function y in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?
77. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form P =
an2 nRT  2, V  nb V
in which a, b, n, and R are constants. Find dP>dV . (See accompanying figure.)
74. The Reciprocal Rule a. The Reciprocal Rule says that at any point where the function y(x) is differentiable and different from zero, d 1 1 dy a b =  2 . dx y y dx Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule. 75. Generalizing the Product Rule gives the formula
The Derivative Product Rule
78. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km A(q) = q + cm + , 2
dy du d (uy) = u + y dx dx dx for the derivative of the product uy of two differentiable functions of x. a. What is the analogous formula for the derivative of the product uyw of three differentiable functions of x? b. What is the formula for the derivative of the product u1 u2 u3 u4 of four differentiable functions of x?
where q is the quantity you order when things run low (shoes, TVs, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dA>dq and d 2A>dq2.
c. What is the formula for the derivative of a product u1 u2 u3 gun of a finite number n of differentiable functions of x?
3.4 The Derivative as a Rate of Change In Section 2.1 we introduced average and instantaneous rates of change. In this section we study further applications in which derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.
Instantaneous Rates of Change If we interpret the difference quotient (ƒ(x + h)  ƒ(x))>h as the average rate of change in ƒ over the interval from x to x + h, we can interpret its limit as h S 0 as the rate at which ƒ is changing at the point x.
3.4 The Derivative as a Rate of Change
147
DEFINITION The instantaneous rate of change of ƒ with respect to x at x0 is the derivative ƒ(x0 + h)  ƒ(x0) , h hS0
ƒ′(x0) = lim provided the limit exists.
Thus, instantaneous rates are limits of average rates. It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change.
EXAMPLE 1
The area A of a circle is related to its diameter by the equation A =
p 2 D. 4
How fast does the area change with respect to the diameter when the diameter is 10 m? Solution The rate of change of the area with respect to the diameter is p pD dA = # 2D = . 4 2 dD When D = 10 m, the area is changing with respect to the diameter at the rate of (p>2)10 = 5p m2 >m ≈ 15.71 m2 >m.
Motion Along a Line: Displacement, Velocity, Speed, Acceleration, and Jerk Suppose that an object (or body, considered as a whole mass) is moving along a coordinate line (an saxis), usually horizontal or vertical, so that we know its position s on that line as a function of time t: Position at time t … s = f(t)
Δs
s = ƒ(t).
and at time t + Δt s + Δs = f (t + Δt)
FIGURE 3.14 The positions of a body moving along a coordinate line at time t and shortly later at time t + ∆t. Here the coordinate line is horizontal.
The displacement of the object over the time interval from t to t + ∆t (Figure 3.14) is s
∆s = ƒ(t + ∆t)  ƒ(t), and the average velocity of the object over that time interval is yay =
displacement ∆s ƒ(t + ∆t)  ƒ(t) = . = travel time ∆t ∆t
To find the body’s velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + ∆t as ∆t shrinks to zero. This limit is the derivative of ƒ with respect to t.
DEFINITION Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body’s position at time t is s = ƒ(t), then the body’s velocity at time t is y(t) =
ƒ(t + ∆t)  ƒ(t) ds = lim . dt ∆t ∆t S 0
148
Chapter 3: Derivatives
s s = f (t) ds > 0 dt
t
0 (a) s increasing: positive slope so moving upward
Besides telling how fast an object is moving along the horizontal line in Figure 3.14, its velocity tells the direction of motion. When the object is moving forward (s increasing), the velocity is positive; when the object is moving backward (s decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity. The blue curves in Figure 3.15 represent position along the line over time; they do not portray the path of motion, which lies along the vertical saxis. If we drive to a friend’s house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show 30 on the way back, even though our distance from home is decreasing. The speedometer always shows speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction. DEFINITION
s
Speed = y(t) = `
s = f (t) ds < 0 dt
(b) s decreasing: negative slope so moving downward
FIGURE 3.15 For motion s = ƒ(t) along a straight line (the vertical axis), y = ds>dt is (a) positive when s increases and (b) negative when s decreases.
HISTORICAL BIOGRAPHY Bernard Bolzano (1781–1848)
ds ` dt
EXAMPLE 2
t
0
Speed is the absolute value of velocity.
Figure 3.16 shows the graph of the velocity y = ƒ′(t) of a particle moving along a horizontal line (as opposed to showing a position function s = ƒ(t) such as in Figure 3.15). In the graph of the velocity function, it’s not the slope of the curve that tells us if the particle is moving forward or backward along the line (which is not shown in the figure), but rather the sign of the velocity. Looking at Figure 3.16, we see that the particle moves forward for the first 3 sec (when the velocity is positive), moves backward for the next 2 sec (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at t = 3 sec (when the velocity is zero) and reverses direction as the velocity starts to become negative. The particle is now moving backward and gaining in speed until t = 4 sec, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time t = 4, the particle starts to slow down again until it finally stops at time t = 5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at t = 6 sec, speeding up during the final second of the forward motion indicated in the velocity graph. The rate at which a body’s velocity changes is the body’s acceleration. The acceleration measures how quickly the body picks up or loses speed. In Chapter 13 we will study motion in the plane and in space, where acceleration of an object may also lead to a change in direction. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt. DEFINITIONS Acceleration is the derivative of velocity with respect to time. If a body’s position at time t is s = ƒ(t), then the body’s acceleration at time t is a(t) =
dy d 2s = 2. dt dt
Jerk is the derivative of acceleration with respect to time: j(t) =
da d 3s = 3. dt dt
Near the surface of Earth all bodies fall with the same constant acceleration. Galileo’s experiments with free fall (see Section 2.1) lead to the equation s =
1 2 gt , 2
149
3.4 The Derivative as a Rate of Change y MOVES FORWARD
FORWARD AGAIN
(y > 0)
(y > 0)
Velocity y = f ′(t) Speeds up
Steady (y = const)
Slows down
Speeds up Stands still (y = 0)
0
1
2
3
4
5
6
7
t (sec)
Greatest speed
Speeds up
Slows down
MOVES BACKWARD
(y < 0)
FIGURE 3.16 The velocity graph of a particle moving along a horizontal line, discussed in Example 2.
where s is the distance fallen and g is the acceleration due to Earth’s gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are significant. The value of g in the equation s = (1>2)gt 2 depends on the units used to measure t and s. With t in seconds (the usual unit), the value of g determined by measurement at sea level is approximately 32 ft>sec2 (feet per second squared) in English units, and g = 9.8 m>sec2 (meters per second squared) in metric units. (These gravitational constants depend on the distance from Earth’s center of mass, and are slightly lower on top of Mt. Everest, for example.) The jerk associated with the constant acceleration of gravity (g = 32 ft>sec2) is zero: j = t (seconds) t=0
s (meters)
t=1
5
0
10 15 t=2
20 25 30 35 40
t=3
45
d (g) = 0. dt
An object does not exhibit jerkiness during free fall.
EXAMPLE 3 Figure 3.17 shows the free fall of a heavy ball bearing released from rest at time t = 0 sec. (a) How many meters does the ball fall in the first 3 sec? (b) What is its velocity, speed, and acceleration when t = 3? Solution (a) The metric freefall equation is s = 4.9t 2. During the first 3 sec, the ball falls s(3) = 4.9(3)2 = 44.1 m. (b) At any time t, velocity is the derivative of position:
FIGURE 3.17 A ball bearing falling from rest (Example 3).
y(t) = s′(t) =
d ( 4.9t2 ) = 9.8t. dt
150
Chapter 3: Derivatives
At t = 3, the velocity is y(3) = 29.4 m>sec in the downward (increasing s) direction. The speed at t = 3 is speed = 0 y(3) 0 = 29.4 m>sec.
The acceleration at any time t is a(t) = y′(t) = s″(t) = 9.8 m>sec2. At t = 3, the acceleration is 9.8 m>sec2.
EXAMPLE 4 A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft > sec (about 109 mph) (Figure 3.18a). It reaches a height of s = 160t  16t 2 ft after t sec.
s y=0
Height (ft)
smax
(a) How high does the rock go? (b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down? (c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock hit the ground again?
t=?
256
Solution (a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is 0. To find the maximum height, all we need to do is to find when y = 0 and evaluate s at this time. At any time t during the rock’s motion, its velocity is
s=0 (a) s, y
y =
400
s = 160t − 16t 2
ds d = (160t  16t 2) = 160  32t ft>sec. dt dt
The velocity is zero when 160  32t = 0
160
or
t = 5 sec.
The rock’s height at t = 5 sec is 0 −160
5
10
t
y = ds = 160 − 32t dt (b)
FIGURE 3.18 (a) The rock in Example 4. (b) The graphs of s and y as functions of time; s is largest when y = ds>dt = 0. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock’s velocity, graphed here as a straight line.
smax = s(5) = 160(5)  16(5)2 = 800  400 = 400 ft. See Figure 3.18b. (b) To find the rock’s velocity at 256 ft on the way up and again on the way down, we first find the two values of t for which s(t) = 160t  16t 2 = 256. To solve this equation, we write 16t 2  160t + 256 16(t 2  10t + 16) (t  2)(t  8) t
= = = =
0 0 0 2 sec, t = 8 sec.
The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock’s velocities at these times are y(2) = 160  32(2) = 160  64 = 96 ft>sec. y(8) = 160  32(8) = 160  256 = 96 ft>sec.
3.4 The Derivative as a Rate of Change
151
At both instants, the rock’s speed is 96 ft > sec. Since y(2) 7 0, the rock is moving upward (s is increasing) at t = 2 sec; it is moving downward (s is decreasing) at t = 8 because y(8) 6 0. (c) At any time during its flight following the explosion, the rock’s acceleration is a constant a =
d dy = (160  32t) = 32 ft>sec2. dt dt
The acceleration is always downward and is the effect of gravity on the rock. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = 0. The equation 160t  16t 2 = 0 factors to give 16t(10  t) = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward. It returned to the ground 10 sec later. Cost y (dollars)
Derivatives in Economics
Slope = marginal cost
y = c (x)
x+h x Production (tons/week)
0
x
FIGURE 3.19 Weekly steel production: c(x) is the cost of producing x tons per week. The cost of producing an additional h tons is c(x + h)  c(x).
Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc>dx. Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton: average cost of each of the additional c(x + h)  c(x) = h tons of steel produced. h The limit of this ratio as h S 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.19): c(x + h)  c(x) dc = lim = marginal cost of production. dx h S 0 h
y y = c(x)
Δc
Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit: ∆c c(x + 1)  c(x) , = 1 ∆x
dc dx
which is approximated by the value of dc>dx at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc>dx, which is the rise in the tangent line if ∆x = 1 (Figure 3.20). The approximation works best for large values of x. Economists often represent a total cost function by a cubic polynomial
Δx = 1
c(x) = ax3 + bx2 + gx + d 0
x
x+1
x
FIGURE 3.20 The marginal cost dc>dx is approximately the extra cost ∆c of producing ∆x = 1 more unit.
where d represents fixed costs, such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs, such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.
EXAMPLE 5
Suppose that it costs c(x) = x3  6x2 + 15x
152
Chapter 3: Derivatives
dollars to produce x radiators when 8 to 30 radiators are produced and that r(x) = x3  3x2 + 12x gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling 11 radiators a day? Solution The cost of producing one more radiator a day when 10 are produced is about c′(10): c′(x) =
d 3 1 x  6x2 + 15x 2 = 3x2  12x + 15 dx
c′(10) = 3(100)  12(10) + 15 = 195. The additional cost will be about $195. The marginal revenue is r′(x) =
d 3 (x  3x2 + 12x) = 3x2  6x + 12. dx
The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about r′(10) = 3(100)  6(10) + 12 = +252 if you increase sales to 11 radiators a day.
EXAMPLE 6 To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes T with respect to income is dT>dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.
y 1
Sensitivity to Change
y = 2p − p 2 0
1
p
(a)
When a small change in x produces a large change in the value of a function ƒ(x), we say that the function is relatively sensitive to changes in x. The derivative ƒ′(x) is a measure of this sensitivity.
EXAMPLE 7
dydp
Genetic Data and Sensitivity to Change
The Austrian monk Gregor Johann Mendel (1822–1884), working with garden peas and other plants, provided the first scientific explanation of hybridization. His careful records showed that if p (a number between 0 and 1) is the frequency of the gene for smooth skin in peas (dominant) and (1  p) is the frequency of the gene for wrinkled skin in peas, then the proportion of smoothskinned peas in the next generation will be
2
dy = 2 − 2p dp
y = 2p(1  p) + p2 = 2p  p2. 0
1
p
(b)
FIGURE 3.21 (a) The graph of y = 2p  p2, describing the proportion of smoothskinned peas in the next generation. (b) The graph of dy>dp (Example 7).
The graph of y versus p in Figure 3.21a suggests that the value of y is more sensitive to a change in p when p is small than when p is large. Indeed, this fact is borne out by the derivative graph in Figure 3.21b, which shows that dy>dp is close to 2 when p is near 0 and close to 0 when p is near 1. The implication for genetics is that introducing a few more smooth skin genes into a population where the frequency of wrinkled skin peas is large will have a more dramatic effect on later generations than will a similar increase when the population has a large proportion of smooth skin peas.
3.4 The Derivative as a Rate of Change
Exercises
153
3.4
Motion Along a Coordinate Line Exercises 1–6 give the positions s = ƒ(t) of a body moving on a coordinate line, with s in meters and t in seconds.
a. Find the body’s displacement and average velocity for the given time interval. b. Find the body’s speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? 1. s = t 2  3t + 2, 0 … t … 2
12. Speeding bullet A 45caliber bullet shot straight up from the surface of the moon would reach a height of s = 832t  2.6t 2 ft after t sec. On Earth, in the absence of air, its height would be s = 832t  16t 2 ft after t sec. How long will the bullet be aloft in each case? How high will the bullet go? 13. Free fall from the Tower of Pisa Had Galileo dropped a cannonball from the Tower of Pisa, 179 ft above the ground, the ball’s height above the ground t sec into the fall would have been s = 179  16t 2.
2. s = 6t  t 2, 0 … t … 6
a. What would have been the ball’s velocity, speed, and acceleration at time t?
3. s = t 3 + 3t 2  3t, 0 … t … 3
b. About how long would it have taken the ball to hit the ground?
4. s = ( t 4 >4 )  t 3 + t 2, 0 … t … 3 5. s =
25 5  t, 1 … t … 5 t2
6. s =
25 , 4 … t … 0 t + 5
7. Particle motion At time t, the position of a body moving along the saxis is s = t 3  6t 2 + 9t m. a. Find the body’s acceleration each time the velocity is zero. b. Find the body’s speed each time the acceleration is zero. c. Find the total distance traveled by the body from t = 0 to t = 2. 8. Particle motion At time t Ú 0, the velocity of a body moving along the horizontal saxis is y = t 2  4t + 3. a. Find the body’s acceleration each time the velocity is zero.
c. What would have been the ball’s velocity at the moment of impact? 14. Galileo’s freefall formula Galileo developed a formula for a body’s velocity during free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball’s behavior when the plank was vertical and the ball fell freely; see part (a) of the accompanying figure. He found that, for any given angle of the plank, the ball’s velocity t sec into motion was a constant multiple of t. That is, the velocity was given by a formula of the form y = kt. The value of the constant k depended on the inclination of the plank. In modern notation—part (b) of the figure—with distance in meters and time in seconds, what Galileo determined by experiment was that, for any given angle u, the ball’s velocity t sec into the roll was y = 9.8(sin u)t m>sec.
b. When is the body moving forward? Backward? c. When is the body’s velocity increasing? Decreasing?
Freefall position
FreeFall Applications 9. Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t 2 on Mars and s = 11.44t 2 on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m > sec (about 100 km > h) on each planet?
10. Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m > sec (about 86 km > h) reaches a height of s = 24t  0.8t 2 m in t sec. a. Find the rock’s velocity and acceleration at time t. (The acceleration in this case is the acceleration of gravity on the moon.) b. How long does it take the rock to reach its highest point? c. How high does the rock go? d. How long does it take the rock to reach half its maximum height? e. How long is the rock aloft? 11. Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m > sec. Because the acceleration of gravity at the planet’s surface was gs m>sec2, the explorers expected the ball bearing to reach a height of s = 15t  (1>2)gs t 2 m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gs?
?
u (b)
(a)
a. What is the equation for the ball’s velocity during free fall? b. Building on your work in part (a), what constant acceleration does a freely falling body experience near the surface of Earth? Understanding Motion from Graphs 15. The accompanying figure shows the velocity y = ds>dt = ƒ(t) (m > sec) of a body moving along a coordinate line. y (m/sec) y = f (t)
3 0
2
4
6
8 10
t (sec)
−3
a. When does the body reverse direction? b. When (approximately) is the body moving at a constant speed?
154
Chapter 3: Derivatives
c. Graph the body’s speed for 0 … t … 10.
18. The accompanying figure shows the velocity y = ƒ(t) of a particle moving on a horizontal coordinate line.
d. Graph the acceleration, where defined.
y
16. A particle P moves on the number line shown in part (a) of the accompanying figure. Part (b) shows the position of P as a function of time t. P
s (cm)
0
y = f(t) 0
1 2 3 4 5 6 7 8 9
t (sec)
(a) s (cm)
a. When does the particle move forward? Move backward? Speed up? Slow down?
s = f (t)
2
b. When is the particle’s acceleration positive? Negative? Zero?
0
1
2
3
4
5
6
t (sec)
−2
c. When does the particle move at its greatest speed? d. When does the particle stand still for more than an instant?
(6, −4)
−4 (b)
a. When is P moving to the left? Moving to the right? Standing still?
19. Two falling balls The multiflash photograph in the accompanying figure shows two balls falling from rest. The vertical rulers are marked in centimeters. Use the equation s = 490t 2 (the freefall equation for s in centimeters and t in seconds) to answer the following questions. (Source: PSSC Physics, 2nd ed., Reprinted by permission of Education Development Center, Inc.)
b. Graph the particle’s velocity and speed (where defined). 17. Launching a rocket When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following. a. How fast was the rocket climbing when the engine stopped? b. For how many seconds did the engine burn? 200
Velocity (ftsec)
150 100 50 0 −50 −100 0
2
4 6 8 10 Time after launch (sec)
12
c. When did the rocket reach its highest point? What was its velocity then? d. When did the parachute pop out? How fast was the rocket falling then?
a. How long did it take the balls to fall the first 160 cm? What was their average velocity for the period?
e. How long did the rocket fall before the parachute opened?
b. How fast were the balls falling when they reached the 160cm mark? What was their acceleration then?
f. When was the rocket’s acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)?
c. About how fast was the light flashing (flashes per second)?
3.4 The Derivative as a Rate of Change
20. A traveling truck The accompanying graph shows the position s of a truck traveling on a highway. The truck starts at t = 0 and returns 15 h later at t = 15. a. Use the technique described in Section 3.2, Example 3, to graph the truck’s velocity y = ds>dt for 0 … t … 15. Then repeat the process, with the velocity curve, to graph the truck’s acceleration dy>dt. b. Suppose that s = 15t 2  t 3. Graph ds>dt and d 2s>dt 2 and compare your graphs with those in part (a).
Position, s (km)
500
155
Economics 23. Marginal cost Suppose that the dollar cost of producing x washing machines is c(x) = 2000 + 100x  0.1x2.
a. Find the average cost per machine of producing the first 100 washing machines. b. Find the marginal cost when 100 washing machines are produced. c. Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly.
400
24. Marginal revenue Suppose that the revenue from selling x washing machines is
300
1 r(x) = 20,000a1  x b dollars.
200
a. Find the marginal revenue when 100 machines are produced. 100 0
5 10 Elapsed time, t (hr)
15
21. The graphs in the accompanying figure show the position s, velocity y = ds>dt, and acceleration a = d 2s>dt 2 of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers. y A
B
b. Use the function r′(x) to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week. c. Find the limit of r′(x) as x S q. How would you interpret this number? Additional Applications 25. Bacterium population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time t (hours) was b = 106 + 104t  103t 2. Find the growth rates at
a. t = 0 hours. C
b. t = 5 hours. c. t = 10 hours. t
0
22. The graphs in the accompanying figure show the position s, the velocity y = ds>dt, and the acceleration a = d 2s>dt 2 of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers. y
26. Body surface area A typical male’s body surface area S in 1 square meters is often modeled by the formula S = 60 2wh, where h is the height in cm, and w the weight in kg, of the person. Find the rate of change of body surface area with respect to weight for males of constant height h = 180 cm (roughly 5′9″). Does S increase more rapidly with respect to weight at lower or higher body weights? Explain. T 27. Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula y = 6a1 
A
t 2 b m. 12
a. Find the rate dy>dt (m > h) at which the tank is draining at time t. b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy>dt at these times?
0
t B
c. Graph y and dy>dt together and discuss the behavior of y in relation to the signs and values of dy>dt. 28. Draining a tank The number of gallons of water in a tank t minutes after the tank has started to drain is Q(t) = 200(30  t)2. How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min?
C
156
Chapter 3: Derivatives
29. Vehicular stopping distance Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is
vent in the crater’s floor, which at one point shot lava 1900 ft straight into the air (a Hawaiian record). What was the lava’s exit velocity in feet per second? In miles per hour? (Hint: If y0 is the exit velocity of a particle of lava, its height t sec later will be s = y0 t  16t 2 ft. Begin by finding the time at which ds>dt = 0. Neglect air resistance.)
s = 1.1y + 0.054y2, where s is measured in ft and y in mph. The linear term 1.1y models the distance the car travels during the time the driver perceives a need to stop until the brakes are applied, and the quadratic term 0.054y2 models the additional braking distance once they are applied. Find ds>dy at y = 35 and y = 70 mph, and interpret the meaning of the derivative. 30. Inflating a balloon The volume V = (4>3)pr 3 of a spherical balloon changes with the radius. a. At what rate (ft3 >ft) does the volume change with respect to the radius when r = 2 ft?
Analyzing Motion Using Graphs T Exercises 33–36 give the position function s = ƒ(t) of an object moving along the saxis as a function of time t. Graph ƒ together with the velocity function y(t) = ds>dt = ƒ′(t) and the acceleration function a(t) = d 2s>dt 2 = ƒ″(t). Comment on the object’s behavior in relation to the signs and values of y and a. Include in your commentary such topics as the following:
a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)?
b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 ft? 31. Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = (10>9)t 2, where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km>h. How long will it take to become airborne, and what distance will it travel in that time? 32. Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single
c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? 33. s = 200t  16t 2, 0 … t … 12.5 (a heavy object fired straight up from Earth’s surface at 200 ft > sec) 34. s = t 2  3t + 2, 0 … t … 5 35. s = t 3  6t 2 + 7t, 0 … t … 4 36. s = 4  7t + 6t 2  t 3, 0 … t … 4
3.5 Derivatives of Trigonometric Functions Many phenomena of nature are approximately periodic (electromagnetic fields, heart rhythms, tides, weather). The derivatives of sines and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.
Derivative of the Sine Function To calculate the derivative of ƒ(x) = sin x, for x measured in radians, we combine the limits in Example 5a and Theorem 7 in Section 2.4 with the angle sum identity for the sine function: sin (x + h) = sin x cos h + cos x sin h. If ƒ(x) = sin x, then ƒ(x + h)  ƒ(x) sin (x + h)  sin x = lim h h hS0 (sin x cos h + cos x sin h)  sin x = lim h hS0 sin x (cos h  1) + cos x sin h = lim h hS0
ƒ′(x) = lim
hS0
= lim asin x # hS0
Derivative definition
cos h  1 sin h b + lim acos x # b h h hS0
cos h  1 sin h + cos x # lim = sin x # 0 + cos x # 1 = cos x. = sin x # lim h hS0 hS0 h (++)++* (11)11* Example 5a and limit 0
limit 1
Theorem 7, Section 2.4
3.5 Derivatives of Trigonometric Functions
157
The derivative of the sine function is the cosine function: d (sin x) = cos x. dx
EXAMPLE 1
We find derivatives of the sine function involving differences, products,
and quotients. (a) y = x2  sin x:
dy d = 2x (sin x) dx dx
Difference Rule
(b) y = exsin x:
= 2x  cos x dy d x d = ex (sin x) + (e ) sin x dx dx dx = ex cos x + ex sin x
Product Rule
= ex (cos x + sin x) sin x (c) y = x :
#d # dy x dx (sin x)  sin x 1 = dx x2 =
Quotient Rule
x cos x  sin x x2
Derivative of the Cosine Function With the help of the angle sum formula for the cosine function, cos (x + h) = cos x cos h  sin x sin h, we can compute the limit of the difference quotient: cos (x + h)  cos x d (cos x) = lim dx h hS0 = lim
hS0
(cos x cos h  sin x sin h)  cos x h
Derivative definition Cosine angle sum identity
cos x (cos h  1)  sin x sin h h hS0
= lim
= lim cos x #
cos h  1 sin h  lim sin x # h h hS0
= cos x # lim
cos h  1 sin h  sin x # lim h S h 0 h
hS0
hS0
y y = cos x
1 −p
p
0 −1 y′
0 −1
y′ = −sin x
p
Example 5a and Theorem 7, Section 2.4
x
1 −p
= cos x # 0  sin x # 1 = sin x.
x
FIGURE 3.22 The curve y′ = sin x as the graph of the slopes of the tangents to the curve y = cos x.
The derivative of the cosine function is the negative of the sine function: d (cos x) = sin x. dx
Figure 3.22 shows a way to visualize this result in the same way we did for graphing derivatives in Section 3.2, Figure 3.6.
158
Chapter 3: Derivatives
EXAMPLE 2
We find derivatives of the cosine function in combinations with other
functions. (a) y = 5ex + cos x: dy d ( 5ex ) + d (cos x) = dx dx dx
Sum Rule
= 5ex  sin x (b) y = sin x cos x: dy d d = sin x (cos x) + cos x (sin x) dx dx dx
Product Rule
= sin x (sin x) + cos x (cos x) = cos2 x  sin2 x (c) y =
cos x : 1  sin x d d dy (1  sin x) dx (cos x)  cos x dx (1  sin x) = dx (1  sin x)2 (1  sin x)(sin x)  cos x(0  cos x) (1  sin x)2 1  sin x = (1  sin x)2 1 = 1  sin x
Quotient Rule
=
sin2 x + cos2 x = 1
Simple Harmonic Motion The motion of an object or weight bobbing freely up and down with no resistance on the end of a spring is an example of simple harmonic motion. The motion is periodic and repeats indefinitely, so we represent it using trigonometric functions. The next example describes a case in which there are no opposing forces such as friction to slow the motion.
EXAMPLE 3 A weight hanging from a spring (Figure 3.23) is stretched down 5 units beyond its rest position and released at time t = 0 to bob up and down. Its position at any later time t is s = 5 cos t.
−5
What are its velocity and acceleration at time t? 0
Rest position
5
Position at t=0
Solution We have Position: s = 5 cos t ds d Velocity: y = = (5 cos t) = 5 sin t dt dt
s
FIGURE 3.23 A weight hanging from a vertical spring and then displaced oscillates above and below its rest position (Example 3).
Acceleration: a =
dy d = (5 sin t) = 5 cos t. dt dt
Notice how much we can learn from these equations: 1.
2.
As time passes, the weight moves down and up between s = 5 and s = 5 on the saxis. The amplitude of the motion is 5. The period of the motion is 2p, the period of the cosine function. The velocity y = 5 sin t attains its greatest magnitude, 5, when cos t = 0, as the graphs show in Figure 3.24. Hence, the speed of the weight, 0 y 0 = 5 0 sin t 0 , is greatest
3.5 Derivatives of Trigonometric Functions s, y 5
y = −5 sin t
s = 5 cos t
3.
0
p 2
p
3p 2
2p 5p 2
t
4. −5
FIGURE 3.24 The graphs of the position and velocity of the weight in Example 3.
159
when cos t = 0, that is, when s = 0 (the rest position). The speed of the weight is zero when sin t = 0. This occurs when s = 5 cos t = {5, at the endpoints of the interval of motion. The weight is acted on by the spring and by gravity. When the weight is below the rest position, the combined forces pull it up, and when it is above the rest position, they pull it down. The weight’s acceleration is always proportional to the negative of its displacement. This property of springs is called Hooke’s Law, and is studied further in Section 6.5. The acceleration, a = 5 cos t, is zero only at the rest position, where cos t = 0 and the force of gravity and the force from the spring balance each other. When the weight is anywhere else, the two forces are unequal and acceleration is nonzero. The acceleration is greatest in magnitude at the points farthest from the rest position, where cos t = {1.
EXAMPLE 4
The jerk associated with the simple harmonic motion in Example 3 is j =
da d = (5 cos t) = 5 sin t. dt dt
It has its greatest magnitude when sin t = {1, not at the extremes of the displacement but at the rest position, where the acceleration changes direction and sign.
Derivatives of the Other Basic Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x tan x = cos x ,
cot x =
cos x , sin x
1 sec x = cos x ,
and
csc x =
1 sin x
are differentiable at every value of x at which they are defined. Their derivatives, calculated from the Quotient Rule, are given by the following formulas. Notice the negative signs in the derivative formulas for the cofunctions. The derivatives of the other trigonometric functions: d (tan x) = sec2 x dx d (sec x) = sec x tan x dx
d (cot x) = csc2 x dx d (csc x) = csc x cot x dx
To show a typical calculation, we find the derivative of the tangent function. The other derivations are left to Exercise 60.
EXAMPLE 5
Find d(tan x)>dx.
Solution We use the Derivative Quotient Rule to calculate the derivative: d d sin x (tan x) = a b = dx dx cos x
cos x
d d (sin x)  sin x (cos x) dx dx cos2 x
=
cos x cos x  sin x (sin x) cos2 x
=
cos2 x + sin2 x cos2 x
=
1 = sec2 x. cos2 x
Quotient Rule
160
Chapter 3: Derivatives
EXAMPLE 6
Find y″ if y = sec x.
Solution Finding the second derivative involves a combination of trigonometric derivatives. y = sec x y′ = sec x tan x d y″ = (sec x tan x) dx = sec x
Derivative rule for secant function
d d (tan x) + tan x (sec x) dx dx
= sec x (sec2 x) + tan x (sec x tan x) = sec3 x + sec x tan2 x
Derivative Product Rule Derivative rules
The differentiability of the trigonometric functions throughout their domains gives another proof of their continuity at every point in their domains (Theorem 1, Section 3.2). So we can calculate limits of algebraic combinations and composites of trigonometric functions by direct substitution.
EXAMPLE 7 We can use direct substitution in computing limits provided there is no division by zero, which is algebraically undefined. lim
xS0
Exercises
22 + sec x 22 + sec 0 22 + 1 23 = = = =  23 cos (p  tan x) cos (p  tan 0) cos (p  0) 1
3.5 In Exercises 23–26, find dr>du.
Derivatives In Exercises 1–18, find dy>dx.
23. r = 4  u 2 sin u
24. r = u sin u + cos u
1. y =  10x + 3 cos x
3 2. y = x + 5 sin x
25. r = sec u csc u
26. r = (1 + sec u) sin u
3. y = x2 cos x
4. y = 2x sec x + 3
In Exercises 27–32, find dp>dq. 1 27. p = 5 + cot q 28. p = (1 + csc q) cos q
5. y = csc x  4 1x +
7 ex
7. ƒ(x) = sin x tan x 9. y = xex sec x 11. y =
cot x 1 + cot x
1 4 13. y = cos x + tan x
6. y = x2 cot x 
1 x2
cos x 8. g(x) = sin2 x 10. y = (sin x + cos x) sec x 12. y =
cos x 1 + sin x
x cos x 14. y = x + cos x
15. y = (sec x + tan x) (sec x  tan x) 16. y = x2 cos x  2x sin x  2 cos x
29. p =
sin q + cos q cos q
30. p =
tan q 1 + tan q
31. p =
q sin q q2  1
32. p =
3q + tan q q sec q
33. Find y″ if a. y = csc x. 34. Find y
(4)
b. y = sec x.
= d y>dx if 4
a. y = 2 sin x.
4
b. y = 9 cos x.
18. g(x) = (2  x) tan2 x
Tangent Lines In Exercises 35–38, graph the curves over the given intervals, together with their tangents at the given values of x. Label each curve and tangent with its equation.
19. s = tan t  et
20. s = t 2  sec t + 5et
35. y = sin x, 3p>2 … x … 2p
1 + csc t 21. s = 1  csc t
sin t 22. s = 1  cos t
17. ƒ(x) = x3 sin x cos x In Exercises 19–22, find ds>dt.
x =  p, 0, 3p>2
3.5 Derivatives of Trigonometric Functions
36. y = tan x,  p>2 6 x 6 p>2
Theory and Examples The equations in Exercises 55 and 56 give the position s = ƒ(t) of a body moving on a coordinate line (s in meters, t in seconds). Find the body’s velocity, speed, acceleration, and jerk at time t = p>4 sec.
x =  p>3, 0, p>3 37. y = sec x,  p>2 6 x 6 p>2 x =  p>3, p>4
55. s = 2  2 sin t
38. y = 1 + cos x,  3p>2 … x … 2p
56. s = sin t + cos t
57. Is there a value of c that will make
x =  p>3, 3p>2
sin2 3x , x≠0 2 ƒ(x) = W x c, x = 0
T Do the graphs of the functions in Exercises 39–42 have any horizontal tangents in the interval 0 … x … 2p? If so, where? If not, why not? Visualize your findings by graphing the functions with a grapher.
continuous at x = 0? Give reasons for your answer.
39. y = x + sin x
58. Is there a value of b that will make
40. y = 2x + sin x 41. y = x  cot x
g(x) = e
42. y = x + 2 cos x 43. Find all points on the curve y = tan x, p>2 6 x 6 p>2, where the tangent line is parallel to the line y = 2x. Sketch the curve and tangent(s) together, labeling each with its equation. 44. Find all points on the curve y = cot x, 0 6 x 6 p, where the tangent line is parallel to the line y =  x. Sketch the curve and tangent(s) together, labeling each with its equation. In Exercises 45 and 46, find an equation for (a) the tangent to the curve at P and (b) the horizontal tangent to the curve at Q. 45.
46.
y
x + b, x 6 0 cos x, x Ú 0
continuous at x = 0? Differentiable at x = 0? Give reasons for your answers. 59. By computing the first few derivatives and looking for a pattern, find d 999 >dx999 (cos x). 60. Derive the formula for the derivative with respect to x of a. sec x.
b. csc x.
c. cot x.
61. A weight is attached to a spring and reaches its equilibrium position (x = 0). It is then set in motion resulting in a displacement of x = 10 cos t,
y Q
where x is measured in centimeters and t is measured in seconds. See the accompanying figure.
p P a , 2b 2
2
161
1
p P a , 4b 4
4
−10 0
p 2 2 y = 4 + cot x − 2csc x 1
x Q 0 p1 4
0
2
3
x
y = 1 + " 2 csc x + cot x
Equilibrium position at x = 0
10 x
Trigonometric Limits Find the limits in Exercises 47–54.
a. Find the spring’s displacement when t = 0, t = p>3, and t = 3p>4.
1 1 47. lim sin a x  b 2 xS2 48. 49.
lim
x S  p>6
lim
u S p>6
b. Find the spring’s velocity when t = 0, t = p>3, and t = 3p>4.
21 + cos (p csc x)
sin u u 
p 6
1 2
62. Assume that a particle’s position on the xaxis is given by 50.
lim
u S p>4
tan u  1 u  p4
p b  1d 51. lim secc ex + p tan a 4 sec x xS0 52. lim sin a xS0
tS0
sin t t b
where x is measured in feet and t is measured in seconds. a. Find the particle’s position when t = 0, t = p>2, and t = p.
p + tan x b tan x  2 sec x
53. lim tan a1 
x = 3 cos t + 4 sin t,
b. Find the particle’s velocity when t = 0, t = p>2, and t = p. 54. lim cos a uS0
pu b sin u
162
Chapter 3: Derivatives
T 63. Graph y = cos x for  p … x … 2p. On the same screen, graph y =
sin (x + h)  sin x h
for h = 1, 0.5, 0.3, and 0.1. Then, in a new window, try h =  1, 0.5, and 0.3. What happens as h S 0+? As h S 0? What phenomenon is being illustrated here? T 64. Graph y =  sin x for  p … x … 2p. On the same screen, graph cos (x + h)  cos x y = h
T 65. Centered difference quotients The centered difference quotient ƒ(x + h)  ƒ(x  h) 2h is used to approximate ƒ′(x) in numerical work because (1) its limit as h S 0 equals ƒ′(x) when ƒ′(x) exists, and (2) it usually gives a better approximation of ƒ′(x) for a given value of h than the difference quotient ƒ(x + h)  ƒ(x) . h See the accompanying figure. y Slope = f ′(x) Slope = B
0
x−h
f (x + h) − f (x − h) 2h
h x
x+h
x
sin (x + h)  sin (x  h) 2h
over the interval 3 p, 2p4 for h = 1, 0.5, and 0.3. Compare the results with those obtained in Exercise 63 for the same values of h. b. To see how rapidly the centered difference quotient for ƒ(x) = cos x converges to ƒ′(x) =  sin x, graph y = sin x together with y =
00 + h0  00  h0 2h
.
As you will see, the limit exists even though ƒ(x) = 0 x 0 has no derivative at x = 0. Moral: Before using a centered difference quotient, be sure the derivative exists. T 67. Slopes on the graph of the tangent function Graph y = tan x and its derivative together on ( p>2, p>2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers. T 68. Slopes on the graph of the cotangent function Graph y = cot x and its derivative together for 0 6 x 6 p. Does the graph of the cotangent function appear to have a smallest slope? A largest slope? Is the slope ever positive? Give reasons for your answers. T 69. Exploring (sin kx) , x Graph y = (sin x)>x, y = (sin 2x)>x, and y = (sin 4x)>x together over the interval  2 … x … 2. Where does each graph appear to cross the yaxis? Do the graphs really intersect the axis? What would you expect the graphs of y = (sin 5x)>x and y = (sin ( 3x))>x to do as x S 0? Why? What about the graph of y = (sin kx)>x for other values of k? Give reasons for your answers.
ƒ(h) =
a. To see how rapidly the centered difference quotient for ƒ(x) = sin x converges to ƒ′(x) = cos x, graph y = cos x together with y =
lim
a. With your graphing calculator or computer grapher in degree mode, graph
y = f(x) h
may have a limit as h S 0 when ƒ has no derivative at x. As a case in point, take ƒ(x) = 0 x 0 and calculate
T 70. Radians versus degrees: degree mode derivatives What happens to the derivatives of sin x and cos x if x is measured in degrees instead of radians? To find out, take the following steps.
f (x + h) − f (x) h
A Slope =
ƒ(x + h)  ƒ(x  h) 2h
hS0
for h = 1, 0.5, 0.3, and 0.1. Then, in a new window, try h =  1, 0.5, and 0.3. What happens as h S 0+? As h S 0? What phenomenon is being illustrated here?
C
66. A caution about centered difference quotients (Continuation of Exercise 65.) The quotient
cos (x + h)  cos (x  h) 2h
over the interval 3 p, 2p4 for h = 1, 0.5, and 0.3. Compare the results with those obtained in Exercise 64 for the same values of h.
sin h h
and estimate limh S 0 ƒ(h). Compare your estimate with p>180. Is there any reason to believe the limit should be p>180? b. With your grapher still in degree mode, estimate lim
hS0
cos h  1 . h
c. Now go back to the derivation of the formula for the derivative of sin x in the text and carry out the steps of the derivation using degreemode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos x using degreemode limits. What formula do you obtain for the derivative? e. The disadvantages of the degreemode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degreemode derivatives of sin x and cos x?
3.6 The Chain Rule
163
3.6 The Chain Rule How do we differentiate F(x) = sin (x2  4)? This function is the composite ƒ ∘ g of two functions y = ƒ(u) = sin u and u = g(x) = x2  4 that we know how to differentiate. The answer, given by the Chain Rule, says that the derivative is the product of the derivatives of ƒ and g. We develop the rule in this section. 2
3 1
C: y turns B: u turns A: x turns
FIGURE 3.25 When gear A makes x turns, gear B makes u turns and gear C makes y turns. By comparing circumferences or counting teeth, we see that y = u>2 (C turns onehalf turn for each B turn) and u = 3x (B turns three times for A’s one), so y = 3x>2. Thus, dy>dx = 3>2 = (1>2)(3) = (dy>du)(du>dx).
Derivative of a Composite Function The function y =
3 1 1 x = (3x) is the composite of the functions y = u and u = 3x. 2 2 2
We have dy 3 = , dx 2 Since
dy 1 = , du 2
du = 3. dx
and
3 1# 3, we see in this case that = 2 2 dy dy du # . = dx du dx
If we think of the derivative as a rate of change, our intuition allows us to see that this relationship is reasonable. If y = ƒ(u) changes half as fast as u and u = g(x) changes three times as fast as x, then we expect y to change 3>2 times as fast as x. This effect is much like that of a multiple gear train (Figure 3.25). Let’s look at another example.
EXAMPLE 1
The function y = ( 3x2 + 1 ) 2
is the composite of y = ƒ(u) = u2 and u = g(x) = 3x2 + 1. Calculating derivatives, we see that dy du # = 2u # 6x du dx
= 2(3x2 + 1) # 6x = 36x3 + 12x.
Substitute for u
Calculating the derivative from the expanded formula (3x2 + 1)2 = 9x4 + 6x2 + 1 gives the same result: dy d ( 9x4 + 6x2 + 1 ) = dx dx = 36x3 + 12x. The derivative of the composite function ƒ(g(x)) at x is the derivative of ƒ at g(x) times the derivative of g at x. This is known as the Chain Rule (Figure 3.26). Composite f ˚ g Rate of change at x is f ′(g(x)) · g′(x).
x
g
f
Rate of change at x is g′(x).
Rate of change at g(x) is f ′( g(x)).
u = g(x)
y = f (u) = f(g(x))
FIGURE 3.26 Rates of change multiply: The derivative of ƒ ∘ g at x is the derivative of ƒ at g(x) times the derivative of g at x.
164
Chapter 3: Derivatives
THEOREM 2—The Chain Rule If ƒ(u) is differentiable at the point u = g(x) and g(x) is differentiable at x, then the composite function (ƒ ∘ g) (x) = ƒ(g(x)) is differentiable at x, and (ƒ ∘ g)′(x) = ƒ′(g(x)) # g′(x).
In Leibniz’s notation, if y = ƒ(u) and u = g(x), then dy dy du # , = dx du dx where dy>du is evaluated at u = g(x).
A Proof of One Case of the Chain Rule: Let ∆u be the change in u when x changes by ∆x, so that ∆u = g(x + ∆x)  g(x). Then the corresponding change in y is ∆y = ƒ(u + ∆u)  ƒ(u). If ∆u ≠ 0, we can write the fraction ∆y> ∆x as the product ∆y ∆y ∆u # = ∆x ∆u ∆x
(1)
and take the limit as ∆x S 0: ∆y dy = lim dx S ∆x ∆x 0 ∆y ∆u # ∆x S 0 ∆u ∆x
= lim
∆y # lim ∆u ∆x S 0 ∆u ∆x S 0 ∆x
= lim
= lim
∆y
#
lim
∆u S 0 ∆u ∆x S 0
=
∆u ∆x
(Note that ∆u S 0 as ∆x S 0 since g is continuous.)
dy du # . du dx
The problem with this argument is that if the function g(x) oscillates rapidly near x, then ∆u can be zero even when ∆x ≠ 0, so the cancelation of ∆u in Equation (1) would be invalid. A complete proof requires a different approach that avoids this problem, and we give one such proof in Section 3.11.
EXAMPLE 2
An object moves along the xaxis so that its position at any time t Ú 0 is given by x(t) = cos (t 2 + 1). Find the velocity of the object as a function of t. Solution We know that the velocity is dx>dt. In this instance, x is a composite function: x = cos(u) and u = t 2 + 1. We have dx = sin(u) du
x = cos(u)
du = 2t. dt
u = t2 + 1
3.6 The Chain Rule
165
By the Chain Rule, dx dx # du = dt du dt
= sin (u) # 2t = sin (t 2 + 1) # 2t = 2t sin (t 2 + 1).
Ways to Write the Chain Rule (ƒ ∘ g)′(x) = ƒ′(g(x)) # g′(x) dy dy du # = dx du dx
“OutsideInside” Rule A difficulty with the Leibniz notation is that it doesn’t state specifically where the derivatives in the Chain Rule are supposed to be evaluated. So it sometimes helps to think about the Chain Rule using functional notation. If y = ƒ(g(x)), then dy = ƒ′(g(x)) # g′(x). dx
dy = ƒ′(g(x)) # g′(x) dx du d ƒ(u) = ƒ′(u) dx dx
dx evaluated at u du
In words, differentiate the “outside” function ƒ and evaluate it at the “inside” function g(x) left alone; then multiply by the derivative of the “inside function.”
EXAMPLE 3
Differentiate sin ( x2 + ex ) with respect to x.
Solution We apply the Chain Rule directly and find d sin ( x2 + ex ) = cos ( x2 + ex ) # ( 2x + ex ) . dx (1)1* (1)1* (1)1* inside
EXAMPLE 4
inside left alone
derivative of the inside
Differentiate y = ecos x.
Solution Here the inside function is u = g(x) = cos x and the outside function is the exponential function ƒ(x) = ex. Applying the Chain Rule, we get dy d cos x ( e ) = ecos x d (cos x) = ecos x(sin x) = ecos x sin x. = dx dx dx Generalizing Example 4, we see that the Chain Rule gives the formula
du d u e = eu . dx dx
For example, d kx d (e ) = ekx # (kx) = kekx, dx dx
for any constant k
and 2 d x2 d 2 1 e 2 = ex2 # dx (x ) = 2xex . dx
Repeated Use of the Chain Rule We sometimes have to use the Chain Rule two or more times to find a derivative.
166
Chapter 3: Derivatives
HISTORICAL BIOGRAPHY Johann Bernoulli (1667–1748)
EXAMPLE 5
Find the derivative of g(t) = tan (5  sin 2t).
Solution Notice here that the tangent is a function of 5  sin 2t, whereas the sine is a function of 2t, which is itself a function of t. Therefore, by the Chain Rule, g′(t) =
d (tan (5  sin 2t)) dt
= sec2 (5  sin 2t) #
Derivative of tan u with u = 5  sin 2t
d (5  sin 2t) dt
= sec2 (5  sin 2t) # a0  cos 2t # = sec2 (5  sin 2t) # (cos 2t) # 2 = 2(cos 2t) sec2 (5  sin 2t).
d (2t)b dt
Derivative of 5  sin u with u = 2t
The Chain Rule with Powers of a Function If ƒ is a differentiable function of u and if u is a differentiable function of x, then substituting y = ƒ(u) into the Chain Rule formula dy dy du # = dx du dx leads to the formula du d ƒ(u) = ƒ′(u) . dx dx If n is any real number and ƒ is a power function, ƒ(u) = un, the Power Rule tells us that ƒ′(u) = nun  1. If u is a differentiable function of x, then we can use the Chain Rule to extend this to the Power Chain Rule: d n du (u ) = nun  1 . dx dx
d n (u ) = nun  1 du
EXAMPLE 6 The Power Chain Rule simplifies computing the derivative of a power of an expression. Power Chain Rule with d ( 5x3  x4 ) 7 = 7 ( 5x3  x4 ) 6 d ( 5x3  x4 ) (a) u = 5x3  x4, n = 7 dx dx = 7 ( 5x3  x4 ) 6 ( 5 # 3x2  4x3 ) = 7 ( 5x3  x4 ) 6 ( 15x2  4x3 )
(b)
d d 1 a b = (3x  2)1 dx 3x  2 dx d (3x  2) dx = 1(3x  2)2(3) 3 = (3x  2)2 = 1(3x  2)2
Power Chain Rule with u = 3x  2, n =  1
In part (b) we could also find the derivative with the Derivative Quotient Rule. (c)
d ( sin5 x ) = 5 sin4 x # d sin x dx dx = 5 sin4 x cos x
Power Chain Rule with u = sin x, n = 5, because sinn x means (sin x)n, n ≠  1.
3.6 The Chain Rule
(d)
167
d 23x + 1 d 1e 2 = e23x + 1 # dx 1 23x + 1 2 dx 1 = e23x + 1 # (3x + 1)1>2 # 3 2 =
Power Chain Rule with u = 3x + 1, n = 1>2
3 e23x + 1 2 23x + 1
In Section 3.2, we saw that the absolute value function y = 0 x 0 is not differentiable at x = 0. However, the function is differentiable at all other real numbers, as we now show. Since 0 x 0 = 2x2 , we can derive the following formula:
EXAMPLE 7
d ( 0 x 0 ) = d 2x2 dx dx
Derivative of the Absolute Value Function d x (0x0) = , x≠0 dx 0x0 = e
1
=
#
d 2 (x ) dx
2 2x 1 # = 2x 20x0
1, x 7 0  1, x 6 0
=
EXAMPLE 8
2
x
0x0
Power Chain Rule with u = x2, n = 1>2, x ≠ 0 2x 2 = 0 x 0
, x ≠ 0.
Show that the slope of every line tangent to the curve y = 1>(1  2x)3
is positive. Solution We find the derivative: dy d = (1  2x)3 dx dx = 3(1  2x)4 #
d (1  2x) dx
Power Chain Rule with u = (1  2x), n =  3
= 3(1  2x)4 # (2) 6 = . (1  2x)4
At any point (x, y) on the curve, the coordinate x is not 1>2 and the slope of the tangent line is dy 6 = , dx (1  2x)4 which is the quotient of two positive numbers.
EXAMPLE 9 The formulas for the derivatives of both sin x and cos x were obtained under the assumption that x is measured in radians, not degrees. The Chain Rule gives us new insight into the difference between the two. Since 180° = p radians, x° = px>180 radians where x° is the size of the angle measured in degrees. By the Chain Rule, d d p p px px sin (x°) = sin a b = cos a b = cos (x°). 180 180 180 180 dx dx See Figure 3.27. Similarly, the derivative of cos (x°) is (p>180) sin (x°). The factor p>180 would compound with repeated differentiation, showing an advantage for the use of radian measure in computations.
168
Chapter 3: Derivatives
y
y = sin(x°) = sin px 180
1
x
180
y = sin x
FIGURE 3.27 The function sin (x°) oscillates only p>180 times as often as sin x oscillates. Its maximum slope is p>180 at x = 0 (Example 9).
Exercises
3.6
Derivative Calculations In Exercises 1–8, given y = ƒ(u) and u = g(x), find dy>dx = ƒ′(g(x))g′(x).
1. y = 6u  9, u = (1>2)x4
2. y = 2u3, u = 8x  1
3. y = sin u, u = 3x + 1
4. y = cos u, u = ex
5. y = 2u, u = sin x
6. y = sin u, u = x  cos x
7. y = tan u, u = px2
1 8. y = sec u, u = x + 7x
In Exercises 9–22, write the function in the form y = ƒ(u) and u = g(x). Then find dy>dx as a function of x. 10. y = (4  3x)9
9. y = (2x + 1)5 11. y = a1 
x 7 b 7
12. y = a
2x
2
 1b
10
14. y = 23x2  4x + 6
15. y = sec (tan x)
1 16. y = cot ap  x b
17. y = tan3 x
18. y = 5 cos4 x
19. y = e5x
20. y = e2x>3
27. r = (csc u + cot u)1
3pt 3pt b + cos a b 2 2
x 1 30. y = x sin5 x  cos3 x 3 1 1 1 (3x  2)6 + a4  2 b 31. y = 18 2x 32. y = (5  2x)
4
33. y = (4x + 3)4(x + 1)3
34. y = (2x  5)1 ( x2  5x ) 6
35. y = xex + ex
36. y = (1 + 2x)e2x
3
37. y = ( x2  2x + 2 ) e5x>2
t 2t + 1
b
49. y = cos 1 eu 2
tan 3x (x + 7)4
44. g(t) = a
1 + sin 3t 1 b 3  2t
1 46. r = sec 2u tan a b u sin t 48. q = cota t b 50. y = u 3e2u cos 5u
51. y = sin2 (pt  2)
52. y = sec2 pt
53. y = (1 + cos 2t)4
54. y = (1 + cot (t>2))2
55. y = (t tan t)
56. y = (t 3>4 sin t)4>3
10
(pt  1)
t b t  4t 2
58. y = ( esin (t>2) ) 3 60. y = a
3
3
63. y = a1 + tan4 a
28. r = 6 (sec u  tan u)3>2
1 2 + a x + 1b 8
47. q = sin a
61. y = sin (cos (2t  5))
29. y = x2 sin4 x + x cos2 x
3
2 sin u b 1 + cos u
45. r = sin (u 2) cos (2u)
59. y = a
3 24. q = 2 2r  r 2
26. s = sin a
43. ƒ(u) = a
57. y = ecos
2
4 4 sin 3t + cos 5t 3p 5p
42. g(x) =
2
Find the derivatives of the functions in Exercises 23–50.
25. s =
41. ƒ(x) = 27 + x sec x
In Exercises 51–70, find dy>dt.
22. y = e142x + x 2
23. p = 23  t
1 40. k(x) = x2 sec a x b
2
x2 1 4 13. y = a + x  x b 8
21. y = e5  7x
39. h(x) = x tan 1 2 1x 2 + 7
38. y = (9x2  6x + 2)ex
3
3 t bb 12
3t  4 5 b 5t + 2
t 62. y = cos a5 sin a b b 3 64. y =
1 1 1 + cos2 (7t) 23 6
65. y = 21 + cos (t 2)
66. y = 4 sin 1 21 + 1t 2
67. y = tan2 ( sin3 t )
68. y = cos4 ( sec2 3t )
69. y = 3t ( 2t 2  5 ) 4
70. y = 43t + 32 + 21  t
Second Derivatives Find y″ in Exercises 71–78.
1 3 71. y = a1 + x b 73. y =
1 cot (3x  1) 9
72. y = 1 1  1x 21 x 74. y = 9 tan a b 3
75. y = x (2x + 1)4
76. y = x2 ( x3  1 ) 5
77. y = ex + 5x
78. y = sin ( x2ex )
2
3.6 The Chain Rule Finding Derivative Values In Exercises 79–84, find the value of (ƒ ∘ g)′ at the given value of x.
79. ƒ(u) = u5 + 1, u = g(x) = 1x, x = 1 1 1 80. ƒ(u) = 1  u , u = g(x) = , x = 1 1  x 81. ƒ(u) = cot
83. ƒ(u) =
a. y = (u>5) + 7 and u = 5x  35
1 , u = g(x) = px, x = 1>4 cos2 u
2u , u = g(x) = 10x2 + x + 1, x = 0 u2 + 1
84. ƒ(u) = a
u  1 2 1 b , u = g(x) = 2  1, x =  1 u + 1 x
85. Assume that ƒ′(3) =  1, g′(2) = 5, g(2) = 3, and y = ƒ(g(x)). What is y′ at x = 2? 86. If r = sin (ƒ(t)), ƒ(0) = p>3, and ƒ′(0) = 4, then what is dr>dt at t = 0? 87. Suppose that functions ƒ and g and their derivatives with respect to x have the following values at x = 2 and x = 3. x
ƒ(x)
g(x)
ƒ′(x)
g′(x)
2 3
8 3
2 4
1>3 2p
3 5
Find the derivatives with respect to x of the following combinations at the given value of x. a. 2ƒ(x), x = 2
c. ƒ(x) # g(x), x = 3
b. ƒ(x) + g(x), x = 3 d. ƒ(x)>g(x), x = 2
e. ƒ(g(x)), x = 2
f. 2ƒ(x), x = 2
g. 1>g2(x), x = 3
h. 2ƒ2(x) + g2(x), x = 2
88. Suppose that the functions ƒ and g and their derivatives with respect to x have the following values at x = 0 and x = 1.
b. y = 1 + (1>u) and u = 1>(x  1). 92. Find dy>dx if y = x3>2 by using the Chain Rule with y as a composite of a. y = u3 and u = 1x b. y = 1u and u = x3. 93. Find the tangent to y = ((x  1)>(x + 1))2 at x = 0. 94. Find the tangent to y = 2x2  x + 7 at x = 2. 95. a. Find the tangent to the curve y = 2 tan (px>4) at x = 1. b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval 2 6 x 6 2? Give reasons for your answer. 96. Slopes on sine curves a. Find equations for the tangents to the curves y = sin 2x and y = sin (x>2) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer. b. Can anything be said about the tangents to the curves y = sin mx and y = sin (x>m) at the origin (m a constant ≠ 0)? Give reasons for your answer. c. For a given m, what are the largest values the slopes of the curves y = sin mx and y = sin (x>m) can ever have? Give reasons for your answer. d. The function y = sin x completes one period on the interval 30, 2p4, the function y = sin 2x completes two periods, the function y = sin (x>2) completes half a period, and so on. Is there any relation between the number of periods y = sin mx completes on 30, 2p4 and the slope of the curve y = sin mx at the origin? Give reasons for your answer.
x
ƒ(x)
g(x)
ƒ′(x)
g′(x)
97. Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time t sec is
0 1
1 3
1 4
5  1>3
1>3  8>3
s = A cos (2pbt),
Find the derivatives with respect to x of the following combinations at the given value of x. a. 5ƒ(x)  g(x), x = 1 c.
Theory and Examples What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 91 and 92.
91. Find dy>dx if y = x by using the Chain Rule with y as a compsite of
pu , u = g(x) = 5 1x, x = 1 10
82. ƒ(u) = u +
169
ƒ(x) , x = 1 g(x) + 1
e. g(ƒ(x)), x = 0
b. ƒ(x)g3(x), x = 0 d. ƒ(g(x)), x = 0 f. (x11 + ƒ(x))2, x = 1
g. ƒ(x + g(x)), x = 0 89. Find ds>dt when u = 3p>2 if s = cos u and du>dt = 5. 90. Find dy>dt when x = 1 if y = x2 + 7x  5 and dx>dt = 1>3.
with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston’s velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.) 98. Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365day year. The equation that approximates the temperature on day x is y = 37 sin c
2p (x  101) d + 25 365
and is graphed in the accompanying figure.
170
Chapter 3: Derivatives
a. On what day is the temperature increasing the fastest? b. About how many degrees per day is the temperature increasing when it is increasing at its fastest?
T 105. The derivative of sin 2x Graph the function y = 2 cos 2x for 2 … x … 3.5. Then, on the same screen, graph y =
y
Temperature (°F)
60 40 20 0
for h = 1.0, 0.5, and 0.2. Experiment with other values of h, including negative values. What do you see happening as h S 0? Explain this behavior. 106. The derivative of cos (x2) Graph y =  2x sin (x2) for 2 … x … 3. Then, on the same screen, graph
. ... .... .... .... ............ ..... ..
x
Ja n Fe b M ar A pr M ay Ju n Ju l A ug Se p O ct N ov D ec Ja n Fe b M ar
−20
... .... .... .. ......
.. .. ...
....... .... . ........ .. . ... . . .... . .... .. ... ... ... ... ... ... ..
99. Particle motion The position of a particle moving along a coordinate line is s = 21 + 4t, with s in meters and t in seconds. Find the particle’s velocity and acceleration at t = 6 sec. 100. Constant acceleration Suppose that the velocity of a falling body is y = k 1s m>sec (k a constant) at the instant the body has fallen s m from its starting point. Show that the body’s acceleration is constant. 101. Falling meteorite The velocity of a heavy meteorite entering Earth’s atmosphere is inversely proportional to 2s when it is s km from Earth’s center. Show that the meteorite’s acceleration is inversely proportional to s2. 102. Particle acceleration A particle moves along the xaxis with velocity dx>dt = ƒ(x). Show that the particle’s acceleration is ƒ(x)ƒ′(x). 103. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation T = 2p
L , Ag
where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant, dL = kL. du Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT>2. 104. Chain Rule composites
sin 2(x + h)  sin 2x h
Suppose that ƒ(x) = x2 and g(x) = x . Then the
(ƒ ∘ g)(x) = x 2 = x2 and (g ∘ ƒ)(x) = x2 = x2 are both differentiable at x = 0 even though g itself is not differentiable at x = 0. Does this contradict the Chain Rule? Explain.
y =
cos ((x + h)2)  cos (x2) h
for h = 1.0, 0.7, and 0.3. Experiment with other values of h. What do you see happening as h S 0? Explain this behavior. Using the Chain Rule, show that the Power Rule (d>dx)xn = nxn  1 holds for the functions xn in Exercises 107 and 108. 107. x1>4 = 2 1x
108. x3>4 = 2x 1x
COMPUTER EXPLORATIONS Trigonometric Polynomials 109. As the accompanying figure shows, the trigonometric “polynomial”
s = ƒ(t) = 0.78540  0.63662 cos 2t  0.07074 cos 6t  0.02546 cos 10t  0.01299 cos 14t gives a good approximation of the sawtooth function s = g(t) on the interval 3 p, p4. How well does the derivative of ƒ approximate the derivative of g at the points where dg>dt is defined? To find out, carry out the following steps. a. Graph dg>dt (where defined) over 3 p, p4.
b. Find dƒ>dt. c. Graph dƒ>dt. Where does the approximation of dg>dt by dƒ>dt seem to be best? Least good? Approximations by trigonometric polynomials are important in the theories of heat and oscillation, but we must not expect too much of them, as we see in the next exercise. s p 2
−p
0
s = g(t) s = f (t)
p
t
110. (Continuation of Exercise 109.) In Exercise 109, the trigonometric polynomial ƒ(t) that approximated the sawtooth function g(t) on 3 p, p4 had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polynomial to approximate a function in a reasonable way without its derivative approximating the function’s derivative at all well. As a case in point, the trigonometric “polynomial” s = h(t) = 1.2732 sin 2t + 0.4244 sin 6t + 0.25465 sin 10t + 0.18189 sin 14t + 0.14147 sin 18t
3.7 Implicit Differentiation
a. Graph dk>dt (where defined) over 3 p, p4.
graphed in the accompanying figure approximates the step function s = k(t) shown there. Yet the derivative of h is nothing like the derivative of k. s
p − 2
b. Find dh>dt. c. Graph dh>dt to see how badly the graph fits the graph of dk>dt. Comment on what you see.
s = k(t) s = h(t)
1 −p
171
p 2
0
p
t
−1
3.7 Implicit Differentiation Most of the functions we have dealt with so far have been described by an equation of the form y = ƒ(x) that expresses y explicitly in terms of the variable x. We have learned rules for differentiating functions defined in this way. Another situation occurs when we encounter equations like x3 + y3  9xy = 0,
y2  x = 0,
or
x2 + y2  25 = 0.
(See Figures 3.28, 3.29, and 3.30.) These equations define an implicit relation between the variables x and y. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation F(x, y) = 0 in the form y = ƒ(x) to differentiate it in the usual way, we may still be able to find dy>dx by implicit differentiation. This section describes the technique.
Implicitly Defined Functions We begin with examples involving familiar equations that we can solve for y as a function of x to calculate dy>dx in the usual way. Then we differentiate the equations implicitly, and find the derivative to compare the two methods. Following the examples, we summarize the steps involved in the new method. In the examples and exercises, it is always assumed that the given equation determines y implicitly as a differentiable function of x so that dy>dx exists.
y
5
y = f1(x)
EXAMPLE 1
(x 0, y 1) A
Solution The equation y2 = x defines two differentiable functions of x that we can actually find, namely y1 = 2x and y2 =  2x (Figure 3.29). We know how to calculate the derivative of each of these for x 7 0:
x 3 + y 3 − 9xy = 0 y = f2(x)
(x 0, y 2) x0
0
(x 0, y3)
5
Find dy>dx if y2 = x.
x
y = f3 (x)
FIGURE 3.28 The curve x3 + y3  9xy = 0 is not the graph of any one function of x. The curve can, however, be divided into separate arcs that are the graphs of functions of x. This particular curve, called a folium, dates to Descartes in 1638.
dy1 1 = dx 2 1x
and
dy2 1 = . dx 2 1x
But suppose that we knew only that the equation y2 = x defined y as one or more differentiable functions of x for x 7 0 without knowing exactly what these functions were. Could we still find dy>dx? The answer is yes. To find dy>dx, we simply differentiate both sides of the equation y2 = x with respect to x, treating y = ƒ(x) as a differentiable function of x: y2 = x dy 2y = 1 dx dy 1 = . dx 2y
The Chain Rule gives
d 2 (y ) = dx
dy d 3 ƒ(x) 4 2 = 2ƒ(x)ƒ′(x) = 2y . dx dx
172
Chapter 3: Derivatives
y
y
y2 = x Slope = 1 = 1 2y1 2 "x
y1 = "25 − x 2
y1 = "x
P(x, "x ) x
0
−5
0
5
x
Q(x, − "x ) y 2 = − "x
Slope = 1 = − 1 2y 2 2 "x
(3, −4) Slope = − xy = 3 4
y2 = −"25 − x 2
FIGURE 3.29 The equation y  x = 0, or y2 = x as it is usually written, defines two differentiable functions of x on the interval x 7 0. Example 1 shows how to find the derivatives of these functions without solving the equation y2 = x for y. 2
FIGURE 3.30 The circle combines the graphs of two functions. The graph of y2 is the lower semicircle and passes through (3,  4).
This one formula gives the derivatives we calculated for both explicit solutions y1 = 2x and y2 =  2x: dy1 1 1 = = 2y1 2 1x dx
EXAMPLE 2
and
dy2 1 1 1 = = = . 2y2 21  1x 2 dx 2 1x
Find the slope of the circle x2 + y2 = 25 at the point (3, 4).
Solution The circle is not the graph of a single function of x. Rather, it is the combined graphs of two differentiable functions, y1 = 225  x2 and y2 =  225  x2 (Figure 3.30). The point (3, 4) lies on the graph of y2, so we can find the slope by calculating the derivative directly, using the Power Chain Rule: d 1  ( 25  x2 ) 1>2 2 = dx dy2 3 2x 6 ` = ` = = . 1 dx x = 3  ( 25  x2 ) 1>2( 2x) 2 225  x2 x = 3 2 225  9 4 2
We can solve this problem more easily by differentiating the given equation of the circle implicitly with respect to x: d 2 ( x ) + d ( y2 ) = d (25) dx dx dx 2x + 2y
dy = 0 dx
See Example 1.
dy x =  y. dx x The slope at (3, 4) is  y `
(3, 4)
= 
3 3 = . 4 4
Notice that unlike the slope formula for dy2 >dx, which applies only to points below the xaxis, the formula dy>dx = x>y applies everywhere the circle has a slope; that is, at all circle points (x, y) where y ≠ 0. Notice also that the derivative involves both variables x and y, not just the independent variable x. To calculate the derivatives of other implicitly defined functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defining equation.
3.7 Implicit Differentiation
173
Implicit Differentiation 1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy>dx on one side of the equation and solve for dy>dx.
EXAMPLE 3
y 4
y2 = x2 + sin xy
Solution We differentiate the equation implicitly. y2 = x2 + sin xy d 2 d d 2 1 y 2 = dx 1 x 2 + dx 1 sin xy 2 dx
2
−4
−2
0
2
4
Find dy>dx if y2 = x2 + sin xy (Figure 3.31).
x
2y
dy d = 2x + (cos xy) (xy) dx dx
ctreating y as a function of x and using the Chain Rule.
2y
dy dy = 2x + (cos xy)ay + x b dx dx
Treat xy as a product.
−2 −4
2y FIGURE 3.31 in Example 3.
The graph of the equation
Differentiate both sides with respect to x c
dy dy  (cos xy) ax b = 2x + (cos xy)y dx dx (2y  x cos xy)
Collect terms with dy>dx.
dy = 2x + y cos xy dx dy 2x + y cos xy = dx 2y  x cos xy
Solve for dy>dx.
Notice that the formula for dy>dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x and y, not just the independent variable x.
Derivatives of Higher Order Implicit differentiation can also be used to find higher derivatives.
EXAMPLE 4
Find d 2y>dx2 if 2x3  3y2 = 8.
Solution To start, we differentiate both sides of the equation with respect to x in order to find y′ = dy>dx. d ( 2x3  3y2 ) = d (8) dx dx 2 6x  6yy′ = 0 2
x y′ = y ,
Treat y as a function of x.
when y ≠ 0
Solve for y′.
We now apply the Quotient Rule to find y″. y″ =
2xy  x2y′ 2x x2 d x2 = y  2 # y′ ayb = dx y2 y
Finally, we substitute y′ = x2 >y to express y″ in terms of x and y. 2x x2 x2 2x x4 y″ = y  2 a y b = y  3 , y y
when y ≠ 0
174
Chapter 3: Derivatives
Lenses, Tangents, and Normal Lines
Tangent
Light ray
Curve of lens surface
Normal line A
Point of entry P B
In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular to the surface of the lens at the point of entry (angles A and B in Figure 3.32). This line is called the normal to the surface at the point of entry. In a profile view of a lens like the one in Figure 3.32, the normal is the line perpendicular (also said to be orthogonal) to the tangent of the profile curve at the point of entry.
EXAMPLE 5
Show that the point (2, 4) lies on the curve x3 + y3  9xy = 0. Then find the tangent and normal to the curve there (Figure 3.33).
FIGURE 3.32 The profile of a lens, showing the bending (refraction) of a ray of light as it passes through the lens surface.
y
Solution The point (2, 4) lies on the curve because its coordinates satisfy the equation given for the curve: 23 + 43  9(2) (4) = 8 + 64  72 = 0. To find the slope of the curve at (2, 4), we first use implicit differentiation to find a formula for dy>dx: x3 + y3  9xy = 0
t
en
g an
d 3 d 3 ( y )  d (9xy) = d (0) (x ) + dx dx dx dx
T
4 No
3x2 + 3y2
al
rm
x 3 + y 3 − 9xy = 0
dy dy dx  9ax + y b = 0 dx dx dx dy
0
2
x
( 3y2  9x ) + 3x2  9y = 0 dx 3 ( y2  3x )
FIGURE 3.33 Example 5 shows how to find equations for the tangent and normal to the folium of Descartes at (2, 4).
Differentiate both sides with respect to x. Treat xy as a product and y as a function of x.
dy = 9y  3x2 dx dy 3y  x2 . = dx y2  3x
Solve for dy>dx.
We then evaluate the derivative at (x, y) = (2, 4): 3y  x2 dy 3(4)  22 8 4 = ` = 2 = 2 ` = . dx (2, 4) y  3x (2, 4) 4  3(2) 10 5 The tangent at (2, 4) is the line through (2, 4) with slope 4>5: y = 4 + y =
4 (x  2) 5
4 12 x + . 5 5
The normal to the curve at (2, 4) is the line perpendicular to the tangent there, the line through (2, 4) with slope 5>4: y = 4 
5 (x  2) 4
5 13 y =  x + . 4 2
3.7 Implicit Differentiation
Exercises
3.7
Differentiating Implicitly Use implicit differentiation to find dy>dx in Exercises 1–16.
1. x2y + xy2 = 6
2. x3 + y3 = 18xy
3. 2xy + y = x + y
4. x3  xy + y3 = 1
5. x2(x  y)2 = x2  y2
6. (3xy + 7)2 = 6y
2
7. y2 =
175
x  1 x + 1
8. x3 =
9. x = sec y
2x  y x + 3y
41. Parallel tangents Find the two points where the curve x2 + xy + y2 = 7 crosses the xaxis, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents? 42. Normals parallel to a line Find the normals to the curve xy + 2x  y = 0 that are parallel to the line 2x + y = 0. 43. The eight curve Find the slopes of the curve y4 = y2  x2 at the two points shown here.
10. xy = cot (xy)
y
12. x4 + sin y = x3y2
11. x + tan (xy) = 0 1 13. y sin a y b = 1  xy
14. x cos (2x + 3y) = y sin x
15. e2x = sin (x + 3y)
16. ex y = 2x + 2y
a" 3 , " 3b 4 2
1
2
a" 3 , 1b 4 2
Find dr>du in Exercises 17–20. 18. r  2 2u =
1 19. sin (r u) = 2
20. cos r + cot u = er u
x
0
3 2>3 4 3>4 u + u 2 3
17. u 1>2 + r 1>2 = 1
y4 = y2 − x2 −1
Second Derivatives In Exercises 21–26, use implicit differentiation to find dy>dx and then d 2y>dx2.
21. x2 + y2 = 1
22. x2>3 + y2>3 = 1
23. y2 = ex + 2x
24. y2  2x = 1  2y
25. 2 1y = x  y
26. xy + y = 1
2
44. The cissoid of Diocles (from about 200 b.c.) Find equations for the tangent and normal to the cissoid of Diocles y2(2  x) = x3 at (1, 1). y y 2(2 − x) = x 3
2
27. If x3 + y3 = 16, find the value of d 2y>dx2 at the point (2, 2). 28. If xy + y2 = 1, find the value of d 2y>dx2 at the point (0,  1).
(1, 1)
1
In Exercises 29 and 30, find the slope of the curve at the given points. 29. y2 + x2 = y4  2x at ( 2, 1) and (2,  1) 30. (x + y ) = (x  y) 2
2 2
2
0
at (1, 0) and (1,  1)
Slopes, Tangents, and Normals In Exercises 31–40, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
31. x2 + xy  y2 = 1, (2, 3)
1
x
45. The devil’s curve (Gabriel Cramer, 1750) Find the slopes of the devil’s curve y4  4y2 = x4  9x2 at the four indicated points. y
32. x2 + y2 = 25, (3,  4)
y 4 − 4y 2 = x 4 − 9x 2
33. x2y2 = 9, ( 1, 3) 34. y2  2x  4y  1 = 0, ( 2, 1) 35. 6x2 + 3xy + 2y2 + 17y  6 = 0, ( 1, 0) 36. x  23xy + 2y = 5, 2
2
1 23, 2 2
37. 2xy + p sin y = 2p, (1, p>2) 38. x sin 2y = y cos 2x, (p>4, p>2) 39. y = 2 sin (px  y), (1, 0) 40. x2 cos2 y  sin y = 0, (0, p)
(−3, 2) −3 (−3, −2)
2
−2
(3, 2) x 3 (3, −2)
176
Chapter 3: Derivatives
46. The folium of Descartes
(See Figure 3.28.)
a. Find the slope of the folium of Descartes x + y  9xy = 0 at the points (4, 2) and (2, 4). 3
3
52. The graph of y2 = x3 is called a semicubical parabola and is shown in the accompanying figure. Determine the constant b so that the line y =  13 x + b meets this graph orthogonally.
b. At what point other than the origin does the folium have a horizontal tangent?
y
c. Find the coordinates of the point A in Figure 3.28 where the folium has a vertical tangent. Theory and Examples 47. Intersecting normal The line that is normal to the curve x2 + 2xy  3y2 = 0 at (1, 1) intersects the curve at what other point?
y2 = x3
1 y=− x+b 3 0
x
48. Power rule for rational exponents Let p and q be integers with q 7 0. If y = x p>q, differentiate the equivalent equation yq = xp implicitly and show that, for y ≠ 0, d p>q p (p>q)  1 x = qx . dx 49. Normals to a parabola Show that if it is possible to draw three normals from the point (a, 0) to the parabola x = y2 shown in the accompanying diagram, then a must be greater than 1>2. One of the normals is the xaxis. For what value of a are the other two normals perpendicular? y x = y2
T In Exercises 53 and 54, find both dy>dx (treating y as a differentiable function of x) and dx>dy (treating x as a differentiable function of y). How do dy>dx and dx>dy seem to be related? Explain the relationship geometrically in terms of the graphs. 53. xy3 + x2y = 6 54. x3 + y2 = sin2 y 55. Derivative of arcsine Assume that y = sin1 x is a differentiable function of x. By differentiating the equation x = sin y implicitly, show that dy>dx = 1> 21  x2 . 56. Use the formula in Exercise 55 to find dy>dx if a. y = ( sin1 x ) 2
x
(a, 0)
0
1 b. y = sin1 a x b .
COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises 57–64.
a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation. 50. Is there anything special about the tangents to the curves y2 = x3 and 2x2 + 3y2 = 5 at the points (1, {1)? Give reasons for your answer. y y2 = x3
58. x5 + y3x + yx2 + y4 = 4, P (1, 1) (1, 1)
59. y2 + y = x
0 (1, −1)
51. Verify that the following pairs of curves meet orthogonally. 1 b. x = 1  y , x = y2 3 2
c. Use the slope found in part (b) to find an equation for the tangent line to the curve at P. Then plot the implicit curve and tangent line together on a single graph. 57. x3  xy + y3 = 7, P (2, 1)
2x 2 + 3y 2 = 5
a. x2 + y2 = 4, x2 = 3y2
b. Using implicit differentiation, find a formula for the derivative dy>dx and evaluate it at the given point P.
2 + x , P (0, 1) 1  x
60. y3 + cos xy = x2, P (1, 0) y p 61. x + tan a x b = 2, P a1, b 4 p 62. xy3 + tan (x + y) = 1, P a , 0b 4 63. 2y2 + (xy)1>3 = x2 + 2, P (1, 1) 64. x 21 + 2y + y = x2, P (1, 0)
3.8 Derivatives of Inverse Functions and Logarithms
177
3.8 Derivatives of Inverse Functions and Logarithms In Section 1.6 we saw how the inverse of a function undoes, or inverts, the effect of that function. We defined there the natural logarithm function ƒ1(x) = ln x as the inverse of the natural exponential function ƒ(x) = ex. This is one of the most important functioninverse pairs in mathematics and science. We learned how to differentiate the exponential function in Section 3.3. Here we learn a rule for differentiating the inverse of a differentiable function and we apply the rule to find the derivative of the natural logarithm function. y
y = 2x − 2
y=x
y= 1x+1 2
Derivatives of Inverses of Differentiable Functions We calculated the inverse of the function ƒ(x) = (1>2)x + 1 as ƒ1(x) = 2x  2 in Example 3 of Section 1.6. Figure 3.34 shows again the graphs of both functions. If we calculate their derivatives, we see that d d 1 1 ƒ(x) = a x + 1b = 2 dx dx 2
1 −2
1
x
d 1 d ƒ (x) = (2x  2) = 2. dx dx
−2
FIGURE 3.34 Graphing a line and its inverse together shows the graphs’ symmetry with respect to the line y = x. The slopes are reciprocals of each other.
The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. (See Figure 3.34.) This is not a special case. Reflecting any nonhorizontal or nonvertical line across the line y = x always inverts the line’s slope. If the original line has slope m ≠ 0, the reflected line has slope 1 > m. y
y y = f (x)
b = f (a)
(a, b)
a = f –1(b)
0
a
x
(b, a)
0
The slopes are reciprocal: ( f –1)′(b) =
y = f –1(x) b
x
1 or ( f –1)′(b) = 1 f ′(a) f ′( f –1(b))
FIGURE 3.35 The graphs of inverse functions have reciprocal slopes at corresponding points.
The reciprocal relationship between the slopes of ƒ and ƒ 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = ƒ(x) at the point (a, ƒ(a)) is ƒ′(a) and ƒ′(a) ≠ 0, then the slope of y = ƒ 1(x) at the point (ƒ(a), a) is the reciprocal 1>ƒ′(a) (Figure 3.35). If we set b = ƒ(a), then (ƒ 1)′(b) =
1 1 . = ƒ′(a) ƒ′(ƒ 1(b))
If y = ƒ(x) has a horizontal tangent line at (a, ƒ(a)), then the inverse function ƒ 1 has a vertical tangent line at (ƒ(a), a), and this infinite slope implies that ƒ 1 is not differentiable at ƒ(a). Theorem 3 gives the conditions under which ƒ 1 is differentiable in its domain (which is the same as the range of ƒ).
178
Chapter 3: Derivatives
THEOREM 3—The Derivative Rule for Inverses If ƒ has an interval I as domain and ƒ′(x) exists and is never zero on I, then ƒ 1 is differentiable at every point in its domain (the range of ƒ). The value of (ƒ 1)′ at a point b in the domain of ƒ 1 is the reciprocal of the value of ƒ′ at the point a = ƒ 1(b):
( ƒ 1 ) ′(b) =
1 ƒ′ ( ƒ1 ( b ) )
(1)
or dƒ 1 1 2 = . dx x = b dƒ 2 dx x = ƒ 1(b)
Theorem 3 makes two assertions. The first of these has to do with the conditions under which ƒ 1 is differentiable; the second assertion is a formula for the derivative of ƒ 1 when it exists. While we omit the proof of the first assertion, the second one is proved in the following way: ƒ ( ƒ 1 ( x ) ) = x
Inverse function relationship
d ƒ ( ƒ 1 ( x ) ) = 1 dx ƒ′ ( ƒ 1 ( x ) ) #
Differentiating both sides
d 1 ƒ (x) = 1 dx d 1 1 ƒ (x) = . 1 dx ƒ′ ( ƒ ( x ) )
Chain Rule Solving for the derivative
EXAMPLE 1
The function ƒ(x) = x2, x 7 0 and its inverse ƒ 1(x) = 2x have derivatives ƒ′(x) = 2x and (ƒ 1)′(x) = 1> 1 2 2x 2. Let’s verify that Theorem 3 gives the same formula for the derivative of ƒ 1(x): 1 ƒ′ ( ƒ 1 ( x ) ) 1 = 2 ( ƒ 1 ( x ) ) 1 = . 2 ( 1x )
( ƒ 1 ) ′(x) = y y = x 2, x > 0
4
Slope 4 (2, 4)
3
Slope 1– 4
2
(4, 2)
Theorem 3 gives a derivative that agrees with the known derivative of the square root function. Let’s examine Theorem 3 at a specific point. We pick x = 2 (the number a) and ƒ(2) = 4 (the value b). Theorem 3 says that the derivative of ƒ at 2, which is ƒ′(2) = 4, and the derivative of ƒ 1 at ƒ(2), which is ( ƒ 1 ) ′(4), are reciprocals. It states that
y = "x
1 0
1
2
3
4
ƒ′(x) = 2x with x replaced by ƒ 1(x)
( ƒ 1 ) ′(4) =
x
1 1 1 1 = ` = . = ƒ′(2) 2x x = 2 4 ƒ′ ( ƒ 1 ( 4 ) )
See Figure 3.36. FIGURE 3.36 The derivative of ƒ1(x) = 1x at the point (4, 2) is the reciprocal of the derivative of ƒ(x) = x2 at (2, 4) (Example 1).
We will use the procedure illustrated in Example 1 to calculate formulas for the derivatives of many inverse functions throughout this chapter. Equation (1) sometimes enables us to find specific values of dƒ 1 >dx without knowing a formula for ƒ 1.
3.8 Derivatives of Inverse Functions and Logarithms
y 6
(2, 6)
y = x3 − 2 Slope 3x 2 = 3(2)2 = 12
179
Let ƒ(x) = x3  2, x 7 0. Find the value of dƒ 1 >dx at x = 6 = ƒ(2) without finding a formula for ƒ 1(x).
EXAMPLE 2
Solution We apply Theorem 3 to obtain the value of the derivative of ƒ 1 at x = 6: dƒ 2 = 3x2 ` = 12 dx x = 2 x=2
Reciprocal slope: 1 12 (6, 2)
−2
0
6
dƒ 1 1 1 2 = = . 12 dx x = ƒ(2) dƒ 2 dx x = 2
x
−2
FIGURE 3.37 The derivative of ƒ(x) = x3  2 at x = 2 tells us the derivative of ƒ 1 at x = 6 (Example 2).
Eq. (1)
See Figure 3.37.
Derivative of the Natural Logarithm Function Since we know the exponential function ƒ(x) = ex is differentiable everywhere, we can apply Theorem 3 to find the derivative of its inverse ƒ 1(x) = ln x:
( ƒ 1 ) ′(x) = = =
1 ƒ′ ( ƒ 1 ( x ) ) 1 eƒ
1
(x)
1 eln x
1 = x.
Theorem 3
ƒ′(u) = eu x 7 0
Inverse function relationship
Alternate Derivation Instead of applying Theorem 3 directly, we can find the derivative of y = ln x using implicit differentiation, as follows: y = ln x ey = x d y d (e ) = (x) dx dx ey
x 7 0 Inverse function relationship Differentiate implicitly.
dy = 1 dx
Chain Rule
dy 1 1 = = x. dx ey
ey = x
No matter which derivation we use, the derivative of y = ln x with respect to x is d 1 (ln x) = x , x 7 0. dx The Chain Rule extends this formula to positive functions u(x):
d 1 du ln u = u , dx dx
u 7 0.
(2)
180
Chapter 3: Derivatives
EXAMPLE 3 (a)
We use Equation (2) to find derivatives.
d 1 1 d 1 (2) = x , x 7 0 ln 2x = (2x) = 2x dx 2x dx
(b) Equation (2) with u = x2 + 3gives d 2x 1 # d 2 1 # (x + 3) = 2 ln (x2 + 3) = 2 . 2x = 2 dx x + 3 dx x + 3 x + 3
(c) Equation (2) with u = 0 x 0 gives an important derivative: du d d ln 0 x 0 = ln u # dx du dx 1# x = u 0x0 =
Derivative of ln ∣ x ∣
1 # x 0x0 0x0
u = 0x0, x ≠ 0 d ( 0x0 ) = x dx 0x0 Substitute for u.
x x2 1 = x.
d 1 ln 0 x 0 = x , x ≠ 0 dx
=
So 1>x is the derivative of ln x on the domain x 7 0, and the derivative of ln (x) on the domain x 6 0. Notice from Example 3a that the function y = ln 2x has the same derivative as the function y = ln x. This is true of y = ln bx for any constant b, provided that bx 7 0: d 1 # d 1 1 ln bx = (bx) = (b) = x . dx bx dx bx
d 1 ln bx = x , bx 7 0 dx
(3)
EXAMPLE 4 A line with slope m passes through the origin and is tangent to the graph of y = ln x. What is the value of m? Solution Suppose the point of tangency occurs at the unknown point x = a 7 0. Then we know that the point (a, ln a) lies on the graph and that the tangent line at that point has slope m = 1>a (Figure 3.38). Since the tangent line passes through the origin, its slope is y
m = 2 1
0
ln a  0 ln a = a . a  0
Setting these two formulas for m equal to each other, we have (a, ln a)
1 2 y = ln x
1 Slope = a 3
4
5
ln a a ln a eln a a
1 = a = 1 = e1 = e 1 m = e.
x
FIGURE 3.38 The tangent line intersects the curve at some point (a, ln a), where the slope of the curve is 1>a (Example 4).
The Derivatives of au and log a u We start with the equation ax = eln (a ) = e x ln a , a 7 0, which was seen in Section 1.6: x
d x d x ln a a = e dx dx = e x ln a #
d (x ln a) dx
= ax ln a.
d u du e = eu dx dx
3.8 Derivatives of Inverse Functions and Logarithms
181
That is, if a 7 0, then ax is differentiable and d x a = ax ln a. dx
(4)
This equation shows why ex is the preferred exponential function in calculus. If a = e, then ln a = 1 and the derivative of ax simplifies to d x e = ex ln e = ex. dx With the Chain Rule, we get a more general form for the derivative of a general exponential function au.
If a 7 0 and u is a differentiable function of x, then au is a differentiable function of x and d u du a = au ln a . dx dx
EXAMPLE 5
(5)
Here are some derivatives of general exponential functions.
d x 3 = 3x ln 3 dx d x d (b) 3 = 3x(ln 3) (x) = 3x ln 3 dx dx d sin x d (c) = 3sin x(ln 3) (sin x) = 3sin x(ln 3) cos x 3 dx dx
(a)
Eq. (5) with a = 3, u = x Eq. (5) with a = 3, u =  x c, u = sin x
In Section 3.3 we looked at the derivative ƒ′(0) for the exponential functions ƒ(x) = ax at various values of the base a. The number ƒ′(0) is the limit, limh S 0 (ah  1)>h, and gives the slope of the graph of ax when it crosses the yaxis at the point (0, 1). We now see from Equation (4) that the value of this slope is lim
hS0
ah  1 = ln a. h
(6)
In particular, when a = e we obtain eh  1 = ln e = 1. h hS0 lim
However, we have not fully justified that these limits actually exist. While all of the arguments given in deriving the derivatives of the exponential and logarithmic functions are correct, they do assume the existence of these limits. In Chapter 7 we will give another development of the theory of logarithmic and exponential functions which fully justifies that both limits do in fact exist and have the values derived above. To find the derivative of loga u for an arbitrary base (a 7 0, a ≠ 1), we start with the changeofbase formula for logarithms (reviewed in Section 1.6) and express loga u in terms of natural logarithms, loga x =
ln x . ln a
182
Chapter 3: Derivatives
Taking derivatives, we have d d ln x loga x = a b dx dx ln a 1 # d ln x = ln a dx =
1 #1 ln a x
=
1 . x ln a
ln a is a constant.
If u is a differentiable function of x and u 7 0, the Chain Rule gives a more general formula.
For a 7 0 and a ≠ 1, d 1 du loga u = . dx u ln a dx
(7)
Logarithmic Differentiation The derivatives of positive functions given by formulas that involve products, quotients, and powers can often be found more quickly if we take the natural logarithm of both sides before differentiating. This enables us to use the laws of logarithms to simplify the formulas before differentiating. The process, called logarithmic differentiation, is illustrated in the next example.
EXAMPLE 6
Find dy>dx if y =
(x2 + 1)(x + 3)1>2 , x  1
x 7 1.
Solution We take the natural logarithm of both sides and simplify the result with the algebraic properties of logarithms from Theorem 1 in Section 1.6: ln y = ln
(x2 + 1)(x + 3)1>2 x  1
= ln ((x2 + 1)(x + 3)1>2)  ln (x  1)
Rule 2
= ln (x + 1) + ln (x + 3)
 ln (x  1)
Rule 1
1 ln (x + 3)  ln (x  1). 2
Rule 4
2
= ln (x2 + 1) +
1>2
We then take derivatives of both sides with respect to x, using Equation (2) on the left: 1 1 dy 1# 1 1 # y dx = x2 + 1 2x + 2 x + 3  x  1 . Next we solve for dy>dx: dy 2x 1 1 = ya 2 + b. dx x + 1 2x + 6 x  1
3.8 Derivatives of Inverse Functions and Logarithms
183
Finally, we substitute for y: dy (x2 + 1)(x + 3)1>2 2x 1 1 = + a 2 b. x  1 dx x + 1 2x + 6 x  1
Irrational Exponents and the Power Rule (General Version) The definition of the general exponential function enables us to raise any positive number to any real power n, rational or irrational. That is, we can define the power function y = xn for any exponent n.
DEFINITION
For any x 7 0 and for any real number n, xn = en ln x.
Because the logarithm and exponential functions are inverses of each other, the definition gives ln xn = n ln x, for all real numbers n. That is, the rule for taking the natural logarithm of any power holds for all real exponents n, not just for rational exponents. The definition of the power function also enables us to establish the derivative Power Rule for any real power n, as stated in Section 3.3.
General Power Rule for Derivatives For x 7 0 and any real number n, d n x = nxn  1. dx If x … 0, then the formula holds whenever the derivative, xn, and xn  1 all exist.
Proof
Differentiating xn with respect to x gives d n d n ln x x = e dx dx = en ln x #
Definition of xn, x 7 0
d (n ln x) dx
Chain Rule for eu
n = xn # x
Definition and derivative of ln x
= nxn  1.
xn # x1 = xn  1
In short, whenever x 7 0, d n x = nxn  1. dx For x 6 0, if y = xn, y′, and xn  1 all exist, then
ln 0 y 0 = ln 0 x 0 n = n ln 0 x 0 .
184
Chapter 3: Derivatives
Using implicit differentiation (which assumes the existence of the derivative y′) and Example 3(c), we have y′ n y = x. Solving for the derivative, y xn y′ = n x = n x = nxn  1.
y = xn
It can be shown directly from the definition of the derivative that the derivative equals 0 when x = 0 and n Ú 1 (see Exercise 99). This completes the proof of the general version of the Power Rule for all values of x.
EXAMPLE 7
Differentiate ƒ(x) = xx, x 7 0.
Solution We note that ƒ(x) = xx = ex ln x, so differentiation gives ƒ′(x) =
d x ln x (e ) dx
= ex ln x
d (x ln x) dx
d u dx e ,
u = x ln x
1 = ex ln x aln x + x # x b = xx (ln x + 1).
x 7 0
We can also find the derivative of y = xx using logarithmic differentiation, assuming y′ exists.
The Number e Expressed as a Limit In Section 1.5 we defined the number e as the base value for which the exponential function y = ax has slope 1 when it crosses the yaxis at (0, 1). Thus e is the constant that satisfies the equation eh  1 = ln e = 1. h hS0 lim
Slope equals ln e from Eq. (6).
We now prove that e can be calculated as a certain limit.
THEOREM 4—The Number e as a Limit The number e can be calculated as the limit e = lim (1 + x)1>x.
y
xS0
e 3 2
Proof
y = (1 + x)1x
ƒ(1 + h)  ƒ(1) ƒ(1 + x)  ƒ(1) = lim x h hS0 xS0
ƒ′(1) = lim
1
0
If ƒ(x) = ln x, then ƒ′(x) = 1>x, so ƒ′(1) = 1. But, by the definition of derivative,
x
FIGURE 3.39 The number e is the limit of the function graphed here as x S 0.
ln (1 + x)  ln 1 1 = lim x ln (1 + x) x xS0 xS0
= lim
ln 1 = 0
= lim ln (1 + x)1>x = ln c lim (1 + x)1>x d .
ln is continuous, Theorem 10 in Chapter 2.
xS0
xS0
185
3.8 Derivatives of Inverse Functions and Logarithms
Because ƒ′(1) = 1, we have ln c lim (1 + x)1>x d = 1. xS0
Therefore, exponentiating both sides we get lim (1 + x)1>x = e.
xS0
See Figure 3.39 on the previous page. Approximating the limit in Theorem 4 by taking x very small gives approximations to e. Its value is e ≈ 2.718281828459045 to 15 decimal places.
Exercises
3.8 13. y = ln ( t 2 )
Derivatives of Inverse Functions In Exercises 1–4:
14. y = ln ( t 3>2 ) + 2t
3 15. y = ln x
a. Find ƒ 1(x).
16. y = ln (sin x) u
b. Graph ƒ and ƒ 1 together.
c. Evaluate dƒ>dx at x = a and dƒ >dx at x = ƒ(a) to show that at these points dƒ 1 >dx = 1>(dƒ>dx). 1
1. ƒ(x) = 2x + 3, a = 1
2. ƒ(x) = (1>5)x + 7, a = 1
3. ƒ(x) = 5  4x, a = 1>2
4. ƒ(x) = 2x2, x Ú 0, a = 5
3 5. a. Show that ƒ(x) = x3 and g(x) = 1 x are inverses of one another.
b. Graph ƒ and g over an xinterval large enough to show the graphs intersecting at (1, 1) and ( 1,  1). Be sure the picture shows the required symmetry about the line y = x.
17. y = ln (u + 1)  e
18. y = (cos u) ln (2u + 2)
19. y = ln x3
20. y = (ln x)3
21. y = t (ln t)
22. y = t ln 2t
2
x4 x4 ln x 23. y = 4 16
24. y = (x2 ln x)4
ln t 25. y = t
26. y =
27. y =
ln x 1 + ln x
28. y =
c. Find the slopes of the tangents to the graphs of ƒ and g at (1, 1) and ( 1, 1) (four tangents in all).
29. y = ln (ln x)
d. What lines are tangent to the curves at the origin?
32. y = ln (sec u + tan u)
6. a. Show that h(x) = x3 >4 and k(x) = (4x)1>3 are inverses of one another. b. Graph h and k over an xinterval large enough to show the graphs intersecting at (2, 2) and ( 2, 2). Be sure the picture shows the required symmetry about the line y = x.
x ln x 1 + ln x
30. y = ln (ln (ln x))
31. y = u(sin (ln u) + cos (ln u)) 33. y = ln 35. y =
1 x 2x + 1
34. y =
1 + ln t 1  ln t
c. Find the slopes of the tangents to the graphs at h and k at (2, 2) and (2,  2).
37. y = ln (sec (ln u))
d. What lines are tangent to the curves at the origin?
39. y = ln a
7. Let ƒ(x) = x3  3x2  1, x Ú 2. Find the value of dƒ 1 >dx at the point x =  1 = ƒ(3).
t 2ln t
(x2 + 1)5 21  x
1 1 + x ln 2 1  x
36. y = 2ln 1t 38. y = ln a b
40. y = ln
2sin u cos u
1 + 2 ln u
b
(x + 1)5 C(x + 2)20
8. Let ƒ(x) = x2  4x  5, x 7 2. Find the value of dƒ 1 >dx at the point x = 0 = ƒ(5).
Logarithmic Differentiation In Exercises 41–54, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
9. Suppose that the differentiable function y = ƒ(x) has an inverse and that the graph of ƒ passes through the point (2, 4) and has a slope of 1>3 there. Find the value of dƒ 1 >dx at x = 4.
41. y = 2x(x + 1)
10. Suppose that the differentiable function y = g(x) has an inverse and that the graph of g passes through the origin with slope 2. Find the slope of the graph of g1 at the origin. Derivatives of Logarithms In Exercises 11–40, find the derivative of y with respect to x, t, or u, as appropriate. 1 12. y = 11. y = ln 3x + x ln 3x
43. y =
t At + 1
42. y = 2(x2 + 1)(x  1)2 44. y =
1 A t(t + 1)
45. y = (sin u) 2u + 3
46. y = (tan u) 22u + 1
47. y = t(t + 1)(t + 2)
48. y =
49. y =
u + 5 u cos u
50. y =
51. y =
x 2x2 + 1 (x + 1)2>3
52. y =
1 t(t + 1)(t + 2) u sin u 2sec u
(x + 1)10 C(2x + 1)5
186
53. y =
Chapter 3: Derivatives
x(x  2) B x2 + 1 3
54. y =
x(x + 1)(x  2) B (x2 + 1)(2x + 3) 3
Finding Derivatives In Exercises 55–62, find the derivative of y with respect to x, t, or u, as appropriate.
56. y = ln (3ueu)
55. y = ln (cos2 u) t
58. y = ln (2et sin t)
57. y = ln (3te ) 59. y = ln a
eu b 1 + eu
60. y = ln a
61. y = e(cos t + ln t)
2u b 1 + 2u
sides of this equation with respect to x, using the Chain Rule to express (g ∘ ƒ)′(x) as a product of derivatives of g and ƒ. What do you find? (This is not a proof of Theorem 3 because we assume here the theorem’s conclusion that g = ƒ 1 is differentiable.) x n 98. Show that limn S q a1 + n b = ex for any x 7 0. 99. If ƒ(x) = xn, n Ú 1, show from the definition of the derivative that ƒ′(0) = 0. 100. Using mathematical induction, show that for n 7 1 (n  1)! dn ln x = (1)n  1 . xn dxn
62. y = esin t (ln t 2 + 1)
In Exercises 63–66, find dy>dx. 63. ln y = ey sin x
64. ln xy = ex + y
65. xy = yx
66. tan y = ex + ln x
In Exercises 67–88, find the derivative of y with respect to the given independent variable. 67. y = 2x
68. y = 3x
69. y = 52s
70. y = 2(s )
71. y = xp
72. y = t 1  e
2
73. y = log2 5u
74. y = log3 (1 + u ln 3)
75. y = log4 x + log4 x2
76. y = log25 ex  log5 1x
x + 1 ln 3 b b 79. y = log3 a a x  1
ln 5 7x b 80. y = log5 a B 3x + 2
77. y = log2 r # log4 r
78. y = log3 r # log9 r
81. y = u sin (log7 u)
82. y = log7 a
sin u cos u b eu 2u
83. y = log5 ex
84. y = log2 a
x2e2 b 2 2x + 1
85. y = 3log2 t
86. y = 3 log8 (log2 t)
87. y = log2 (8t ln 2)
88. y = t log3 1 e(sin t)(ln 3) 2
Logarithmic Differentiation with Exponentials In Exercises 89–96, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
89. y = (x + 1)x
90. y = x(x + 1)
91. y =
92. y = t 2t
( 1t ) t
93. y = (sin x)x
94. y = xsin x
95. y = x
96. y = (ln x)
ln x
a. Plot the function y = ƒ(x) together with its derivative over the given interval. Explain why you know that ƒ is onetoone over the interval. b. Solve the equation y = ƒ(x) for x as a function of y, and name the resulting inverse function g. c. Find the equation for the tangent line to ƒ at the specified point (x0, ƒ(x0)). d. Find the equation for the tangent line to g at the point (ƒ(x0), x0) located symmetrically across the 45° line y = x (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions ƒ and g, the identity, the two tangent lines, and the line segment joining the points (x0, ƒ(x0)) and (ƒ(x0), x0). Discuss the symmetries you see across the main diagonal. 101. y = 23x  2,
2 … x … 4, x0 = 3 3
102. y =
3x + 2 , 2 … x … 2, x0 = 1>2 2x  11
103. y =
4x , 1 … x … 1, x0 = 1>2 x2 + 1
104. y =
x3 , 1 … x … 1, x0 = 1>2 x + 1 2
105. y = x3  3x2  1, 2 … x … 5, x0 =
27 10
106. y = 2  x  x3,  2 … x … 2, x0 =
3 2
ln x
Theory and Applications 97. If we write g(x) for ƒ1(x), Equation (1) can be written as
g′(ƒ(a)) =
COMPUTER EXPLORATIONS In Exercises 101–108, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:
1 , or g′(ƒ(a)) # ƒ′(a) = 1. ƒ′(a)
107. y = ex,  3 … x … 5, x0 = 1 108. y = sin x, 
p p … x … , x0 = 1 2 2
If we then write x for a, we get
In Exercises 109 and 110, repeat the steps above to solve for the functions y = ƒ(x) and x = ƒ 1(y) defined implicitly by the given equations over the interval.
The latter equation may remind you of the Chain Rule, and indeed there is a connection. Assume that ƒ and g are differentiable functions that are inverses of one another, so that (g ∘ ƒ)(x) = x. Differentiate both
109. y1>3  1 = (x + 2)3,  5 … x … 5, x0 =  3>2
g′(ƒ(x)) # ƒ′(x) = 1.
110. cos y = x1>5, 0 … x … 1, x0 = 1>2
187
3.9 Inverse Trigonometric Functions
3.9 Inverse Trigonometric Functions We introduced the six basic inverse trigonometric functions in Section 1.6, but focused there on the arcsine and arccosine functions. Here we complete the study of how all six inverse trigonometric functions are defined, graphed, and evaluated, and how their derivatives are computed.
Inverses of tan x, cot x, sec x, and csc x The graphs of these four basic inverse trigonometric functions are shown again in Figure 3.40. We obtain these graphs by reflecting the graphs of the restricted trigonometric functions (as discussed in Section 1.6) through the line y = x. Let’s take a closer look at the arctangent, arccotangent, arcsecant, and arccosecant functions.
Domain: −∞ < x < ∞ Range: − p < y < p 2 2 y
−
p 2
y
y = tan 1
2
Domain: x ≤ −1 or x ≥ 1 Range: − p ≤ y ≤ p , y ≠ 0 2 2 y p 2
p
–1x
p 2
x
−2 −1
y=
1 (b)
(a)
FIGURE 3.40
Domain: x ≤ −1 or x ≥ 1 Range: 0 ≤ y ≤ p, y ≠ p 2 y
p
p 2 −2 −1
Domain: −∞ < x < ∞ 02, p>2) for which tan y = x. cot−1 x is the number in (0, p) for which cot y = x. sec−1 x is the number in 3 0, p/2) ∪ (p/2, p4 for which sec y = x. csc−1 x is the number in 3 p/2, 0) ∪ (0, p/24 for which csc y = x.
We use open or halfopen intervals to avoid values for which the tangent, cotangent, secant, and cosecant functions are undefined. (See Figure 3.40.) The graph of y = tan1 x is symmetric about the origin because it is a branch of the graph x = tan y that is symmetric about the origin (Figure 3.40a). Algebraically this means that tan1 (x) = tan1 x; the arctangent is an odd function. The graph of y = cot1 x has no such symmetry (Figure 3.40b). Notice from Figure 3.40a that the graph of the arctangent function has two horizontal asymptotes: one at y = p>2 and the other at y = p>2.
188
Chapter 3: Derivatives
The inverses of the restricted forms of sec x and csc x are chosen to be the functions graphed in Figures 3.40c and 3.40d.
Domain: 0 x 0 ≥ 1 Range: 0 ≤ y ≤ p, y ≠ p 2 y 3p 2 B
p A y=
p 2
sec –1x
−1
0
x
1
−
Caution There is no general agreement about how to define sec1 x for negative values of x. We chose angles in the second quadrant between p>2 and p. This choice makes sec1 x = cos1 (1>x). It also makes sec1 x an increasing function on each interval of its domain. Some tables choose sec1 x to lie in 3 p, p>2) for x 6 0 and some texts choose it to lie in 3 p, 3p>2) (Figure 3.41). These choices simplify the formula for the derivative (our formula needs absolute value signs) but fail to satisfy the computational equation sec1 x = cos1 (1>x). From this, we can derive the identity p 1 1 sec1 x = cos1 a x b =  sin1 a x b 2
p 2
(1)
C −p
by applying Equation (5) in Section 1.6. −
3p 2
FIGURE 3.41 There are several logical choices for the lefthand branch of y = sec1 x. With choice A, sec1 x = cos1 (1>x), a useful identity employed by many calculators.
EXAMPLE 1
The accompanying figures show two values of tan1 x.
y
tan–1 1 = tan–1 " 3 = p 3 6 "3 p 6 2 1 x 0 "3
y
tan–1 a−" 3b = − p 3 − 1
0 2
23
1 23>3  23>3 1  23
y
y = sin–1x Domain: −1 ≤ x ≤ 1 Range: −p2 ≤ y ≤ p2
p 2
−1
1 −
x
p 2
FIGURE 3.42 The graph of y = sin1 x has vertical tangents at x = 1 and x = 1.
x −" 3
tan a− pb = −" 3 3
tan p = 1 6 "3
x
p 3
tan1 x p>3 p>4 p>6 p>6 p>4 p>3
The angles come from the first and fourth quadrants because the range of tan1 x is (p>2, p>2).
The Derivative of y = sin−1 u We know that the function x = sin y is differentiable in the interval p>2 6 y 6 p>2 and that its derivative, the cosine, is positive there. Theorem 3 in Section 3.8 therefore assures us that the inverse function y = sin1 x is differentiable throughout the interval 1 6 x 6 1. We cannot expect it to be differentiable at x = 1 or x = 1 because the tangents to the graph are vertical at these points (see Figure 3.42).
3.9 Inverse Trigonometric Functions
189
We find the derivative of y = sin1 x by applying Theorem 3 with ƒ(x) = sin x and ƒ (x) = sin1 x: 1
( ƒ 1 ) ′(x) =
1 ƒ′ ( ƒ 1(x) )
Theorem 3
1 cos ( sin1 x ) 1 = 21  sin2 ( sin1 x ) 1 = . 21  x2 =
ƒ′(u) = cos u cos u = 21  sin2 u sin (sin1 x) = x
If u is a differentiable function of x with 0 u 0 6 1, we apply the Chain Rule to get the general formula
d du 1 (sin1 u) = , 2 dx dx 21  u
EXAMPLE 2
0 u 0 6 1.
Using the Chain Rule, we calculate the derivative d 1 # d (x2) = 2x 4 . (sin1 x2) = dx 21  (x2)2 dx 21  x
The Derivative of y = tan−1 u We find the derivative of y = tan1 x by applying Theorem 3 with ƒ(x) = tan x and ƒ 1(x) = tan1 x. Theorem 3 can be applied because the derivative of tan x is positive for p>2 6 x 6 p>2: 1 ƒ′ ( ƒ 1 ( x ) )
Theorem 3
=
1 sec2 (tan1 x)
ƒ′(u) = sec2 u
=
1 1 + tan2 (tan1 x)
sec2 u = 1 + tan2 u
=
1 . 1 + x2
tan (tan1 x) = x
( ƒ 1 ) ′(x) =
The derivative is defined for all real numbers. If u is a differentiable function of x, we get the Chain Rule form:
d ( tan1 u ) = 1 2 du . dx 1 + u dx
The Derivative of y = sec−1 u Since the derivative of sec x is positive for 0 6 x 6 p>2 and p>2 6 x 6 p, Theorem 3 says that the inverse function y = sec1 x is differentiable. Instead of applying the formula
190
Chapter 3: Derivatives
in Theorem 3 directly, we find the derivative of y = sec1 x, 0 x 0 7 1, using implicit differentiation and the Chain Rule as follows: y = sec1 x sec y = x d d (sec y) = x dx dx sec y tan y
dy = 1 dx
Inverse function relationship Differentiate both sides.
Chain Rule
dy 1 = . dx sec y tan y
Since 0 x 0 7 1, y lies in (0, p>2) ∪ (p>2, p) and sec y tan y ≠ 0.
To express the result in terms of x, we use the relationships sec y = x
tan y = { 2sec2 y  1 = { 2x2  1
and
to get dy 1 = { . dx x 2x2  1 y
Can we do anything about the { sign? A glance at Figure 3.43 shows that the slope of the graph y = sec1 x is always positive. Thus,
y = sec –1x
p
1 d x 2x2  1 sec1 x = d dx 1 x 2x2  1
p 2 −1
0
1
+
x
FIGURE 3.43 The slope of the curve y = sec1 x is positive for both x 6  1 and x 7 1.
if x 7 1 if x 6 1.
With the absolute value symbol, we can write a single expression that eliminates the “ {” ambiguity: d 1 . sec1 x = dx 0 x 0 2x2  1 If u is a differentiable function of x with 0 u 0 7 1, we have the formula
d du 1 ( sec1 u ) = , 2 dx 0 u 0 2u  1 dx
EXAMPLE 3
0 u 0 7 1.
Using the Chain Rule and derivative of the arcsecant function, we find d d 1 ( 5x4 ) sec1 ( 5x4 ) = 4 4 2 dx dx ( ) 0 5x 0 2 5x  1 =
1 ( 20x3 ) 5x 225x8  1
=
4 . x 225x8  1
4
5x4 7 1 7 0
3.9 Inverse Trigonometric Functions
191
Derivatives of the Other Three Inverse Trigonometric Functions We could use the same techniques to find the derivatives of the other three inverse trigonometric functions—arccosine, arccotangent, and arccosecant—but there is an easier way, thanks to the following identities.
Inverse Function–Inverse Cofunction Identities cos1 x = p>2  sin1 x cot1 x = p>2  tan1 x csc1 x = p>2  sec1 x
We saw the first of these identities in Equation (5) of Section 1.6. The others are derived in a similar way. It follows easily that the derivatives of the inverse cofunctions are the negatives of the derivatives of the corresponding inverse functions. For example, the derivative of cos1 x is calculated as follows: d ( cos1 x ) = d a p  sin1 xb dx dx 2 = 
Identity
d ( sin1 x ) dx
= 
1 21  x2
.
Derivative of arcsine
The derivatives of the inverse trigonometric functions are summarized in Table 3.1.
TABLE 3.1 Derivatives of the inverse trigonometric functions
0u0 6 1
1.
d ( sin1 u ) du 1 = , dx 21  u2 dx
2.
d ( cos1 u ) du 1 = , dx 21  u2 dx
3.
d ( tan1 u ) 1 du = dx 1 + u2 dx
4.
d ( cot1 u ) 1 du = dx 1 + u2 dx
5.
d ( sec1 u ) du 1 = , dx 0 u 0 2u2  1 dx
6.
d ( csc1 u ) du 1 = , 2 dx 0 u 0 2u  1 dx
0u0 6 1
0u0 7 1 0u0 7 1
Chapter 3: Derivatives
Exercises
3.9
Common Values Use reference triangles in an appropriate quadrant, as in Example 1, to find the angles in Exercises 1–8. 1 1. a. tan1 1 b. tan11  23 2 c. tan1 a b 23 1
1
b. tan 23
2. a. tan ( 1) 3. a. sin1 a
1 c. tan a b 23
1
1 b 2
b. sin1 a
1 22
b
c. sin1 a
 23 b 2
1 4. a. sin1 a b 2
b. sin1 a
1 b 22
c. sin1 a
23
1 5. a. cos1 a b 2
b. cos1 a
1 b 22
c. cos1 a
23
6. a. csc1 22
b. csc1 a
c. csc1 2
7. a. sec11  22 2
2 b 23
b. sec1 a
8. a. cot1 ( 1)
b. cot1 1 23 2
2 23
b
2 2
b
37. y = s 21  s2 + cos1 s
39. y = tan1 2x2  1 + csc1 x, x 7 1 1 40. y = cot1 x  tan1 x
9. sin acos a
22
2
bb
1 11. tan asin1 a b b 2
Theory and Examples 43. You are sitting in a classroom next to the wall looking at the blackboard at the front of the room. The blackboard is 12 ft long and starts 3 ft from the wall you are sitting next to. Show that your viewing angle is
b
a = cot1
c. sec1( 2) c. cot1 a
10. sec acos
1 1
2
1 b 23
b
12. cot asin1 a
13. lim sin1 x
14.
15. lim tan1 x
16. lim tan1 x
17. lim sec1 x
18. lim sec1 x
12′
a
3′
23
2
bb
Wall x
44. Find the angle a.
19. lim csc1 x
20. lim csc1 x x S q
x Sq
Finding Derivatives In Exercises 21–42, find the derivative of y with respect to the appropriate variable.
21. y = cos1 (x2) 1
22. y = cos1 (1>x)
23. y = sin 22 t
24. y = sin1 (1  t)
25. y = sec1 (2s + 1)
26. y = sec1 5s
1
65°
lim cos1 x
x S q
x Sq
You
x S 1+ x S q
x Sq
x x  cot1 3 15
if you are x ft from the front wall.
Limits Find the limits in Exercises 13–20. (If in doubt, look at the function’s graph.) xS1
41. y = x sin1 x + 21  x2
x 42. y = ln (x2 + 4)  x tan1 a b 2
Evaluations Find the values in Exercises 9–12. 1
38. y = 2s2  1  sec1 s
Blackboard
192
27. y = csc (x + 1), x 7 0 x 28. y = csc1 2 3 1 29. y = sec1 t , 0 6 t 6 1 30. y = sin1 2 t 31. y = cot1 2t 32. y = cot1 2t  1 2
33. y = ln (tan1 x)
34. y = tan1 (ln x)
35. y = csc1 (et)
36. y = cos1 (et)
a
21 50
b
45. Here is an informal proof that tan1 1 + tan1 2 + tan1 3 = p. Explain what is going on.
3.10 Related Rates
46. Two derivations of the identity sec−1 ( −x) = P − sec−1 x 1
a. (Geometric) Here is a pictorial proof that sec ( x) = p  sec1 x. See if you can tell what is going on. y
x
1
x
b. (Algebraic) Derive the identity sec1 ( x) = p  sec1 x by combining the following two equations from the text: sec1 x = cos1 (1>x)
Eq. (1)
1
Explain. b. csc1 ( 1.5)
c. cot1 2
T 58. Find the values of a. sec1( 3)
b. csc1 1.7
c. cot1 ( 2)
T In Exercises 59–61, find the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. 1
48. a. csc (1>2)
b. csc
49. a. sec1 0
b. sin1 22
1
1 and g(x) = tan1 x?
T 57. Find the values of a. sec1 1.5
59. a. y = tan1 (tan x)
b. cos1 2
1
1 2x2 + 1
Eq. (4), Section 1.6
Which of the expressions in Exercises 47–50 are defined, and which are not? Give reasons for your answers. 47. a. tan1 2
x  1 , x Ú 0, and g(x) = 2 tan1 1x? x + 1
ƒ(x) = sin1
0
cos1 ( x) = p  cos1 x
ƒ(x) = sin1
56. What is special about the functions
p 2 −1
55. What is special about the functions
Explain.
y = sec –1x
p
−x
193
2
b. y = tan (tan1 x)
60. a. y = sin (sin x)
b. y = sin (sin1 x)
61. a. y = cos1 (cos x)
b. y = cos (cos1 x)
1
50. a. cot ( 1>2)
b. cos ( 5)
51. Use the identity csc1 u =
p  sec1 u 2
to derive the formula for the derivative of csc1 u in Table 3.1 from the formula for the derivative of sec1 u.
63. Newton’s serpentine Graph Newton’s serpentine, y = 4x>(x2 + 1). Then graph y = 2 sin (2 tan1 x) in the same graphing window. What do you see? Explain. 64. Graph the rational function y = (2  x2)>x2. Then graph y = cos (2 sec1 x) in the same graphing window. What do you see? Explain.
52. Derive the formula dy 1 = dx 1 + x2 for the derivative of y = tan1 x by differentiating both sides of the equivalent equation tan y = x. 53. Use the Derivative Rule in Section 3.8, Theorem 3, to derive d 1 sec1 x = , dx 0 x 0 2x2  1
T Use your graphing utility for Exercises 62–66. 62. Graph y = sec (sec1 x) = sec (cos1(1>x)). Explain what you see.
0 x 0 7 1.
65. Graph ƒ(x) = sin1 x together with its first two derivatives. Comment on the behavior of ƒ and the shape of its graph in relation to the signs and values of ƒ′ and ƒ″. 66. Graph ƒ(x) = tan1 x together with its first two derivatives. Comment on the behavior of ƒ and the shape of its graph in relation to the signs and values of ƒ′ and ƒ″.
54. Use the identity cot1 u =
p  tan1 u 2
to derive the formula for the derivative of cot1 u in Table 3.1 from the formula for the derivative of tan1 u.
3.10 Related Rates In this section we look at problems that ask for the rate at which some variable changes when it is known how the rate of some other related variable (or perhaps several variables) changes. The problem of finding a rate of change from other known rates of change is called a related rates problem.
194
Chapter 3: Derivatives
Related Rates Equations Suppose we are pumping air into a spherical balloon. Both the volume and radius of the balloon are increasing over time. If V is the volume and r is the radius of the balloon at an instant of time, then V =
4 3 pr . 3
Using the Chain Rule, we differentiate both sides with respect to t to find an equation relating the rates of change of V and r, dV dV dr dr = = 4pr 2 . dt dr dt dt So if we know the radius r of the balloon and the rate dV>dt at which the volume is increasing at a given instant of time, then we can solve this last equation for dr>dt to find how fast the radius is increasing at that instant. Note that it is easier to directly measure the rate of increase of the volume (the rate at which air is being pumped into the balloon) than it is to measure the increase in the radius. The related rates equation allows us to calculate dr>dt from dV>dt. Very often the key to relating the variables in a related rates problem is drawing a picture that shows the geometric relations between them, as illustrated in the following example. Water runs into a conical tank at the rate of 9 ft3 >min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?
EXAMPLE 1 dV = 9 ft3min dt 5 ft
Solution Figure 3.44 shows a partially filled conical tank. The variables in the problem are V = volume (ft3) of the water in the tank at time t (min) x = radius (ft) of the surface of the water at time t y = depth (ft) of the water in the tank at time t.
x dy =? dt when y = 6 ft
10 ft y
FIGURE 3.44 The geometry of the conical tank and the rate at which water fills the tank determine how fast the water level rises (Example 1).
We assume that V, x, and y are differentiable functions of t. The constants are the dimensions of the tank. We are asked for dy>dt when y = 6 ft
dV = 9 ft3 >min. dt
and
The water forms a cone with volume V =
1 2 px y. 3
This equation involves x as well as V and y. Because no information is given about x and dx>dt at the time in question, we need to eliminate x. The similar triangles in Figure 3.44 give us a way to express x in terms of y: x 5 y = 10
or
x =
y . 2
Therefore, we find V =
y 2 p 3 1 pa b y = y 3 2 12
to give the derivative dV p # 2 dy p 2 dy = 3y = y . 12 4 dt dt dt
3.10 Related Rates
195
Finally, use y = 6 and dV>dt = 9 to solve for dy>dt. dy p 9 = (6)2 4 dt dy 1 = p ≈ 0.32 dt At the moment in question, the water level is rising at about 0.32 ft>min.
Related Rates Problem Strategy 1. Draw a picture and name the variables and constants. Use t for time. Assume that all variables are differentiable functions of t. 2. Write down the numerical information (in terms of the symbols you have chosen). 3. Write down what you are asked to find (usually a rate, expressed as a derivative). 4. Write an equation that relates the variables. You may have to combine two or more equations to get a single equation that relates the variable whose rate you want to the variables whose rates you know. 5. Differentiate with respect to t. Then express the rate you want in terms of the rates and variables whose values you know. 6. Evaluate. Use known values to find the unknown rate.
EXAMPLE 2 A hot air balloon rising straight up from a level field is tracked by a range finder 150 m from the liftoff point. At the moment the range finder’s elevation angle is p>4, the angle is increasing at the rate of 0.14 rad > min. How fast is the balloon rising at that moment?
Balloon
du = 0.14 radmin dt when u = p4
Range finder
dy =? y dt when u = p4
Solution We answer the question in the six strategy steps. 1.
u 150 m
FIGURE 3.45 The rate of change of the balloon’s height is related to the rate of change of the angle the range finder makes with the ground (Example 2).
Draw a picture and name the variables and constants (Figure 3.45). The variables in the picture are u = the angle in radians the range finder makes with the ground. y = the height in meters of the balloon above the ground. We let t represent time in minutes and assume that u and y are differentiable functions of t. The one constant in the picture is the distance from the range finder to the liftoff point (150 m). There is no need to give it a special symbol. 2. Write down the additional numerical information. du = 0.14 rad>min dt
3. 4.
5.
when
u =
p 4
Write down what we are to find. We want dy>dt when u = p>4. Write an equation that relates the variables y and u. y = tan u or y = 150 tan u 150 Differentiate with respect to t using the Chain Rule. The result tells how dy>dt (which we want) is related to du>dt (which we know). dy du = 150 (sec2 u) dt dt
6.
Evaluate with u = p>4 and du>dt = 0.14 to find dy>dt. dy = 1501 22 22(0.14) = 42 dt
sec
p = 22 4
At the moment in question, the balloon is rising at the rate of 42 m>min.
196
Chapter 3: Derivatives
EXAMPLE 3 A police cruiser, approaching a rightangled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 mi north of the intersection and the car is 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is the speed of the car?
y Situation when x = 0.8, y = 0.6 y dy = −60 dt
ds = 20 dt
0
dx = ? dt
x
x
Solution We picture the car and cruiser in the coordinate plane, using the positive xaxis as the eastbound highway and the positive yaxis as the southbound highway (Figure 3.46). We let t represent time and set x = position of car at time t y = position of cruiser at time t s = distance between car and cruiser at time t.
FIGURE 3.46 The speed of the car is related to the speed of the police cruiser and the rate of change of the distance s between them (Example 3).
We assume that x, y, and s are differentiable functions of t. We want to find dx>dt when x = 0.8 mi,
dy = 60 mph, dt
y = 0.6 mi,
ds = 20 mph. dt
Note that dy>dt is negative because y is decreasing. We differentiate the distance equation between the car and the cruiser, s2 = x2 + y2 (we could also use s = 2x2 + y2 ), and obtain 2s
dy ds dx = 2x + 2y dt dt dt dy ds 1 dx = s ax + y b dt dt dt =
dy dx 1 + y b. ax 2 dt dt 2x + y 2
Finally, we use x = 0.8, y = 0.6, dy>dt = 60, ds>dt = 20, and solve for dx>dt. 20 =
dx 1 + (0.6)(60)b a0.8 2 dt 2(0.8) + (0.6) 2
2 2 dx 20 2(0.8) + (0.6) + (0.6)(60) = 70 = 0.8 dt
At the moment in question, the car’s speed is 70 mph. y P 10 u 0
Q (x, 0)
FIGURE 3.47 The particle P travels clockwise along the circle (Example 4).
x
EXAMPLE 4 A particle P moves clockwise at a constant rate along a circle of radius 10 m centered at the origin. The particle’s initial position is (0, 10) on the yaxis, and its final destination is the point (10, 0) on the xaxis. Once the particle is in motion, the tangent line at P intersects the xaxis at a point Q (which moves over time). If it takes the particle 30 sec to travel from start to finish, how fast is the point Q moving along the xaxis when it is 20 m from the center of the circle? Solution We picture the situation in the coordinate plane with the circle centered at the origin (see Figure 3.47). We let t represent time and let u denote the angle from the xaxis to the radial line joining the origin to P. Since the particle travels from start to finish in 30 sec, it is traveling along the circle at a constant rate of p>2 radians in 1>2 min, or p rad>min. In other words, du>dt = p, with t being measured in minutes. The negative sign appears because u is decreasing over time.
3.10 Related Rates
197
Setting x(t) to be the distance at time t from the point Q to the origin, we want to find dx>dt when x = 20 m
du = p rad>min. dt
and
To relate the variables x and u, we see from Figure 3.47 that x cos u = 10, or x = 10 sec u. Differentiation of this last equation gives du dx = 10 sec u tan u = 10p sec u tan u. dt dt Note that dx>dt is negative because x is decreasing (Q is moving toward the origin). When x = 20, cos u = 1>2 and sec u = 2. Also, tan u = 2sec2 u  1 = 23. It follows that dx = (10p)(2)1 23 2 = 20 23p. dt At the moment in question, the point Q is moving toward the origin at the speed of 20 23p ≈ 109 m>min. A
12,000 u R
x
FIGURE 3.48 Jet airliner A traveling at constant altitude toward radar station R (Example 5).
EXAMPLE 5 A jet airliner is flying at a constant altitude of 12,000 ft above sea level as it approaches a Pacific island. The aircraft comes within the direct line of sight of a radar station located on the island, and the radar indicates the initial angle between sea level and its line of sight to the aircraft is 30°. How fast (in miles per hour) is the aircraft approaching the island when first detected by the radar instrument if it is turning upward (counterclockwise) at the rate of 2>3 deg>sec in order to keep the aircraft within its direct line of sight? Solution The aircraft A and radar station R are pictured in the coordinate plane, using the positive xaxis as the horizontal distance at sea level from R to A, and the positive yaxis as the vertical altitude above sea level. We let t represent time and observe that y = 12,000 is a constant. The general situation and lineofsight angle u are depicted in Figure 3.48. We want to find dx>dt when u = p>6 rad and du>dt = 2>3 deg>sec. From Figure 3.48, we see that 12,000 = tan u x
or
x = 12,000 cot u.
Using miles instead of feet for our distance units, the last equation translates to x =
12,000 cot u. 5280
Differentiation with respect to t gives du dx 1200 = csc2 u . dt dt 528 When u = p>6, sin2 u = 1>4, so csc2 u = 4. Converting du>dt = 2>3 deg>sec to radians per hour, we find du 2 p = a b (3600) rad>hr. 3 180 dt
1 hr = 3600 sec, 1 deg = p>180 rad
Substitution into the equation for dx>dt then gives dx p 1200 2 = ab (4)a b a b (3600) ≈ 380. 3 180 dt 528 The negative sign appears because the distance x is decreasing, so the aircraft is approaching the island at a speed of approximately 380 mi>hr when first detected by the radar.
198
Chapter 3: Derivatives
P
W
dx = 4 ftsec dt M
O
5 ft
x (a)
P
EXAMPLE 6 Figure 3.49a shows a rope running through a pulley at P and bearing a weight W at one end. The other end is held 5 ft above the ground in the hand M of a worker. Suppose the pulley is 25 ft above ground, the rope is 45 ft long, and the worker is walking rapidly away from the vertical line PW at the rate of 4 ft>sec. How fast is the weight being raised when the worker’s hand is 21 ft away from PW? Solution We let OM be the horizontal line of length x ft from a point O directly below the pulley to the worker’s hand M at any instant of time (Figure 3.49). Let h be the height of the weight W above O, and let z denote the length of rope from the pulley P to the worker’s hand. We want to know dh>dt when x = 21 given that dx>dt = 4. Note that the height of P above O is 20 ft because O is 5 ft above the ground. We assume the angle at O is a right angle. At any instant of time t we have the following relationships (see Figure 3.49b): 20  h + z = 45 202 + x2 = z2.
z
20 ft W h
dh = ? dt
O
x
M
(b)
Total length of rope is 45 ft. Angle at O is a right angle.
If we solve for z = 25 + h in the first equation, and substitute into the second equation, we have 202 + x2 = (25 + h)2.
(1)
Differentiating both sides with respect to t gives
FIGURE 3.49 A worker at M walks to the right, pulling the weight W upward as the rope moves through the pulley P (Example 6).
2x
dx dh = 2(25 + h) , dt dt
and solving this last equation for dh>dt we find dh x dx = . dt 25 + h dt
(2)
Since we know dx>dt, it remains only to find 25 + h at the instant when x = 21. From Equation (1), 202 + 212 = (25 + h)2 so that (25 + h)2 = 841,
or
25 + h = 29.
Equation (2) now gives 84 dh 21 # = 4 = ≈ 2.9 ft>sec 29 29 dt as the rate at which the weight is being raised when x = 21 ft.
Exercises
3.10
1. Area Suppose that the radius r and area A = pr 2 of a circle are differentiable functions of t. Write an equation that relates dA>dt to dr>dt. 2. Surface area Suppose that the radius r and surface area S = 4pr 2 of a sphere are differentiable functions of t. Write an equation that relates dS>dt to dr>dt. 3. Assume that y = 5x and dx>dt = 2. Find dy>dt. 4. Assume that 2x + 3y = 12 and dy>dt = 2. Find dx>dt. 5. If y = x and dx>dt = 3, then what is dy>dt when x = 1? 2
6. If x = y3  y and dy>dt = 5, then what is dx>dt when y = 2? 7. If x2 + y2 = 25 and dx>dt = 2, then what is dy>dt when x = 3 and y = 4? 8. If x2y3 = 4>27 and dy>dt = 1>2, then what is dx>dt when x = 2? 9. If L = 2x2 + y2, dx>dt =  1, and dy>dt = 3, find dL>dt when x = 5 and y = 12. 10. If r + s2 + y3 = 12, dr>dt = 4, and ds>dt =  3, find dy>dt when r = 3 and s = 1.
3.10 Related Rates
11. If the original 24 m edge length x of a cube decreases at the rate of 5 m>min, when x = 3 m at what rate does the cube’s
199
a. Assuming that x, y, and z are differentiable functions of t, how is ds>dt related to dx>dt, dy>dt, and dz>dt?
a. surface area change?
b. How is ds>dt related to dy>dt and dz>dt if x is constant?
b. volume change?
c. How are dx>dt, dy>dt, and dz>dt related if s is constant?
12. A cube’s surface area increases at the rate of 72 in2 >sec. At what rate is the cube’s volume changing when the edge length is x = 3 in?
19. Area The area A of a triangle with sides of lengths a and b enclosing an angle of measure u is
13. Volume The radius r and height h of a right circular cylinder are related to the cylinder’s volume V by the formula V = pr 2h.
A =
1 ab sin u. 2
a. How is dV>dt related to dh>dt if r is constant?
a. How is dA>dt related to du>dt if a and b are constant?
b. How is dV>dt related to dr>dt if h is constant?
b. How is dA>dt related to du>dt and da>dt if only b is constant?
c. How is dV>dt related to dr>dt and dh>dt if neither r nor h is constant?
c. How is dA>dt related to du>dt, da>dt, and db>dt if none of a, b, and u are constant?
14. Volume The radius r and height h of a right circular cone are related to the cone’s volume V by the equation V = (1>3)pr 2h. a. How is dV>dt related to dh>dt if r is constant? b. How is dV>dt related to dr>dt if h is constant? c. How is dV>dt related to dr>dt and dh>dt if neither r nor h is constant? 15. Changing voltage The voltage V (volts), current I (amperes), and resistance R (ohms) of an electric circuit like the one shown here are related by the equation V = IR. Suppose that V is increasing at the rate of 1 volt>sec while I is decreasing at the rate of 1>3 amp>sec. Let t denote time in seconds. + V− I
R
a. What is the value of dV>dt? b. What is the value of dI>dt? c. What equation relates dR>dt to dV>dt and dI>dt? d. Find the rate at which R is changing when V = 12 volts and I = 2 amps. Is R increasing, or decreasing? 16. Electrical power The power P (watts) of an electric circuit is related to the circuit’s resistance R (ohms) and current I (amperes) by the equation P = RI 2.
20. Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 cm > min. At what rate is the plate’s area increasing when the radius is 50 cm? 21. Changing dimensions in a rectangle The length l of a rectangle is decreasing at the rate of 2 cm>sec while the width w is increasing at the rate of 2 cm>sec. When l = 12 cm and w = 5 cm, find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing? 22. Changing dimensions in a rectangular box Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates: dy dz dx = 1 m>sec, = 2 m>sec, = 1 m>sec. dt dt dt Find the rates at which the box’s (a) volume, (b) surface area, and (c) diagonal length s = 2x2 + y2 + z2 are changing at the instant when x = 4, y = 3, and z = 2. 23. A sliding ladder A 13ft ladder is leaning against a house when its base starts to slide away (see accompanying figure). By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft>sec. a. How fast is the top of the ladder sliding down the wall then? b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle u between the ladder and the ground changing then? y
a. How are dP>dt, dR>dt, and dI>dt related if none of P, R, and I are constant? b. How is dR>dt related to dI>dt if P is constant?
y(t)
17. Distance Let x and y be differentiable functions of t and let s = 2x2 + y2 be the distance between the points (x, 0) and (0, y) in the xyplane.
13ft ladder
a. How is ds>dt related to dx>dt if y is constant? b. How is ds>dt related to dx>dt and dy>dt if neither x nor y is constant? c. How is dx>dt related to dy>dt if s is constant? 18. Diagonals If x, y, and z are lengths of the edges of a rectangular box, the common length of the box’s diagonals is s = 2x2 + y2 + z2.
u 0
x(t)
x
24. Commercial air traffic Two commercial airplanes are flying at an altitude of 40,000 ft along straightline courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd). Plane B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when A is 5
200
Chapter 3: Derivatives
nautical miles from the intersection point and B is 12 nautical miles from the intersection point?
Ring at edge of dock
25. Flying a kite A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft>sec. How fast must she let out the string when the kite is 500 ft away from her? 26. Boring a cylinder The mechanics at Lincoln Automotive are reboring a 6in.deep cylinder to fit a new piston. The machine they are using increases the cylinder’s radius onethousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.? 27. A growing sand pile Sand falls from a conveyor belt at the rate of 10 m3 >min onto the top of a conical pile. The height of the pile is always threeeighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute.
u 6'
33. A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft>sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft>sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later? y
28. A draining conical reservoir Water is flowing at the rate of 50 m3 >min from a shallow concrete conical reservoir (vertex down) of base radius 45 m and height 6 m. a. How fast (centimeters per minute) is the water level falling when the water is 5 m deep?
y(t)
b. How fast is the radius of the water’s surface changing then? Answer in centimeters per minute. 29. A draining hemispherical reservoir Water is flowing at the rate of 6 m3 >min from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius R is V = (p>3)y2(3R  y) when the water is y meters deep.
s(t)
Center of sphere 13 Water level r y
x(t)
0
x
34. Making coffee Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in3 >min. a. How fast is the level in the pot rising when the coffee in the cone is 5 in. deep? b. How fast is the level in the cone falling then?
a. At what rate is the water level changing when the water is 8 m deep? 6″
b. What is the radius r of the water’s surface when the water is y m deep? c. At what rate is the radius r changing when the water is 8 m deep? 30. A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop’s radius increases at a constant rate.
6″ How fast is this level falling?
31. The radius of an inflating balloon A spherical balloon is inflated with helium at the rate of 100p ft3 >min. How fast is the balloon’s radius increasing at the instant the radius is 5 ft? How fast is the surface area increasing? 32. Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft>sec. a. How fast is the boat approaching the dock when 10 ft of rope are out? b. At what rate is the angle u changing at this instant (see the figure)?
How fast is this level rising? 6″
3.10 Related Rates
35. Cardiac output In the late 1860s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Würzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 L>min. At rest it is likely to be a bit under 6 L>min. If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 L>min. Your cardiac output can be calculated with the formula Q y = , D where Q is the number of milliliters of CO2 you exhale in a minute and D is the difference between the CO2 concentration (ml>L) in the blood pumped to the lungs and the CO2 concentration in the blood returning from the lungs. With Q = 233 ml>min and D = 97  56 = 41 ml>L, y =
233 ml>min ≈ 5.68 L>min, 41 ml>L
fairly close to the 6 L>min that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when Q = 233 and D = 41, we also know that D is decreasing at the rate of 2 units a minute but that Q remains unchanged. What is happening to the cardiac output?
201
39. A moving shadow A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft away from the light. (See accompanying figure.) How fast is the shadow of the ball moving along the ground 1>2 sec later? (Assume the ball falls a distance s = 16t 2 ft in t sec.) Light
Ball at time t = 0 1/2 sec later
50ft pole
0
Shadow x x(t)
30
40. A building’s shadow On a morning of a day when the sun will pass directly overhead, the shadow of an 80ft building on level ground is 60 ft long. At the moment in question, the angle u the sun makes with the ground is increasing at the rate of 0.27°>min. At what rate is the shadow decreasing? (Remember to use radians. Express your answer in inches per minute, to the nearest tenth.)
36. Moving along a parabola A particle moves along the parabola y = x2 in the first quadrant in such a way that its xcoordinate (measured in meters) increases at a steady 10 m>sec. How fast is the angle of inclination u of the line joining the particle to the origin changing when x = 3 m? 37. Motion in the plane The coordinates of a particle in the metric xyplane are differentiable functions of time t with dx>dt =  1 m>sec and dy>dt =  5 m>sec. How fast is the particle’s distance from the origin changing as it passes through the point (5, 12)? 38. Videotaping a moving car You are videotaping a race from a stand 132 ft from the track, following a car that is moving at 180 mi>h (264 ft>sec), as shown in the accompanying figure. How fast will your camera angle u be changing when the car is right in front of you? A half second later? Camera
80′ u
41. A melting ice layer A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in3 >min, how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing? 42. Highway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 mi>h. The pilot sees an oncoming car and with radar determines that at the instant the lineofsight distance from plane to car is 5 mi, the lineofsight distance is decreasing at the rate of 160 mi>h. Find the car’s speed along the highway.
u
132′
43. Baseball players A baseball diamond is a square 90 ft on a side. A player runs from first base to second at a rate of 16 ft>sec. a. At what rate is the player’s distance from third base changing when the player is 30 ft from first base?
Car
b. At what rates are angles u1 and u2 (see the figure) changing at that time?
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Chapter 3: Derivatives
44. Ships Two ships are steaming straight away from a point O along routes that make a 120° angle. Ship A moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship B moves at 21 knots. How fast are the ships moving apart when OA = 5 and OB = 3 nautical miles?
c. The player slides into second base at the rate of 15 ft>sec. At what rates are angles u1 and u2 changing as the player touches base? Second base 90′ Third base
u1
u2
45. Clock’s moving hands At what rate is the angle between a clock’s minute and hour hands changing at 4 o’clock in the afternoon?
Player 30′
46. Oil spill An explosion at an oil rig located in gulf waters causes an elliptical oil slick to spread on the surface from the rig. The slick is a constant 9 in. thick. After several days, when the major axis of the slick is 2 mi long and the minor axis is 3/4 mi wide, it is determined that its length is increasing at the rate of 30 ft/hr, and its width is increasing at the rate of 10 ft/hr. At what rate (in cubic feet per hour) is oil flowing from the site of the rig at that time?
First base
Home
3.11 Linearization and Differentials Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions discussed in this section are called linearizations, and they are based on tangent lines. Other approximating functions, such as polynomials, are discussed in Chapter 10. We introduce new variables dx and dy, called differentials, and define them in a way that makes Leibniz’s notation for the derivative dy>dx a true ratio. We use dy to estimate error in measurement, which then provides for a precise proof of the Chain Rule (Section 3.6).
Linearization As you can see in Figure 3.50, the tangent to the curve y = x2 lies close to the curve near the point of tangency. For a brief interval to either side, the yvalues along the tangent line 4
2 y = x2
y = x2 y = 2x − 1
y = 2x − 1
(1, 1)
(1, 1) −1
3 0
0
2 0
y = x 2 and its tangent y = 2x − 1 at (1, 1). 1.2
Tangent and curve very close near (1, 1). 1.003
y = x2
y = x2 y = 2x − 1
(1, 1)
(1, 1)
y = 2x − 1 0.8
1.2 0.8
Tangent and curve very close throughout entire xinterval shown.
0.997 0.997
1.003
Tangent and curve closer still. Computer screen cannot distinguish tangent from curve on this xinterval.
FIGURE 3.50 The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent.
3.11 Linearization and Differentials
y
y = f (x) Slope = f ′(a) (a, f(a))
0
y = L(x)
a
x
FIGURE 3.51 The tangent to the curve y = ƒ(x) at x = a is the line L(x) = ƒ(a) + ƒ′(a)(x  a).
203
give good approximations to the yvalues on the curve. We observe this phenomenon by zooming in on the two graphs at the point of tangency or by looking at tables of values for the difference between ƒ(x) and its tangent line near the xcoordinate of the point of tangency. The phenomenon is true not just for parabolas; every differentiable curve behaves locally like its tangent line. In general, the tangent to y = ƒ(x) at a point x = a, where ƒ is differentiable (Figure 3.51), passes through the point (a, ƒ(a)), so its pointslope equation is y = ƒ(a) + ƒ′(a)(x  a). Thus, this tangent line is the graph of the linear function L(x) = ƒ(a) + ƒ′(a)(x  a). For as long as this line remains close to the graph of ƒ as we move off the point of tangency, L(x) gives a good approximation to ƒ(x).
DEFINITIONS
If ƒ is differentiable at x = a, then the approximating function L(x) = ƒ(a) + ƒ′(a)(x  a)
is the linearization of ƒ at a. The approximation ƒ(x) ≈ L(x) of ƒ by L is the standard linear approximation of ƒ at a. The point x = a is the center of the approximation.
EXAMPLE 1
Find the linearization of ƒ(x) = 21 + x at x = 0 (Figure 3.52).
y
y=
x y= 1+ 2
5 x + 4 4
y=1+ x 2
1.1
2 y = "1 + x
y = "1 + x
1.0
1
−1
0
1
2
3
4
x
FIGURE 3.52 The graph of y = 21 + x and its linearizations at x = 0 and x = 3. Figure 3.53 shows a magnified view of the small window about 1 on the yaxis.
Solution
0.9 −0.1
0
0.1
0.2
FIGURE 3.53 Magnified view of the window in Figure 3.52.
Since ƒ′(x) =
1 (1 + x)1>2, 2
we have ƒ(0) = 1 and ƒ′(0) = 1>2, giving the linearization L(x) = ƒ(a) + ƒ′(a)(x  a) = 1 +
x 1 (x  0) = 1 + . 2 2
See Figure 3.53. The following table shows how accurate the approximation 21 + x ≈ 1 + (x>2) from Example 1 is for some values of x near 0. As we move away from zero, we lose
204
Chapter 3: Derivatives
accuracy. For example, for x = 2, the linearization gives 2 as the approximation for 23, which is not even accurate to one decimal place. True value
True value − approximation
= 1.10
1.095445
0.004555 6 102
0.05 = 1.025 2
1.024695
0.000305 6 103
0.005 = 1.00250 2
1.002497
0.000003 6 105
Approximation 21.2 ≈ 1 + 21.05 ≈ 1 + 21.005 ≈ 1 +
0.2 2
Do not be misled by the preceding calculations into thinking that whatever we do with a linearization is better done with a calculator. In practice, we would never use a linearization to find a particular square root. The utility of a linearization is its ability to replace a complicated formula by a simpler one over an entire interval of values. If we have to work with 21 + x for x close to 0 and can tolerate the small amount of error involved, we can work with 1 + (x>2) instead. Of course, we then need to know how much error there is. We further examine the estimation of error in Chapter 10. A linear approximation normally loses accuracy away from its center. As Figure 3.52 suggests, the approximation 21 + x ≈ 1 + (x>2) will probably be too crude to be useful near x = 3. There, we need the linearization at x = 3.
EXAMPLE 2
Find the linearization of ƒ(x) = 21 + x at x = 3.
Solution We evaluate the equation defining L(x) at a = 3. With ƒ(3) = 2,
ƒ′(3) =
1 1 (1 + x)1>2 ` = , 2 4 x=3
we have L(x) = 2 +
5 x 1 (x  3) = + . 4 4 4
At x = 3.2, the linearization in Example 2 gives 21 + x = 21 + 3.2 ≈
5 3.2 + = 1.250 + 0.800 = 2.050, 4 4
which differs from the true value 24.2 ≈ 2.04939 by less than one onethousandth. The linearization in Example 1 gives 21 + x = 21 + 3.2 ≈ 1 +
y
3.2 = 1 + 1.6 = 2.6, 2
a result that is off by more than 25%.
EXAMPLE 3 0
p 2
x y = cos x
y = −x + p 2
FIGURE 3.54 The graph of ƒ(x) = cos x and its linearization at x = p>2. Near x = p>2, cos x ≈  x + (p>2) (Example 3).
Find the linearization of ƒ(x) = cos x at x = p>2 (Figure 3.54).
Solution Since ƒ(p>2) = cos (p>2) = 0, ƒ′(x) = sin x, and ƒ′(p>2) = sin (p>2) = 1, we find the linearization at a = p>2 to be L(x) = ƒ(a) + ƒ′(a)(x  a) = 0 + (1)ax = x +
p . 2
p b 2
3.11 Linearization and Differentials
205
An important linear approximation for roots and powers is (1 + x)k ≈ 1 + kx
(x near 0; any number k)
(Exercise 15). This approximation, good for values of x sufficiently close to zero, has broad application. For example, when x is small, 21 + x ≈ 1 + Approximations Near x = 0 21 + x ≈ 1 +
x 2
1 ≈ 1 + x 1  x 1 21  x2
≈ 1 +
x 2
k = 1>2
1 = (1  x)1 ≈ 1 + (1)(x) = 1 + x 1  x 5 1 3 21 + 5x4 = (1 + 5x4)1>3 ≈ 1 + (5x4) = 1 + x4 3 3 1
2
1 x 2
21  x2
1 1 = (1  x2)1>2 ≈ 1 + a b (x2) = 1 + x2 2 2
k =  1; replace x by  x. k = 1>3; replace x by 5x4. k =  1>2; replace x by  x2 .
Differentials We sometimes use the Leibniz notation dy>dx to represent the derivative of y with respect to x. Contrary to its appearance, it is not a ratio. We now introduce two new variables dx and dy with the property that when their ratio exists, it is equal to the derivative.
DEFINITION Let y = ƒ(x) be a differentiable function. The differential dx is an independent variable. The differential dy is dy = ƒ′(x) dx.
Unlike the independent variable dx, the variable dy is always a dependent variable. It depends on both x and dx. If dx is given a specific value and x is a particular number in the domain of the function ƒ, then these values determine the numerical value of dy. Often the variable dx is chosen to be ∆x, the change in x.
EXAMPLE 4 (a) Find dy if y = x5 + 37x. (b) Find the value of dy when x = 1 and dx = 0.2. Solution (a) dy = (5x4 + 37) dx (b) Substituting x = 1 and dx = 0.2 in the expression for dy, we have dy = (5 # 14 + 37) 0.2 = 8.4.
The geometric meaning of differentials is shown in Figure 3.55. Let x = a and set dx = ∆x. The corresponding change in y = ƒ(x) is ∆y = ƒ(a + dx)  ƒ(a). The corresponding change in the tangent line L is ∆L = L(a + dx)  L(a) = ƒ(a) + ƒ′(a) 3 (a + dx)  a4  ƒ(a) (++++1+)+++++1* ()* L (a + dx)
= ƒ′(a) dx.
L (a)
206
Chapter 3: Derivatives y
y = f (x) (a + dx, f (a + dx)) Δy = f (a + dx) − f (a) ΔL = f ′(a)dx (a, f (a)) dx = Δx When dx is a small change in x, the corresponding change in the linearization is precisely dy.
Tangent line
x
a + dx
a
0
FIGURE 3.55 Geometrically, the differential dy is the change ∆L in the linearization of ƒ when x = a changes by an amount dx = ∆x.
That is, the change in the linearization of ƒ is precisely the value of the differential dy when x = a and dx = ∆x. Therefore, dy represents the amount the tangent line rises or falls when x changes by an amount dx = ∆x. If dx ≠ 0, then the quotient of the differential dy by the differential dx is equal to the derivative ƒ′(x) because dy , dx =
dy ƒ′(x) dx = ƒ′(x) = . dx dx
We sometimes write dƒ = ƒ′(x) dx in place of dy = ƒ′(x) dx, calling dƒ the differential of ƒ. For instance, if ƒ(x) = 3x2  6, then dƒ = d(3x2  6) = 6x dx. Every differentiation formula like d (u + y) du dy = + dx dx dx
or
d (sin u) du = cos u dx dx
or
d(sin u) = cos u du.
has a corresponding differential form like d(u + y) = du + dy
EXAMPLE 5 We can use the Chain Rule and other differentiation rules to find differentials of functions. (a) d (tan 2x) = sec2 (2x) d (2x) = 2 sec2 2x dx (b) d a
(x + 1) dx  x d (x + 1) x dx + dx  x dx dx x b = = = x + 1 (x + 1)2 (x + 1)2 (x + 1)2
Estimating with Differentials Suppose we know the value of a differentiable function ƒ(x) at a point a and want to estimate how much this value will change if we move to a nearby point a + dx. If dx = ∆x is small, then we can see from Figure 3.55 that ∆y is approximately equal to the differential dy. Since ƒ(a + dx) = ƒ(a) + ∆y,
∆x = dx
3.11 Linearization and Differentials
207
the differential approximation gives ƒ(a + dx) ≈ ƒ(a) + dy when dx = ∆x. Thus the approximation ∆y ≈ dy can be used to estimate ƒ(a + dx) when ƒ(a) is known, dx is small, and dy = ƒ′(a) dx.
dr = 0.1
a = 10
EXAMPLE 6 The radius r of a circle increases from a = 10 m to 10.1 m (Figure 3.56). Use dA to estimate the increase in the circle’s area A. Estimate the area of the enlarged circle and compare your estimate to the true area found by direct calculation. Solution Since A = pr 2, the estimated increase is dA = A′(a) dr = 2pa dr = 2p(10)(0.1) = 2p m2.
ΔA ≈ dA = 2 pa dr
FIGURE 3.56 When dr is small compared with a, the differential dA gives the estimate A(a + dr) = pa2 + dA (Example 6).
Thus, since A(r + ∆r) ≈ A(r) + dA, we have A(10 + 0.1) ≈ A(10) + 2p = p(10)2 + 2p = 102p. The area of a circle of radius 10.1 m is approximately 102p m2. The true area is A(10.1) = p(10.1)2 = 102.01p m2. The error in our estimate is 0.01p m2, which is the difference ∆A  dA.
EXAMPLE 7
Use differentials to estimate
(a) 7.971>3 (b) sin (p>6 + 0.01). Solution (a) The differential associated with the cube root function y = x1>3 is dy =
1 dx. 3x2>3
We set a = 8, the closest number near 7.97 where we can easily compute ƒ(a) and ƒ′(a). To arrange that a + dx = 7.97, we choose dx = 0.03. Approximating with the differential gives ƒ(7.97) = ƒ(a + dx) ≈ ƒ(a) + dy = 81>3 + = 2 +
1 (0.03) 3(8)2>3
1 (0.03) = 1.9975 12
This gives an approximation to the true value of 7.971>3, which is 1.997497 to 6 decimals. (b) The differential associated with y = sin x is dy = cos x dx.
208
Chapter 3: Derivatives
To estimate sin (p>6 + 0.01), we set a = p>6 and dx = 0.01. Then ƒ(p>6 + 0.01) = ƒ(a + dx) ≈ ƒ(a) + dy sin (a + dx) ≈ sin a + (cos a) dx
= sin =
p p + acos b (0.01) 6 6
23 1 + (0.01) ≈ 0.5087 2 2
For comparison, the true value of sin (p>6 + 0.01) to 6 decimals is 0.508635.
The method in part (b) of Example 7 is used by some calculator and computer algorithms to give values of trigonometric functions. The algorithms store a large table of sine and cosine values between 0 and p>4. Values between these stored values are computed using differentials as in Example 7b. Values outside of 3 0, p>44 are computed from values in this interval using trigonometric identities.
Error in Differential Approximation Let ƒ(x) be differentiable at x = a and suppose that dx = ∆x is an increment of x. We have two ways to describe the change in ƒ as x changes from a to a + ∆x: The true change: The differential estimate:
∆ƒ = ƒ(a + ∆x)  ƒ(a) dƒ = ƒ′(a) ∆x.
How well does dƒ approximate ∆ƒ? We measure the approximation error by subtracting dƒ from ∆ƒ: Approximation error = ∆ƒ  dƒ = ∆ƒ  ƒ′(a)∆x = ƒ(a + ∆x)  ƒ(a)  ƒ′(a)∆x (+++)+++* ∆ƒ
ƒ(a + ∆x)  ƒ(a) = a  ƒ′(a)b # ∆x ∆x (++++1+)+++++1* = P # ∆x.
Call this part P.
As ∆x S 0, the difference quotient ƒ(a + ∆x)  ƒ(a) ∆x approaches ƒ′(a) (remember the definition of ƒ′(a)), so the quantity in parentheses becomes a very small number (which is why we called it P). In fact, P S 0 as ∆x S 0. When ∆x is small, the approximation error P ∆x is smaller still. ∆ƒ = ƒ′(a)∆x + P ∆x ()* (+)+* ()* true change
estimated change
error
Although we do not know the exact size of the error, it is the product P # ∆x of two small quantities that both approach zero as ∆x S 0. For many common functions, whenever ∆x is small, the error is still smaller.
3.11 Linearization and Differentials
209
Change in y = ƒ(x) near x = a If y = ƒ(x) is differentiable at x = a and x changes from a to a + ∆x, the change ∆y in ƒ is given by ∆y = ƒ′(a) ∆x + P ∆x
(1)
in which P S 0 as ∆x S 0.
In Example 6 we found that 6
∆A = p(10.1)2  p(10)2 = (102.01  100)p = (2p + 0.01p) m2 ()* dA
error
so the approximation error is ∆A  dA = P∆r = 0.01p and P = 0.01p> ∆r = 0.01p>0.1 = 0.1p m.
Proof of the Chain Rule Equation (1) enables us to prove the Chain Rule correctly. Our goal is to show that if ƒ(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then the composite y = ƒ(g(x)) is a differentiable function of x. Since a function is differentiable if and only if it has a derivative at each point in its domain, we must show that whenever g is differentiable at x0 and ƒ is differentiable at g(x0), then the composite is differentiable at x0 and the derivative of the composite satisfies the equation dy 2 = ƒ′(g(x0)) # g′(x0). dx x = x0 Let ∆x be an increment in x and let ∆u and ∆y be the corresponding increments in u and y. Applying Equation (1) we have ∆u = g′(x0)∆x + P1 ∆x = (g′(x0) + P1)∆x, where P1 S 0 as ∆x S 0. Similarly, ∆y = ƒ′(u0)∆u + P2 ∆u = (ƒ′(u0) + P2)∆u, where P2 S 0 as ∆u S 0. Notice also that ∆u S 0 as ∆x S 0. Combining the equations for ∆u and ∆y gives ∆y = (ƒ′(u0) + P2)(g′(x0) + P1)∆x, so ∆y = ƒ′(u0)g′(x0) + P2 g′(x0) + ƒ′(u0)P1 + P2P1. ∆x Since P1 and P2 go to zero as ∆x goes to zero, the last three terms on the right vanish in the limit, leaving dy ∆y 2 = lim = ƒ′(u0)g′(x0) = ƒ′(g(x0)) # g′(x0). dx x = x0 ∆x S 0 ∆x
210
Chapter 3: Derivatives
Sensitivity to Change The equation df = ƒ′(x) dx tells how sensitive the output of ƒ is to a change in input at different values of x. The larger the value of ƒ′ at x, the greater the effect of a given change dx. As we move from a to a nearby point a + dx, we can describe the change in ƒ in three ways:
Absolute change Relative change Percentage change
True
Estimated
∆ƒ = ƒ(a + dx)  ƒ(a) ∆ƒ ƒ(a)
dƒ = ƒ′(a) dx dƒ ƒ(a)
∆ƒ * 100 ƒ(a)
dƒ * 100 ƒ(a)
EXAMPLE 8
You want to calculate the depth of a well from the equation s = 16t 2 by timing how long it takes a heavy stone you drop to splash into the water below. How sensitive will your calculations be to a 0.1sec error in measuring the time? Solution The size of ds in the equation ds = 32t dt
depends on how big t is. If t = 2 sec, the change caused by dt = 0.1 is about ds = 32(2)(0.1) = 6.4 ft. Three seconds later at t = 5 sec, the change caused by the same dt is ds = 32(5)(0.1) = 16 ft. For a fixed error in the time measurement, the error in using ds to estimate the depth is larger when it takes a longer time before the stone splashes into the water. That is, the estimate is more sensitive to the effect of the error for larger values of t.
EXAMPLE 9
Newton’s second law, F =
d dy (my) = m = ma, dt dt
is stated with the assumption that mass is constant, but we know this is not strictly true because the mass of an object increases with velocity. In Einstein’s corrected formula, mass has the value m =
m0
21  y2 >c2
,
where the “rest mass” m0 represents the mass of an object that is not moving and c is the speed of light, which is about 300,000 km>sec. Use the approximation 1 1 ≈ 1 + x2 2 2 21  x
(2)
to estimate the increase ∆m in mass resulting from the added velocity y. Solution When y is very small compared with c, y2 >c2 is close to zero and it is safe to use the approximation 1 1 y2 ≈ 1 + a 2b 2 2 2 c 21  y >c
y Eq. (2) with x = c
3.11 Linearization and Differentials
211
to obtain m =
m0
21  y2 >c2
≈ m0 c 1 +
1 y2 1 1 a b d = m0 + m0 y2 a 2 b , 2 c2 2 c
m ≈ m0 +
1 1 m y2 a 2 b . 2 0 c
or (3)
Equation (3) expresses the increase in mass that results from the added velocity y.
Converting Mass to Energy Equation (3) derived in Example 9 has an important interpretation. In Newtonian physics, (1>2) m0y2 is the kinetic energy (KE) of the object, and if we rewrite Equation (3) in the form (m  m0) c2 ≈
1 m y2, 2 0
we see that (m  m0)c2 ≈
1 1 1 m y2 = m0 y2  m0 (0)2 = ∆(KE), 2 0 2 2
or (∆m)c2 ≈ ∆(KE). So the change in kinetic energy ∆(KE) in going from velocity 0 to velocity y is approximately equal to (∆m) c2, the change in mass times the square of the speed of light. Using c ≈ 3 * 108 m>sec, we see that a small change in mass can create a large change in energy.
Exercises
3.11
Finding Linearizations In Exercises 1–5, find the linearization L(x) of ƒ(x) at x = a.
12. ƒ(x) =
x , a = 1.3 x + 1
1. ƒ(x) = x3  2x + 3, a = 2
13. ƒ(x) = ex, a =  0.1
2. ƒ(x) = 2x2 + 9, a =  4 1 3. ƒ(x) = x + x , a = 1
15. Show that the linearization of ƒ(x) = (1 + x)k at x = 0 is L(x) = 1 + kx.
14. ƒ(x) = sin1 x, a = p>12
3 4. ƒ(x) = 2x, a =  8
5. ƒ(x) = tan x, a = p 6. Common linear approximations at x = 0 tions of the following functions at x = 0. a. sin x
b. cos x
c. tan x
d. ex
Find the linearizae. ln (1 + x)
Linearization for Approximation In Exercises 7–14, find a linearization at a suitably chosen integer near a at which the given function and its derivative are easy to evaluate.
7. ƒ(x) = x + 2x, a = 0.1 2
1
8. ƒ(x) = x , a = 0.9 9. ƒ(x) = 2x2 + 3x  3, a = 0.9 10. ƒ(x) = 1 + x, a = 8.1 3
11. ƒ(x) = 2x, a = 8.5
16. Use the linear approximation (1 + x)k ≈ 1 + kx to find an approximation for the function ƒ(x) for values of x near zero. 2 a. ƒ(x) = (1  x)6 b. ƒ(x) = 1  x c. ƒ(x) =
1 21 + x
e. ƒ(x) = (4 + 3x)1>3
d. ƒ(x) = 22 + x2 f. ƒ(x) =
3
B
a1 
2 x b 2 + x
17. Faster than a calculator Use the approximation (1 + x)k ≈ 1 + kx to estimate the following. a. (1.0002)50
3 b. 2 1.009
18. Find the linearization of ƒ(x) = 2x + 1 + sin x at x = 0. How is it related to the individual linearizations of 2x + 1 and sin x at x = 0?
212
Chapter 3: Derivatives
49. The change in the volume V = pr 2h of a right circular cylinder when the radius changes from r0 to r0 + dr and the height does not change
Derivatives in Differential Form In Exercises 19–38, find dy.
19. y = x3  3 2x 21. y =
20. y = x 21  x2
2x 1 + x2
22. y =
2 1x 3(1 + 1x)
23. 2y3>2 + xy  x = 0
24. xy2  4x3>2  y = 0
25. y = sin (5 1x)
26. y = cos (x2)
27. y = 4 tan (x3 >3)
29. y = 3 csc 1 1  2 2x 2
Applications 51. The radius of a circle is increased from 2.00 to 2.02 m.
28. y = sec (x2  1)
a. Estimate the resulting change in area.
1 30. y = 2 cot a b 1x
b. Express the estimate as a percentage of the circle’s original area.
31. y = e2x
32. y = xex
33. y = ln (1 + x2)
34. y = ln a
35. y = tan1 (ex )
1 36. y = cot1 a 2 b + cos1 2x x
37. y = sec1 (ex)
38. y = etan
2
x + 1 2x  1
1
b
52. The diameter of a tree was 10 in. During the following year, the circumference increased 2 in. About how much did the tree’s diameter increase? The tree’s crosssectional area? 53. Estimating volume Estimate the volume of material in a cylindrical shell with length 30 in., radius 6 in., and shell thickness 0.5 in. 0.5 in.
2x2 + 1
Approximation Error In Exercises 39–44, each function ƒ(x) changes value when x changes from x0 to x0 + dx. Find a. the change ∆ƒ = ƒ(x0 + dx)  ƒ(x0);
b. the value of the estimate dƒ = ƒ′(x0) dx; and c. the approximation error 0 ∆ƒ  dƒ 0 . y
50. The change in the lateral surface area S = 2prh of a right circular cylinder when the height changes from h0 to h0 + dh and the radius does not change
6 in. 30 in.
54. Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%? 55. The radius r of a circle is measured with an error of at most 2%. What is the maximum corresponding percentage error in computing the circle’s
y = f (x)
a. circumference? Δf = f (x 0 + dx) − f (x 0) df = f ′(x 0 ) dx
(x 0, f(x 0 )) dx
a. surface area?
Tangent
b. area?
56. The edge x of a cube is measured with an error of at most 0.5%. What is the maximum corresponding percentage error in computing the cube’s b. volume?
40. ƒ(x) = 2x2 + 4x  3, x0 =  1, dx = 0.1
57. Tolerance The height and radius of a right circular cylinder are equal, so the cylinder’s volume is V = ph3. The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h.
41. ƒ(x) = x3  x, x0 = 1, dx = 0.1
58. Tolerance
0
x0
x 0 + dx
x
39. ƒ(x) = x2 + 2x, x0 = 1, dx = 0.1
42. ƒ(x) = x4, x0 = 1, dx = 0.1 43. ƒ(x) = x1, x0 = 0.5, dx = 0.1 44. ƒ(x) = x3  2x + 3, x0 = 2, dx = 0.1 Differential Estimates of Change In Exercises 45–50, write a differential formula that estimates the given change in volume or surface area.
a. About how accurately must the interior diameter of a 10mhigh cylindrical storage tank be measured to calculate the tank’s volume to within 1% of its true value? b. About how accurately must the tank’s exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5% of the true amount?
45. The change in the volume V = (4>3)pr 3 of a sphere when the radius changes from r0 to r0 + dr
59. The diameter of a sphere is measured as 100 { 1 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation.
46. The change in the volume V = x3 of a cube when the edge lengths change from x0 to x0 + dx
60. Estimate the allowable percentage error in measuring the diameter D of a sphere if the volume is to be calculated correctly to within 3%.
47. The change in the surface area S = 6x2 of a cube when the edge lengths change from x0 to x0 + dx
61. The effect of flight maneuvers on the heart The amount of work done by the heart’s main pumping chamber, the left ventricle, is given by the equation
48. The change in the lateral surface area S = pr 2r 2 + h2 of a right circular cone when the radius changes from r0 to r0 + dr and the height does not change
W = PV +
Vdy2 , 2g
3.11 Linearization and Differentials
where W is the work per unit time, P is the average blood pressure, V is the volume of blood pumped out during the unit of time, d (“delta”) is the weight density of the blood, y is the average velocity of the exiting blood, and g is the acceleration of gravity. When P, V, d, and y remain constant, W becomes a function of g, and the equation takes the simplified form b W = a + g (a, b constant). As a member of NASA’s medical team, you want to know how sensitive W is to apparent changes in g caused by flight maneuvers, and this depends on the initial value of g. As part of your investigation, you decide to compare the effect on W of a given change dg on the moon, where g = 5.2 ft>sec2, with the effect the same change dg would have on Earth, where g = 32 ft>sec2. Use the simplified equation above to find the ratio of dWmoon to dWEarth. 62. Drug concentration The concentration C in milligrams per milliliter (mg>ml) of a certain drug in a person’s bloodstream t hrs after a pill is swallowed is modeled by the approximation C (t) = 1 +
4t  e0.06t. 1 + t3
Estimate the change in concentration when t changes from 20 to 30 min. 63. Unclogging arteries The formula V = kr 4, discovered by the physiologist Jean Poiseuille (1797–1869), allows us to predict how much the radius of a partially clogged artery has to be expanded in order to restore normal blood flow. The formula says that the volume V of blood flowing through the artery in a unit of time at a fixed pressure is a constant k times the radius of the artery to the fourth power. How will a 10% increase in r affect V? 64. Measuring acceleration of gravity When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum’s period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth’s surface, depending on the change in g. By keeping track of ∆T, we can estimate the variation in g from the equation T = 2p(L>g)1>2 that relates T, g, and L. a. With L held constant and g as the independent variable, calculate dT and use it to answer parts (b) and (c). b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 100cm pendulum is moved from a location where g = 980 cm>sec2 to a new location. This increases the period by dT = 0.001 sec. Find dg and estimate the value of g at the new location. 65. Quadratic approximations a. Let Q(x) = b0 + b1(x  a) + b2(x  a)2 be a quadratic approximation to ƒ(x) at x = a with the properties: i) Q(a) = ƒ(a) ii) Q′(a) = ƒ′(a)
T c. Graph ƒ(x) = 1>(1  x) and its quadratic approximation at x = 0. Then zoom in on the two graphs at the point (0, 1). Comment on what you see. T d. Find the quadratic approximation to g(x) = 1>x at x = 1. Graph g and its quadratic approximation together. Comment on what you see. T e. Find the quadratic approximation to h(x) = 21 + x at x = 0. Graph h and its quadratic approximation together. Comment on what you see. f. What are the linearizations of ƒ, g, and h at the respective points in parts (b), (d), and (e)? 66. The linearization is the best linear approximation Suppose that y = ƒ(x) is differentiable at x = a and that g(x) = m(x  a) + c is a linear function in which m and c are constants. If the error E(x) = ƒ(x)  g(x) were small enough near x = a, we might think of using g as a linear approximation of ƒ instead of the linearization L(x) = ƒ(a) + ƒ′(a)(x  a). Show that if we impose on g the conditions 1. E(a) = 0
The approximation error is zero at x = a.
E(x) 2. lim x  a = 0 xSa
The error is negligible when compared with x  a.
then g(x) = ƒ(a) + ƒ′(a)(x  a). Thus, the linearization L(x) gives the only linear approximation whose error is both zero at x = a and negligible in comparison with x  a. The linearization, L(x): y = f (a) + f ′(a)(x − a)
Some other linear approximation, g(x): y = m(x − a) + c y = f (x)
(a, f (a))
a
x
67. The linearization of 2x a. Find the linearization of ƒ(x) = 2x at x = 0. Then round its coefficients to two decimal places. T b. Graph the linearization and function together for 3 … x … 3 and 1 … x … 1. 68. The linearization of log3 x a. Find the linearization of ƒ(x) = log3 x at x = 3. Then round its coefficients to two decimal places. T b. Graph the linearization and function together in the window 0 … x … 8 and 2 … x … 4. COMPUTER EXPLORATIONS In Exercises 69–74, use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps:
iii) Q″(a) = ƒ″(a).
a. Plot the function ƒ over I.
Determine the coefficients b0, b1, and b2.
b. Find the linearization L of the function at the point a.
b. Find the quadratic approximation to ƒ(x) = 1>(1  x) at x = 0.
213
c. Plot ƒ and L together on a single graph.
214
Chapter 3: Derivatives
d. Plot the absolute error ƒ(x)  L(x) over I and find its maximum value.
70. ƒ(x) =
e. From your graph in part (d), estimate as large a d 7 0 as you can, satisfying
71. ƒ(x) = x2>3(x  2),
0x  a0 6 d
0 ƒ(x)  L(x) 0 6 P
1
for P = 0.5, 0.1, and 0.01. Then check graphically to see if your [email protected] holds true. 69. ƒ(x) = x3 + x2  2x,
Chapter
3
3 2, 34, a = 2
30, 2p4, a = 2
30, 24, a = 1
74. ƒ(x) = 2x sin1 x, 30, 14, a =
1 2
Questions to Guide Your Review
2. What role does the derivative play in defining slopes, tangents, and rates of change? 3. How can you sometimes graph the derivative of a function when all you have is a table of the function’s values? 4. What does it mean for a function to be differentiable on an open interval? On a closed interval? 5. How are derivatives and onesided derivatives related? 6. Describe geometrically when a function typically does not have a derivative at a point. 7. How is a function’s differentiability at a point related to its continuity there, if at all? 8. What rules do you know for calculating derivatives? Give some examples. 9. Explain how the three formulas d n (x ) = nxn  1 dx
3 1 c , 1d, a = 4 2
72. ƒ(x) = 2x  sin x, 73. ƒ(x) = x2x,
3 1, 24, a = 1
1. What is the derivative of a function ƒ? How is its domain related to the domain of ƒ? Give examples.
du d (cu) = c dx dx dun du2 du1 d c. (u + u2 + g + un) = + + + dx 1 dx dx g dx enable us to differentiate any polynomial. a.
x  1 , 4x2 + 1
b.
10. What formula do we need, in addition to the three listed in Question 9, to differentiate rational functions?
18. Once you know the derivatives of sin x and cos x, how can you find the derivatives of tan x, cot x, sec x, and csc x? What are the derivatives of these functions? 19. At what points are the six basic trigonometric functions continuous? How do you know? 20. What is the rule for calculating the derivative of a composite of two differentiable functions? How is such a derivative evaluated? Give examples. 21. If u is a differentiable function of x, how do you find (d>dx)(un) if n is an integer? If n is a real number? Give examples. 22. What is implicit differentiation? When do you need it? Give examples. 23. What is the derivative of the natural logarithm function ln x? How does the domain of the derivative compare with the domain of the function? 24. What is the derivative of the exponential function ax, a 7 0 and a ≠ 1? What is the geometric significance of the limit of (ah  1)>h as h S 0? What is the limit when a is the number e? 25. What is the derivative of loga x? Are there any restrictions on a? 26. What is logarithmic differentiation? Give an example. 27. How can you write any real power of x as a power of e? Are there any restrictions on x? How does this lead to the Power Rule for differentiating arbitrary real powers?
11. What is a second derivative? A third derivative? How many derivatives do the functions you know have? Give examples.
28. What is one way of expressing the special number e as a limit? What is an approximate numerical value of e correct to 7 decimal places?
12. What is the derivative of the exponential function ex ? How does the domain of the derivative compare with the domain of the function?
29. What are the derivatives of the inverse trigonometric functions? How do the domains of the derivatives compare with the domains of the functions?
13. What is the relationship between a function’s average and instantaneous rates of change? Give an example.
30. How do related rates problems arise? Give examples.
14. How do derivatives arise in the study of motion? What can you learn about an object’s motion along a line by examining the derivatives of the object’s position function? Give examples. 15. How can derivatives arise in economics? 16. Give examples of still other applications of derivatives. 17. What do the limits limh S 0 ((sin h)>h) and limh S 0 ((cos h  1)>h) have to do with the derivatives of the sine and cosine functions? What are the derivatives of these functions?
31. Outline a strategy for solving related rates problems. Illustrate with an example. 32. What is the linearization L (x) of a function ƒ(x) at a point x = a? What is required of ƒ at a for the linearization to exist? How are linearizations used? Give examples. 33. If x moves from a to a nearby value a + dx, how do you estimate the corresponding change in the value of a differentiable function ƒ(x)? How do you estimate the relative change? The percentage change? Give an example.
Chapter 3 Practice Exercises
Chapter
3
Practice Exercises 57. y = ln cos1 x
Derivatives of Functions Find the derivatives of the functions in Exercises 1–64.
1. y = x5  0.125x2 + 0.25x
2. y = 3  0.7x3 + 0.3x7
3. y = x3  3(x2 + p2)
4. y = x7 + 27x 
1 p + 1 1
5. y = (x + 1) (x + 2x)
6. y = (2x  5)(4  x)
7. y = (u 2 + sec u + 1)3
u2 2 csc u  b 8. y = a 1 2 4
2
2
1t 9. s = 1 + 1t
1 10. s = 1t  1
58. y = z cos1 z  21  z2 59. y = t tan1 t 
1 ln t 2
60. y = ( 1 + t 2 ) cot1 2t 61. y = z sec1 z  2z2  1, z 7 1 62. y = 2 2x  1 sec1 1x 63. y = csc1 (sec u), 0 6 u 6 p>2 64. y = ( 1 + x2 ) etan
1 2 sin2 x sin x
1
x
Implicit Differentiation In Exercises 65–78, find dy>dx by implicit differentiation.
11. y = 2 tan2 x  sec2 x
12. y =
13. s = cos4 (1  2t)
2 14. s = cot3 a t b
65. xy + 2x + 3y = 1
66. x2 + xy + y2  5x = 2
67. x3 + 4xy  3y4>3 = 2x
68. 5x4>5 + 10y6>5 = 15
15. s = (sec t + tan t)5
16. s = csc5 (1  t + 3t 2)
69. 1xy = 1
70. x2y2 = 1
17. r = 22u sin u
18. r = 2u 2cos u
71. y2 =
19. r = sin 22u
20. r = sin 1 u + 2u + 1 2
1 2 21. y = x2 csc x 2
22. y = 2 2x sin 2x
x x + 1
72. y2 =
74. y2 = 2e1>x
75. ln (x>y) = 1
76. x sin1 y = 1 + x2
1
x
= 2
78. xy = 22
24. y = 2x csc (x + 1)3
77. yetan
25. y = 5 cot x2
26. y = x2 cot 5x
In Exercises 79 and 80, find dp>dq.
28. y = x2 sin2 (x3)
79. p3 + 4pq  3q2 = 2
27. y = x sin (2x ) 29. s = a
2
2
4t 2 b t + 1
30. s =
2x b 1 + x
32. y = a
31. y = a
2
x2 + x 33. y = B x2 35. r = a
sin u b cos u  1
2
3 (5x2 + sin 2x)3>2
41. y = 10ex>5 43. y =
2 2x b 2 2x + 1
1 4x 1 4x xe e 4 16
45. y = ln (sin2 u)
36. r = a
1 + sin u b 1  cos u
81. r cos 2s + sin2 s = p 2
2
38. y = 20 (3x  4)1>4 (3x  4)1>5 40. y = (3 + cos3 3x)1>3
84. a. By differentiating dy>dx = x>y.
x2  y2 = 1
implicitly, show that
b. Then show that d 2y>dx2 =  1>y3. Numerical Values of Derivatives 85. Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.
ƒ(x)
g(x)
ƒ′(x)
g′(x)
44. y = x2e2>x
0 1
1 3
1 5
3 1>2
1>2 4
46. y = ln (sec2 u) 50. y = 92t
51. y = 5x3.6
52. y = 22x 22
53. y = (x + 2)x + 2
54. y = 2 (ln x)x>2
55. y = sin1 21  u2, 0 6 u 6 1 b, y 7 1
2 b. y2 = 1  x
x
49. y = 8t
2y
83. Find d 2y>dx2 by implicit differentiation:
42. y = 22e22x
48. y = log5 (3x  7)
56. y = sin1 a
82. 2rs  r  s + s2 =  3
a. x3 + y3 = 1
47. y = log2 (x2 >2)
1
80. q = (5p2 + 2p)3>2
In Exercises 81 and 82, find dr>ds.
34. y = 4x 2x + 1x
37. y = (2x + 1) 22x + 1 39. y =
1 15(15t  1)3
1 + x A1  x
73. ex + 2y = 1
23. y = x1>2 sec (2x)2 2
215
Find the first derivatives of the following combinations at the given value of x. a. 6ƒ(x)  g(x), x = 1 ƒ(x) , x = 1 c. g(x) + 1 e. g(ƒ(x)), x = 0 g. ƒ(x + g(x)), x = 0
b. ƒ(x)g2(x), x = 0 d. ƒ(g(x)), x = 0 f. (x + ƒ(x))3>2, x = 1
216
Chapter 3: Derivatives
86. Suppose that the function ƒ(x) and its first derivative have the following values at x = 0 and x = 1. x
ƒ(x)
ƒ′(x)
0 1
9 3
2 1>5
a. 1x ƒ(x), x = 1
Give reasons for your answers.
b. 2ƒ(x), x = 0
87. Find the value of dy>dt at t = 0 if y = 3 sin 2x and x = t + p. 2
88. Find the value of ds>du at u = 2 if s = t 2 + 5t and t = (u2 + 2u)1>3. 89. Find the value of dw>ds at s = 0 if w = sin 1 e r = 3 sin (s + p>6).
1r
90. Find the value of dr>dt at t = 0 if r = (u + 7) u 2t + u = 1. 2
2
1>3
92. If x1>3 + y1>3 = 4, find d 2y>dx2 at the point (8, 8). Applying the Derivative Definition In Exercises 93 and 94, find the derivative using the definition.
93. ƒ(t) =
1 2t + 1
94. g(x) = 2x2 + 1 95. a. Graph the function ƒ(x) = e
x 2,  1 … x 6 0 0 … x … 1.  x 2,
b. Is ƒ continuous at x = 0? c. Is ƒ differentiable at x = 0? Give reasons for your answers. 96. a. Graph the function x, 1 … x 6 0 ƒ(x) = e tan x, 0 … x … p>4. b. Is ƒ continuous at x = 0? c. Is ƒ differentiable at x = 0? Give reasons for your answers. 97. a. Graph the function x, 0 … x … 1 ƒ(x) = e 2  x, 1 6 x … 2. b. Is ƒ continuous at x = 1? c. Is ƒ differentiable at x = 1? Give reasons for your answers.
101. Horizontal tangents Find the points on the curve y = 2x3  3x2  12x + 20 where the tangent is parallel to the xaxis.
and
102. Tangent intercepts Find the x and yintercepts of the line that is tangent to the curve y = x3 at the point ( 2,  8).
and
103. Tangents perpendicular or parallel to lines Find the points on the curve y = 2x3  3x2  12x + 20 where the tangent is
91. If y + y = 2 cos x, find the value of d y>dx at the point (0, 1). 2
Slopes, Tangents, and Normals 99. Tangents with specified slope Are there any points on the curve y = (x>2) + 1>(2x  4) where the slope is  3>2? If so, find them.
100. Tangents with specified slope Are there any points on the curve y = x  ex where the slope is 2? If so, find them.
px f. 10 sin a b ƒ2(x), x = 1 2
2
sin 2x, x … 0 mx, x 7 0
b. differentiable at x = 0?
d. ƒ(1  5 tan x), x = 0
3
ƒ(x) = e a. continuous at x = 0?
Find the first derivatives of the following combinations at the given value of x. c. ƒ1 2x 2, x = 1 ƒ(x) , x = 0 e. 2 + cos x
98. For what value or values of the constant m, if any, is
a. perpendicular to the line y = 1  (x>24). b. parallel to the line y = 22  12x. 104. Intersecting tangents Show that the tangents to the curve y = (p sin x)>x at x = p and x =  p intersect at right angles. 105. Normals parallel to a line Find the points on the curve y = tan x, p>2 6 x 6 p>2, where the normal is parallel to the line y = x>2. Sketch the curve and normals together, labeling each with its equation. 106. Tangent and normal lines Find equations for the tangent and normal to the curve y = 1 + cos x at the point (p>2, 1). Sketch the curve, tangent, and normal together, labeling each with its equation. 107. Tangent parabola The parabola y = x2 + C is to be tangent to the line y = x. Find C. 108. Slope of tangent Show that the tangent to the curve y = x3 at any point (a, a3) meets the curve again at a point where the slope is four times the slope at (a, a3). 109. Tangent curve For what value of c is the curve y = c>(x + 1) tangent to the line through the points (0, 3) and (5, 2)? 110. Normal to a circle Show that the normal line at any point of the circle x2 + y2 = a2 passes through the origin. In Exercises 111–116, find equations for the lines that are tangent and normal to the curve at the given point. 111. x2 + 2y2 = 9, (1, 2) 112. ex + y2 = 2, (0, 1) 113. xy + 2x  5y = 2, (3, 2) 114. (y  x)2 = 2x + 4, (6, 2) 115. x + 1xy = 6, (4, 1) 116. x3>2 + 2y3>2 = 17, (1, 4) 117. Find the slope of the curve x3y3 + y2 = x + y at the points (1, 1) and (1, 1).
Chapter 3 Practice Exercises
118. The graph shown suggests that the curve y = sin (x  sin x) might have horizontal tangents at the xaxis. Does it? Give reasons for your answer. y
y = sin (x − sin x)
1
−2p
−p
p
0
123. a. What is the value of the derivative of the rabbit population when the number of rabbits is largest? Smallest? b. What is the size of the rabbit population when its derivative is largest? Smallest (negative value)? 124. In what units should the slopes of the rabbit and fox population curves be measured?
x
2p
Number of rabbits
2000
Initial no. rabbits = 1000 Initial no. foxes = 40
−1
Analyzing Graphs Each of the figures in Exercises 119 and 120 shows two graphs, the graph of a function y = ƒ(x) together with the graph of its derivative ƒ′(x). Which graph is which? How do you know?
119.
120.
y
A
(20, 1700) 1000 Number of foxes
y 4
2
0
A
50
150
200
100 150 Time (days) Derivative of the rabbit population (b)
200
3
0
x
1
+100
B
1 0
1
2
x
50 (20, 40)
0
−1 B
100 Time (days) (a)
2
1
−1
217
−50 −2 −100
0
50
121. Use the following information to graph the function y = ƒ(x) for  1 … x … 6. i) The graph of ƒ is made of line segments joined end to end.
Source: NCPMF “Differentiation” by W.U. Walton et al., Project CALC. Reprinted by permission of Educational Development Center, Inc.
ii) The graph starts at the point (1, 2). iii) The derivative of ƒ, where defined, agrees with the step function shown here.
Trigonometric Limits Find the limits in Exercises 125–132.
y y = f ′(x)
125. lim
sin x 2x2  x
126. lim
3x  tan 7x 2x
127. lim
sin r tan 2r
128. lim
sin (sin u) u
xS0
1 −1
1
2
3
4
5
6
x
rS0
−1 −2
122. Repeat Exercise 121, supposing that the graph starts at (1, 0) instead of ( 1, 2). Exercises 123 and 124 are about the accompanying graphs. The graphs in part (a) show the numbers of rabbits and foxes in a small arctic population. They are plotted as functions of time for 200 days. The number of rabbits increases at first, as the rabbits reproduce. But the foxes prey on rabbits and, as the number of foxes increases, the rabbit population levels off and then drops. Part (b) shows the graph of the derivative of the rabbit population, made by plotting slopes.
129.
lim
u S (p>2)
130. lim+ uS0
131. lim
xS0
xS0
uS0
4 tan2 u + tan u + 1 tan2 u + 5
1  2 cot2 u 5 cot u  7 cot u  8 2
x sin x 2  2 cos x
132. lim
uS0
1  cos u u2
Show how to extend the functions in Exercises 133 and 134 to be continuous at the origin. 133. g(x) =
tan (tan x) tan x
134. ƒ(x) =
tan (tan x) sin (sin x)
218
Chapter 3: Derivatives
Logarithmic Differentiation In Exercises 135–140, use logarithmic differentiation to find the derivative of y with respect to the appropriate variable.
135. y =
2(x2 + 1) 2cos 2x
137. y = a
136. y =
3x + 4 A 2x  4
10
(t + 1)(t  1) 5 b , t 7 2 (t  2)(t + 3)
148. Motion of a particle A particle moves along the curve y = x3>2 in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second. Find dx>dt when x = 3. 149. Draining a tank Water drains from the conical tank shown in the accompanying figure at the rate of 5 ft3 >min.
2u2u 138. y = 2u2 + 1 139. y = (sin u)2u
147. Speed of moving particle The coordinates of a particle moving in the metric xyplane are differentiable functions of time t with dx>dt = 10 m>sec and dy>dt = 5 m>sec. How fast is the particle moving away from the origin as it passes through the point (3,  4)?
140. y = (ln x)1>(ln x)
a. What is the relation between the variables h and r in the figure? b. How fast is the water level dropping when h = 6 ft?
Related Rates 141. Right circular cylinder The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S = 2pr 2 + 2prh.
4′
a. How is dS>dt related to dr>dt if h is constant?
r
b. How is dS>dt related to dh>dt if r is constant? c. How is dS>dt related to dr>dt and dh>dt if neither r nor h is constant?
10′ h
d. How is dr>dt related to dh>dt if S is constant? 142. Right circular cone The lateral surface area S of a right circular cone is related to the base radius r and height h by the equation S = pr 2r 2 + h2. a. How is dS>dt related to dr>dt if h is constant? b. How is dS>dt related to dh>dt if r is constant? c. How is dS>dt related to dr>dt and dh>dt if neither r nor h is constant? 143. Circle’s changing area The radius of a circle is changing at the rate of  2>p m>sec. At what rate is the circle’s area changing when r = 10 m?
Exit rate: 5 ft3min
150. Rotating spool As television cable is pulled from a large spool to be strung from the telephone poles along a street, it unwinds from the spool in layers of constant radius (see accompanying figure). If the truck pulling the cable moves at a steady 6 ft > sec (a touch over 4 mph), use the equation s = r u to find how fast (radians per second) the spool is turning when the layer of radius 1.2 ft is being unwound.
144. Cube’s changing edges The volume of a cube is increasing at the rate of 1200 cm3 >min at the instant its edges are 20 cm long. At what rate are the lengths of the edges changing at that instant? 145. Resistors connected in parallel If two resistors of R1 and R2 ohms are connected in parallel in an electric circuit to make an Rohm resistor, the value of R can be found from the equation
1.2′
1 1 1 + . = R R1 R2
+ −
R1
R2 R
If R1 is decreasing at the rate of 1 ohm > sec and R2 is increasing at the rate of 0.5 ohm > sec, at what rate is R changing when R1 = 75 ohms and R2 = 50 ohms? 146. Impedance in a series circuit The impedance Z (ohms) in a series circuit is related to the resistance R (ohms) and reactance X (ohms) by the equation Z = 2R2 + X 2. If R is increasing at 3 ohms > sec and X is decreasing at 2 ohms > sec, at what rate is Z changing when R = 10 ohms and X = 20 ohms?
151. Moving searchlight beam The figure shows a boat 1 km offshore, sweeping the shore with a searchlight. The light turns at a constant rate, du>dt = 0.6 rad/sec. a. How fast is the light moving along the shore when it reaches point A? b. How many revolutions per minute is 0.6 rad>sec?
x u
A 1 km
Chapter 3 Additional and Advanced Exercises
152. Points moving on coordinate axes Points A and B move along the x and yaxes, respectively, in such a way that the distance r (meters) along the perpendicular from the origin to the line AB remains constant. How fast is OA changing, and is it increasing, or decreasing, when OB = 2r and B is moving toward O at the rate of 0.3r m > sec? Linearization 153. Find the linearizations of
a. tan x at x =  p>4
b. sec x at x = p>4.
Graph the curves and linearizations together. 154. We can obtain a useful linear approximation of the function ƒ(x) = 1>(1 + tan x) at x = 0 by combining the approximations 1 ≈ 1  x 1 + x
and
tan x ≈ x
to get
219
158. Controlling error a. How accurately should you measure the edge of a cube to be reasonably sure of calculating the cube’s surface area with an error of no more than 2%? b. Suppose that the edge is measured with the accuracy required in part (a). About how accurately can the cube’s volume be calculated from the edge measurement? To find out, estimate the percentage error in the volume calculation that might result from using the edge measurement. 159. Compounding error The circumference of the equator of a sphere is measured as 10 cm with a possible error of 0.4 cm. This measurement is used to calculate the radius. The radius is then used to calculate the surface area and volume of the sphere. Estimate the percentage errors in the calculated values of a. the radius. b. the surface area.
1 ≈ 1  x. 1 + tan x Show that this result is the standard linear approximation of 1>(1 + tan x) at x = 0. 155. Find the linearization of ƒ(x) = 21 + x + sin x  0.5 at x = 0. 156. Find the linearization of ƒ(x) = 2>(1  x) + 21 + x  3.1 at x = 0.
c. the volume. 160. Finding height To find the height of a lamppost (see accompanying figure), you stand a 6 ft pole 20 ft from the lamp and measure the length a of its shadow, finding it to be 15 ft, give or take an inch. Calculate the height of the lamppost using the value a = 15 and estimate the possible error in the result.
Differential Estimates of Change 157. Surface area of a cone Write a formula that estimates the change that occurs in the lateral surface area of a right circular cone when the height changes from h0 to h0 + dh and the radius does not change.
h
6 ft 20 ft h
a
r V = 1 pr2h 3 S = pr "r2 + h2 (Lateral surface area)
Chapter
3
Additional and Advanced Exercises
1. An equation like sin2 u + cos2 u = 1 is called an identity because it holds for all values of u. An equation like sin u = 0.5 is not an identity because it holds only for selected values of u, not all. If you differentiate both sides of a trigonometric identity in u with respect to u, the resulting new equation will also be an identity. Differentiate the following to show that the resulting equations hold for all u.
a. sin 2u = 2 sin u cos u b. cos 2u = cos2 u  sin2 u 2. If the identity sin (x + a) = sin x cos a + cos x sin a is differentiated with respect to x, is the resulting equation also an identity? Does this principle apply to the equation x2  2x  8 = 0? Explain.
220
Chapter 3: Derivatives
3. a. Find values for the constants a, b, and c that will make ƒ(x) = cos x and g(x) = a + bx + cx
2
satisfy the conditions ƒ(0) = g(0), ƒ′(0) = g′(0), and ƒ″(0) = g″(0).
8. Designing a gondola The designer of a 30ftdiameter spherical hot air balloon wants to suspend the gondola 8 ft below the bottom of the balloon with cables tangent to the surface of the balloon, as shown. Two of the cables are shown running from the top edges of the gondola to their points of tangency, ( 12,  9) and (12, 9). How wide should the gondola be?
b. Find values for b and c that will make ƒ(x) = sin (x + a) and g(x) = b sin x + c cos x satisfy the conditions
y x 2 + y 2 = 225
ƒ(0) = g(0) and ƒ′(0) = g′(0). c. For the determined values of a, b, and c, what happens for the third and fourth derivatives of ƒ and g in each of parts (a) and (b)? 4. Solutions to differential equations a. Show that y = sin x, y = cos x, and y = a cos x + b sin x (a and b constants) all satisfy the equation y″ + y = 0. b. How would you modify the functions in part (a) to satisfy the equation y″ + 4y = 0? Generalize this result. 5. An osculating circle Find the values of h, k, and a that make the circle (x  h)2 + (y  k)2 = a2 tangent to the parabola y = x2 + 1 at the point (1, 2) and that also make the second derivatives d 2y>dx2 have the same value on both curves there. Circles like this one that are tangent to a curve and have the same second derivative as the curve at the point of tangency are called osculating circles (from the Latin osculari, meaning “to kiss”). We encounter them again in Chapter 13. 6. Marginal revenue A bus will hold 60 people. The number x of people per trip who use the bus is related to the fare charged ( p dollars) by the law p = 33  (x>40) 4 2. Write an expression for the total revenue r(x) per trip received by the bus company. What number of people per trip will make the marginal revenue dr>dx equal to zero? What is the corresponding fare? (This fare is the one that maximizes the revenue.) 7. Industrial production a. Economists often use the expression “rate of growth” in relative rather than absolute terms. For example, let u = ƒ(t) be the number of people in the labor force at time t in a given industry. (We treat this function as though it were differentiable even though it is an integervalued step function.) Let y = g(t) be the average production per person in the labor force at time t. The total production is then y = uy. If the labor force is growing at the rate of 4% per year (du>dt = 0.04u) and the production per worker is growing at the rate of 5% per year (dy>dt = 0.05y), find the rate of growth of the total production, y. b. Suppose that the labor force in part (a) is decreasing at the rate of 2% per year while the production per person is increasing at the rate of 3% per year. Is the total production increasing, or is it decreasing, and at what rate?
x
0 15 ft (12, −9)
(−12, −9) Suspension cables
8 ft Gondola Width NOT TO SCALE
9. Pisa by parachute On August 5, 1988, Mike McCarthy of London jumped from the top of the Tower of Pisa. He then opened his parachute in what he said was a world record lowlevel parachute jump of 179 ft. Make a rough sketch to show the shape of the graph of his speed during the jump. (Source: Boston Globe, Aug. 6, 1988.) 10. Motion of a particle The position at time t Ú 0 of a particle moving along a coordinate line is s = 10 cos (t + p>4). a. What is the particle’s starting position (t = 0)? b. What are the points farthest to the left and right of the origin reached by the particle? c. Find the particle’s velocity and acceleration at the points in part (b). d. When does the particle first reach the origin? What are its velocity, speed, and acceleration then? 11. Shooting a paper clip On Earth, you can easily shoot a paper clip 64 ft straight up into the air with a rubber band. In t sec after firing, the paper clip is s = 64t  16t 2 ft above your hand. a. How long does it take the paper clip to reach its maximum height? With what velocity does it leave your hand? b. On the moon, the same acceleration will send the paper clip to a height of s = 64t  2.6t 2 ft in t sec. About how long will it take the paper clip to reach its maximum height, and how high will it go? 12. Velocities of two particles At time t sec, the positions of two particles on a coordinate line are s1 = 3t 3  12t 2 + 18t + 5 m and s2 =  t 3 + 9t 2  12t m. When do the particles have the same velocities? 13. Velocity of a particle A particle of constant mass m moves along the xaxis. Its velocity y and position x satisfy the equation 1 1 m (y2  y0 2) = k (x0 2  x2), 2 2
Chapter 3 Additional and Advanced Exercises
where k, y0, and x0 are constants. Show that whenever y ≠ 0, m
a. Show that if the position x of a moving point is given by a quadratic function of t, x = At 2 + Bt + C, then the average velocity over any time interval 3t1, t2 4 is equal to the instantaneous velocity at the midpoint of the time interval. b. What is the geometric significance of the result in part (a)? 15. Find all values of the constants m and b for which the function y = e
23. Is the derivative of
dy = kx. dt
14. Average and instantaneous velocity
sin x, x 6 p mx + b, x Ú p
is a. continuous at x = p.
h(x) = e
24. Suppose that a function ƒ satisfies the following conditions for all real values of x and y: i) ƒ(x + y) = ƒ(x) # ƒ(y).
ii) ƒ(x) = 1 + xg(x), where limx S 0 g(x) = 1. Show that the derivative ƒ′(x) exists at every value of x and that ƒ′(x) = ƒ(x). 25. The generalized product rule Use mathematical induction to prove that if y = u1 u2 gun is a finite product of differentiable functions, then y is differentiable on their common domain and dun dy du1 du2 = u + u1 u2 gun  1 . u + u1 u + dx dx 2 g n dx g n g dx
16. Does the function 1  cos x , x ≠ 0 x ƒ(x) = W 0, x = 0
have a derivative at x = 0? Explain. 17. a. For what values of a and b will
ax, x 6 2 ax2  bx + 3, x Ú 2
be differentiable for all values of x? b. Discuss the geometry of the resulting graph of ƒ. 18. a. For what values of a and b will ax + b, x … 1 g(x) = e 3 ax + x + 2b, x 7  1 be differentiable for all values of x? b. Discuss the geometry of the resulting graph of g. 19. Odd differentiable functions Is there anything special about the derivative of an odd differentiable function of x? Give reasons for your answer. 20. Even differentiable functions Is there anything special about the derivative of an even differentiable function of x? Give reasons for your answer. 21. Suppose that the functions ƒ and g are defined throughout an open interval containing the point x0, that ƒ is differentiable at x0, that ƒ(x0) = 0, and that g is continuous at x0. Show that the product ƒg is differentiable at x0. This process shows, for example, that although 0 x 0 is not differentiable at x = 0, the product x 0 x 0 is differentiable at x = 0. 22. (Continuation of Exercise 21.) Use the result of Exercise 21 to show that the following functions are differentiable at x = 0. a. 0 x 0 sin x
b. x2>3 sin x
d. h(x) = e
x2 sin (1>x), x ≠ 0 0, x = 0
x2 sin (1>x), x ≠ 0 0, x = 0
continuous at x = 0? How about the derivative of k(x) = xh (x)? Give reasons for your answers.
b. differentiable at x = p.
ƒ(x) = e
221
3 c. 2 x (1  cos x)
26. Leibniz’s rule for higherorder derivatives of products Leibniz’s rule for higherorder derivatives of products of differentiable functions says that a.
d 2(uy) d 2y du dy d 2u + u 2. = 2y + 2 2 dx dx dx dx dx
b.
d 3(uy) du d 2y d 2u dy d 3y d 3u + 3 = 3y + 3 2 + u 3. 3 2 dx dx dx dx dx dx dx
c.
d n(uy) d n  1u dy d nu + g = n y + n n1 dxn dx dx dx +
n(n  1) g(n  k + 1) d n  ku d ky k! dxn  k dxk
d ny + g + u n. dx The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using m m m! m! a b + a b = + . k!(m  k)! (k + 1)!(m  k  1)! k k + 1 27. The period of a clock pendulum The period T of a clock pendulum (time for one full swing and back) is given by the formula T 2 = 4p2L>g, where T is measured in seconds, g = 32.2 ft>sec2, and L, the length of the pendulum, is measured in feet. Find approximately a. the length of a clock pendulum whose period is T = 1 sec. b. the change dT in T if the pendulum in part (a) is lengthened 0.01 ft. c. the amount the clock gains or loses in a day as a result of the period’s changing by the amount dT found in part (b).
222
Chapter 3: Derivatives
28. The melting ice cube Assume that an ice cube retains its cubical shape as it melts. If we call its edge length s, its volume is V = s3 and its surface area is 6s2. We assume that V and s are differentiable functions of time t. We assume also that the cube’s volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms, dV =  k(6s2), dt
Chapter
3
The minus sign indicates that the volume is decreasing. We assume that the proportionality factor k is constant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperature, and the incidence or absence of sunlight, to name only a few.) Assume a particular set of conditions in which the cube lost 1 > 4 of its volume during the first hour, and that the volume is V0 when t = 0. How long will it take the ice cube to melt?
k 7 0.
Technology Application Projects
Mathematica/Maple Modules: Convergence of Secant Slopes to the Derivative Function You will visualize the secant line between successive points on a curve and observe what happens as the distance between them becomes small. The function, sample points, and secant lines are plotted on a single graph, while a second graph compares the slopes of the secant lines with the derivative function. Derivatives, Slopes, Tangent Lines, and Making Movies Parts I–III. You will visualize the derivative at a point, the linearization of a function, and the derivative of a function. You learn how to plot the function and selected tangents on the same graph. Part IV (Plotting Many Tangents) Part V (Making Movies). Parts IV and V of the module can be used to animate tangent lines as one moves along the graph of a function. Convergence of Secant Slopes to the Derivative Function You will visualize righthand and lefthand derivatives. Motion Along a Straight Line: Position S Velocity S Acceleration Observe dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration functions. Figures in the text can be animated.
4 Applications of Derivatives Overview One of the most important applications of the derivative is its use as a tool for
finding the optimal (best) solutions to problems. Optimization problems abound in mathematics, physical science and engineering, business and economics, and biology and medicine. For example, what are the height and diameter of the cylinder of largest volume that can be inscribed in a given sphere? What are the dimensions of the strongest rectangular wooden beam that can be cut from a cylindrical log of given diameter? Based on production costs and sales revenue, how many items should a manufacturer produce to maximize profit? How much does the trachea (windpipe) contract to expel air at the maximum speed during a cough? What is the branching angle at which blood vessels minimize the energy loss due to friction as blood flows through the branches? In this chapter we use derivatives to find extreme values of functions, to determine and analyze the shapes of graphs, and to solve equations numerically. We also introduce the idea of recovering a function from its derivative. The key to many of these applications is the Mean Value Theorem, which paves the way to integral calculus.
4.1 Extreme Values of Functions This section shows how to locate and identify extreme (maximum or minimum) values of a function from its derivative. Once we can do this, we can solve a variety of optimization problems (see Section 4.6). The domains of the functions we consider are intervals or unions of separate intervals. Definitions Let ƒ be a function with domain D. Then ƒ has an absolute maximum value on D at a point c if y
ƒ(x) … ƒ(c)
1
−
and an absolute minimum value on D at c if
y = sin x
y = cos x
ƒ(x) Ú ƒ(c) 0
p 2
p 2
for all x in D for all x in D.
x
−1
Figure 4.1 Absolute extrema for the sine and cosine functions on 3 p>2, p>24. These values can depend on the domain of a function.
Maximum and minimum values are called extreme values of the function ƒ. Absolute maxima or minima are also referred to as global maxima or minima. For example, on the closed interval 3 p>2, p>24 the function ƒ(x) = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function g(x) = sin x takes on a maximum value of 1 and a minimum value of 1 (Figure 4.1). Functions with the same defining rule or formula can have different extrema (maximum or minimum values), depending on the domain. We see this in the following example.
223
224
Chapter 4: Applications of Derivatives
Example 1 The absolute extrema of the following functions on their domains can be seen in Figure 4.2. Each function has the same defining equation, y = x2, but the domains vary. Notice that a function might not have a maximum or minimum if the domain is unbounded or fails to contain an endpoint.
Function rule
y = x2
Domain D
Absolute extrema on D
(a) y = x2 ( q, q)
No absolute maximum Absolute minimum of 0 at x = 0
(b) y = x2 3 0, 24
Absolute maximum of 4 at x = 2 Absolute minimum of 0 at x = 0
(c) y = x2 (0, 24
Absolute maximum of 4 at x = 2 No absolute minimum
(d) y = x2
No absolute extrema
(0, 2)
y = x2
y
D = (−∞, ∞)
x
2 (b) abs max and min
y = x2
y
D = (0, 2]
D = [0, 2]
2 (a) abs min only
y = x2
y
x
D = (0, 2)
2 (c) abs max only
y
x
2 (d) no max or min
x
Figure 4.2 Graphs for Example 1.
Historical Biography Daniel Bernoulli (1700–1789)
Some of the functions in Example 1 did not have a maximum or a minimum value. The following theorem asserts that a function which is continuous over (or on) a finite closed interval 3 a, b4 has an absolute maximum and an absolute minimum value on the interval. We look for these extreme values when we graph a function.
theorem 1—The Extreme Value Theorem If ƒ is continuous on a closed interval 3 a, b4 , then ƒ attains both an absolute maximum value M and an absolute minimum value m in 3 a, b4 . That is, there are numbers x1 and x2 in 3 a, b4 with ƒ(x1) = m, ƒ(x2) = M, and m … ƒ(x) … M for every other x in 3 a, b4 .
The proof of the Extreme Value Theorem requires a detailed knowledge of the real number system (see Appendix 7) and we will not give it here. Figure 4.3 illustrates possible locations for the absolute extrema of a continuous function on a closed interval 3 a, b4 . As we observed for the function y = cos x, it is possible that an absolute minimum (or absolute maximum) may occur at two or more different points of the interval. The requirements in Theorem 1 that the interval be closed and finite, and that the function be continuous, are key ingredients. Without them, the conclusion of the theorem
225
4.1 Extreme Values of Functions
(x2, M) y = f (x)
y = f (x)
M
M x1
a
x2
b
0m0
m
x
b
a
x
Maximum and minimum at endpoints
(x1, m) Maximum and minimum at interior points
y = f (x) y = f (x)
M
m
m a
x2
b
x
Maximum at interior point, minimum at endpoint
y
a
x1
b
x
Minimum at interior point, maximum at endpoint
Figure 4.3 Some possibilities for a continuous function’s maximum and minimum on a closed interval 3a, b4.
need not hold. Example 1 shows that an absolute extreme value may not exist if the interval fails to be both closed and finite. The exponential function y = ex over ( q, q) shows that neither extreme value need exist on an infinite interval. Figure 4.4 shows that the continuity requirement cannot be omitted.
No largest value
1 y=x 0≤ xx) on the interval (0, 14 . (We graphed this function in Figure 2.40.)
226
Chapter 4: Applications of Derivatives Absolute maximum No greater value of f anywhere. Also a local maximum.
Local maximum No greater value of f nearby.
Local minimum No smaller value of f nearby.
y = f (x) Absolute minimum No smaller value of f anywhere. Also a local minimum.
Local minimum No smaller value of f nearby. a
c
e
d
b
x
Figure 4.5 How to identify types of maxima and minima for a function with domain a … x … b.
An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list of all local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.
Finding Extrema The next theorem explains why we usually need to investigate only a few values to find a function’s extrema.
Local maximum value
y = f (x)
THEOREM 2—The First Derivative Theorem for Local Extreme Values If ƒ has a local maximum or minimum value at an interior point c of its domain, and if ƒ′ is defined at c, then ƒ′(c) = 0. Secant slopes ≥ 0 (never negative)
x
Secant slopes ≤ 0 (never positive)
c
x
x
Figure 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of nonpositive numbers and nonnegative numbers, is zero.
Proof To prove that ƒ′(c) is zero at a local extremum, we show first that ƒ′(c) cannot be positive and second that ƒ′(c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what ƒ′(c) must be. To begin, suppose that ƒ has a local maximum value at x = c (Figure 4.6) so that ƒ(x)  ƒ(c) … 0 for all values of x near enough to c. Since c is an interior point of ƒ’s domain, ƒ′(c) is defined by the twosided limit lim
xSc
ƒ(x)  ƒ(c) x  c .
This means that the righthand and lefthand limits both exist at x = c and equal ƒ′(c). When we examine these limits separately, we find that
ƒ′(c) = lim+
ƒ(x)  ƒ(c) … 0. Because (x x  c
 c) 7 0 and ƒ(x) … ƒ(c)
(1)
ƒ′(c) = lim
ƒ(x)  ƒ(c) Ú 0. Because (x x  c
 c) 6 0 and ƒ(x) … ƒ(c)
(2)
xSc
Similarly,
xSc
Together, Equations (1) and (2) imply ƒ′(c) = 0. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use ƒ(x) Ú ƒ(c), which reverses the inequalities in Equations (1) and (2).
4.1 Extreme Values of Functions
Theorem 2 says that a function’s first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. If we recall that all the domains we consider are intervals or unions of separate intervals, the only places where a function ƒ can possibly have an extreme value (local or global) are
y y = x3 1
−1
1
0
227
x
−1
1. interior points where ƒ′ = 0, 2. interior points where ƒ′ is undefined, 3. endpoints of the domain of ƒ.
At x = c and x = e in Fig. 4.5 At x = d in Fig. 4.5 At x = a and x = b in Fig. 4.5
The following definition helps us to summarize these results. (a)
Definition An interior point of the domain of a function ƒ where ƒ′ is zero or undefined is a critical point of ƒ.
y 1 y = x13 −1
0
1
x
−1 (b)
Figure 4.7 Critical points without extreme values. (a) y′ = 3x2 is 0 at x = 0, but y = x3 has no extremum there. (b) y′ = (1>3)x2>3 is undefined at x = 0, but y = x1>3 has no extremum there.
Thus the only domain points where a function can assume extreme values are critical points and endpoints. However, be careful not to misinterpret what is being said here. A function may have a critical point at x = c without having a local extreme value there. For instance, both of the functions y = x3 and y = x1>3 have critical points at the origin, but neither function has a local extreme value at the origin. Instead, each function has a point of inflection there (see Figure 4.7). We define and explore inflection points in Section 4.4. Most problems that ask for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can simply list these points and calculate the corresponding function values to find what the largest and smallest values are, and where they are located. Of course, if the interval is not closed or not finite (such as a 6 x 6 b or a 6 x 6 q), we have seen that absolute extrema need not exist. If an absolute maximum or minimum value does exist, it must occur at a critical point or at an included right or lefthand endpoint of the interval.
How to Find the Absolute Extrema of a Continuous Function ƒ on a Finite Closed Interval 1. Evaluate ƒ at all critical points and endpoints. 2. Take the largest and smallest of these values.
Example 2 Find the absolute maximum and minimum values of ƒ(x) = x2 on 3 2, 14 .
Solution The function is differentiable over its entire domain, so the only critical point is where ƒ′(x) = 2x = 0, namely x = 0. We need to check the function’s values at x = 0 and at the endpoints x = 2 and x = 1: Critical point value: Endpoint values:
ƒ(0) = 0 ƒ(2) = 4 ƒ(1) = 1.
The function has an absolute maximum value of 4 at x = 2 and an absolute minimum value of 0 at x = 0.
Example 3 Find the absolute maximum and minimum values of ƒ(x) = 10x (2  ln x) on the interval 3 1, e2 4 .
228
Chapter 4: Applications of Derivatives
Solution Figure 4.8 suggests that ƒ has its absolute maximum value near x = 3 and its absolute minimum value of 0 at x = e2. Let’s verify this observation. We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative is
y 30
(e, 10e)
25 20 15
(1, 20)
1 ƒ′(x) = 10(2  ln x)  10xa x b = 10(1  ln x).
10 5
(e 2, 0) 1
0
2
3
4
5
6
7
x
8
Figure 4.8 The extreme values of ƒ(x) = 10x(2  ln x) on 31, e2 4 occur at x = e and x = e2 (Example 3).
The only critical point in the domain 3 1, e2 4 is the point x = e, where ln x = 1. The values of ƒ at this one critical point and at the endpoints are Critical point value:
ƒ(e) = 10e
Endpoint values:
ƒ(1) = 10(2  ln 1) = 20 ƒ(e2) = 10e2(2  2 ln e) = 0.
We can see from this list that the function’s absolute maximum value is 10e ≈ 27.2; it occurs at the critical interior point x = e. The absolute minimum value is 0 and occurs at the right endpoint x = e2.
Example 4 Find the absolute maximum and minimum values of ƒ(x) = x2>3 on the interval 3 2, 34 . y
Solution We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative
y = x 23, −2 ≤ x ≤ 3
Local maximum 2
−1
0
1 2 3 Absolute minimum; also a local minimum
2 1>3 2 x = 3 3 3 2x
has no zeros but is undefined at the interior point x = 0. The values of ƒ at this one critical point and at the endpoints are Critical point value: ƒ(0) = 0
1 −2
ƒ′(x) =
Absolute maximum; also a local maximum
x
Endpoint values:
3 ƒ(2) = (2)2>3 = 2 4 3 ƒ(3) = (3)2>3 = 2 9.
3 We can see from this list that the function’s absolute maximum value is 2 9 ≈ 2.08, and it occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the interior point x = 0 where the graph has a cusp (Figure 4.9).
Figure 4.9 The extreme values of ƒ(x) = x2>3 on 3 2, 34 occur at x = 0 and x = 3 (Example 4).
Exercises 4.1 Finding Extrema from Graphs In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on 3a, b4 . Then explain how your answer is consistent with .Theorem 1
1. y
2.
3. y y = f (x)
y = h(x)
y = f (x) 0
a
c1
c2
b
x
y
y
y = h(x)
0
4.
0
a
c
b
x
a
c
b
x
0
a
c
b
x
4.1 Extreme Values of Functions
5. y
6.
In Exercises 15–20, sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem 1.
y
y = g(x)
229
y = g(x)
15. ƒ(x) = 0 x 0 ,  1 6 x 6 2
6 , 1 6 x 6 1 x2 + 2 x, 0 … x 6 1 17. g(x) = e x  1, 1 … x … 2 16. y =
0
a
c
b
x
0
a
c
x
b
In Exercises 7–10, find the absolute extreme values and where they occur. y
7.
y
8.
1
−2
0
x
2
y (1, 2)
2
x
2
−3
cos x,
0 6 x …
p 2
Absolute Extrema on Finite Closed Intervals In Exercises 21–40, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
21. ƒ(x) =
−1
2 x  5, 2 … x … 3 3
22. ƒ(x) =  x  4, 4 … x … 1
x
2
0 … x … 4
20. ƒ(x) = •
5
0
2x,
19. y = 3 sin x, 0 6 x 6 2p
10.
y
1 … x 6 0
x + 1, 1 … x 6 0
−1
9.
1 x,
2 x
1
−1
18. h(x) = •
23. ƒ(x) = x2  1,  1 … x … 2 24. ƒ(x) = 4  x3,  2 … x … 1
In Exercises 11–14, match the table with a graph. 11. x
12. x
ƒ′(x)
ƒ′(x)
14. x
ƒ′(x)
a does not exist b 0 c 2
a
b c
(a)
3 27. h(x) = 2x,  1 … x … 8
28. h(x) =  3x2>3,  1 … x … 1
29. g(x) = 24  x2 ,  2 … x … 1
ƒ′(x)
30. g(x) =  25  x2 ,  25 … x … 0 p 5p 31. ƒ(u) = sin u,  … u … 2 6 p p 32. ƒ(u) = tan u,  … u … 3 4 p 2p 33. g(x) = csc x, … x … 3 3 p p 34. g(x) = sec x,  … x … 3 6 35. ƒ(t) = 2  0 t 0 ,  1 … t … 3
does not exist does not exist 1.7
a b c
1 , 0.5 … x … 2 x2
1 26. F(x) =  x , 2 … x … 1
a 0 b 0 c 5
a 0 b 0 c 5 13. x
25. F(x) = 
a
b
c
36. ƒ(t) = 0 t  5 0 , 4 … t … 7
(b)
37. g(x) = xex, 1 … x … 1
38. h(x) = ln (x + 1), 0 … x … 3 1 39. ƒ(x) = x + ln x, 0.5 … x … 4 2
a
b (c)
c
a (d)
b
c
40. g(x) = ex ,  2 … x … 1
230
Chapter 4: Applications of Derivatives
In Exercises 41–44, find the function’s absolute maximum and minimum values and say where they are assumed. 41. ƒ(x) = x4>3,  1 … x … 8 42. ƒ(x) = x5>3,  1 … x … 8 43. g(u) = u 3>5,  32 … u … 1 44. h(u) = 3u 2>3,  27 … u … 8 Finding Critical Points In Exercises 45–52, determine all critical points for each function. 2
45. y = x  6x + 7 3
2
46. ƒ(x) = 6x  x
3
47. ƒ(x) = x(4  x)
48. g(x) = (x  1)2(x  3)2
2 49. y = x2 + x
50. ƒ(x) =
51. y = x2  32 2x
52. g(x) = 22x  x2
x2 x  2
Finding Extreme Values In Exercises 53–68, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
53. y = 2x2  8x + 9
54. y = x3  2x + 4
55. y = x3 + x2  8x + 5
56. y = x3(x  5)2
57. y = 2x2  1
58. y = x  4 2x
1 3 21  x2 x 61. y = 2 x + 1 59. y =
60. y = 23 + 2x  x2 x + 1 62. y = 2 x + 2x + 2
Theory and Examples 79. A minimum with no derivative The function ƒ(x) = 0 x 0 has an absolute minimum value at x = 0 even though ƒ is not differentiable at x = 0. Is this consistent with Theorem 2? Give reasons for your answer.
80. Even functions If an even function ƒ(x) has a local maximum value at x = c, can anything be said about the value of ƒ at x = c? Give reasons for your answer. 81. Odd functions If an odd function g(x) has a local minimum value at x = c, can anything be said about the value of g at x = c? Give reasons for your answer. 82. No critical points or endpoints exist We know how to find the extreme values of a continuous function ƒ(x) by investigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasons for your answers. 83. The function V(x) = x(10  2x)(16  2x),
0 6 x 6 5,
models the volume of a box. a. Find the extreme values of V. b. Interpret any values found in part (a) in terms of the volume of the box. 84. Cubic functions Consider the cubic function ƒ(x) = ax3 + bx2 + cx + d.
63. y = ex + ex
64. y = ex  ex
65. y = x ln x
66. y = x2 ln x
a. Show that ƒ can have 0, 1, or 2 critical points. Give examples and graphs to support your argument.
67. y = cos1 (x2)
68. y = sin1(ex)
b. How many local extreme values can ƒ have?
Local Extrema and Critical Points In Exercises 69–76, find the critical points, domain endpoints, and extreme values (absolute and local) for each function.
69. y = x2>3(x + 2)
70. y = x2>3(x2  4)
71. y = x 24  x2
72. y = x2 23  x
73. y = e
74. y = e
75. y = e
4  2x, x … 1 x + 1, x 7 1
3  x, x 6 0 3 + 2x  x2, x Ú 0
x2  2x + 4, x … 1  x2 + 6x  4, x 7 1
15 1 1 , x … 1  x2  x + 2 4 76. y = c 4 x3  6x2 + 8x, x 7 1 In Exercises 77 and 78, give reasons for your answers. 77. Let ƒ(x) = (x  2)2>3. a. Does ƒ′(2) exist? b. Show that the only local extreme value of ƒ occurs at x = 2. c. Does the result in part (b) contradict the Extreme Value Theorem? d. Repeat parts (a) and (b) for ƒ(x) = (x  a)2>3, replacing 2 by a.
85. Maximum height of a vertically moving body The height of a body moving vertically is given by 1 s =  gt 2 + y0 t + s0, 2
g 7 0,
with s in meters and t in seconds. Find the body’s maximum height. 86. Peak alternating current Suppose that at any given time t (in seconds) the current i (in amperes) in an alternating current circuit is i = 2 cos t + 2 sin t. What is the peak current for this circuit (largest magnitude)? T Graph the functions in Exercises 87–90. Then find the extreme values of the function on the interval and say where they occur. 87. ƒ(x) = 0 x  2 0 + 0 x + 3 0 ,  5 … x … 5
88. g(x) = 0 x  1 0  0 x  5 0 ,  2 … x … 7 89. h(x) = 0 x + 2 0  0 x  3 0 ,  q 6 x 6 q
90. k(x) = 0 x + 1 0 + 0 x  3 0 ,  q 6 x 6 q
Computer Explorations In Exercises 91–98, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps.
a. Plot the function over the interval to see its general behavior there.
Does ƒ′(3) exist? a. Does ƒ′(0) exist? b.
b. Find the interior points where ƒ′ = 0. (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot ƒ′ as well.
Determine all extrema of ƒ. c. Does ƒ′( 3) exist? d.
c. Find the interior points where ƒ′ does not exist.
78. Let ƒ(x) = 0 x3  9x 0 .
4.2 The Mean Value Theorem
d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval.
94. ƒ(x) = 2 + 2x  3x2>3,
e. Find the function’s absolute extreme values on the interval and identify where they occur.
3>4
91. ƒ(x) = x4  8x2 + 4x + 2, 4
3
92. ƒ(x) =  x + 4x  4x + 1, 93. ƒ(x) = x2>3(3  x),
3 2, 24
3 20>25, 64>254 3 3>4, 34
231
3 1, 10>34
30, 2p4 1 96. ƒ(x) = x  sin x + , 30, 2p4 2 30, 54 97. ƒ(x) = px2e  3x>2, 95. ƒ(x) = 2x + cos x,
98. ƒ(x) = ln (2x + x sin x),
31, 154
4.2 The Mean Value Theorem We know that constant functions have zero derivatives, but could there be a more complicated function whose derivative is always zero? If two functions have identical derivatives over an interval, how are the functions related? We answer these and other questions in this chapter by applying the Mean Value Theorem. First we introduce a special case, known as Rolle’s Theorem, which is used to prove the Mean Value Theorem.
y f ′(c) = 0
Rolle’s Theorem
y = f (x)
0
a
c
As suggested by its graph, if a differentiable function crosses a horizontal line at two different points, there is at least one point between them where the tangent to the graph is horizontal and the derivative is zero (Figure 4.10). We now state and prove this result.
x
b
(a)
THEOREM 3—Rolle’s Theorem Suppose that y = ƒ(x) is continuous over the closed interval 3 a, b4 and differentiable at every point of its interior (a, b). If ƒ(a) = ƒ(b), then there is at least one number c in (a, b) at which ƒ′(c) = 0.
y f ′(c1 ) = 0
0
a
c1
f ′(c3 ) = 0 f ′(c2 ) = 0
c2
y = f (x)
c3
b
x
(b)
Figure 4.10 Rolle’s Theorem says that a differentiable curve has at least one horizontal tangent between any two points where it crosses a horizontal line. It may have just one (a), or it may have more (b).
Historical Biography Michel Rolle (1652–1719)
Proof Being continuous, ƒ assumes absolute maximum and minimum values on 3 a, b4 by Theorem 1. These can occur only
1. at interior points where ƒ′ is zero, 2. at interior points where ƒ′ does not exist, 3. at endpoints of the function’s domain, in this case a and b.
By hypothesis, ƒ has a derivative at every interior point. That rules out possibility (2), leaving us with interior points where ƒ′ = 0 and with the two endpoints a and b. If either the maximum or the minimum occurs at a point c between a and b, then ƒ′(c) = 0 by Theorem 2 in Section 4.1, and we have found a point for Rolle’s Theorem. If both the absolute maximum and the absolute minimum occur at the endpoints, then because ƒ(a) = ƒ(b) it must be the case that ƒ is a constant function with ƒ(x) = ƒ(a) = ƒ(b) for every x∊ 3 a, b4 . Therefore ƒ′(x) = 0 and the point c can be taken anywhere in the interior (a, b). The hypotheses of Theorem 3 are essential. If they fail at even one point, the graph may not have a horizontal tangent (Figure 4.11). Rolle’s Theorem may be combined with the Intermediate Value Theorem to show when there is only one real solution of an equation ƒ(x) = 0, as we illustrate in the next example.
Example 1 Show that the equation x3 + 3x + 1 = 0 has exactly one real solution.
232
Chapter 4: Applications of Derivatives y
y
y
y = f (x)
a
y = f (x)
b
(a) Discontinuous at an endpoint of [a, b]
y
x
a
x0 b
y = f(x)
x
(b) Discontinuous at an interior point of [a, b]
(1, 5)
a
x0
b
x
(c) Continuous on [a, b] but not differentiable at an interior point
Figure 4.11 There may be no horizontal tangent if the hypotheses of Rolle’s Theorem do not hold.
0
−1
Solution We define the continuous function
y = x 3 + 3x + 1
1
1
ƒ(x) = x3 + 3x + 1.
x
(−1, −3)
Figure 4.12 The only real zero of the polynomial y = x3 + 3x + 1 is the one shown here where the curve crosses the xaxis between  1 and 0 (Example 1).
Since ƒ(1) = 3 and ƒ(0) = 1, the Intermediate Value Theorem tells us that the graph of ƒ crosses the xaxis somewhere in the open interval (1, 0). (See Figure 4.12.) Now, if there were even two points x = a and x = b where ƒ(x) was zero, Rolle’s Theorem would guarantee the existence of a point x = c in between them where ƒ′ was zero. However, the derivative ƒ′(x) = 3x2 + 3 is never zero (because it is always positive). Therefore, ƒ has no more than one zero. Our main use of Rolle’s Theorem is in proving the Mean Value Theorem.
The Mean Value Theorem The Mean Value Theorem, which was first stated by JosephLouis Lagrange, is a slanted version of Rolle’s Theorem (Figure 4.13). The Mean Value Theorem guarantees that there is a point where the tangent line is parallel to the secant joining A and B. Tangent parallel to secant
y Slope f ′(c)
B Slope
THEOREM 4—The Mean Value Theorem Suppose y = ƒ(x) is continuous over a closed interval 3 a, b4 and differentiable on the interval’s interior (a, b). Then there is at least one point c in (a, b) at which
f (b) − f (a) b−a
A 0
a y = f(x)
c
b
x
Figure 4.13 Geometrically, the Mean Value Theorem says that somewhere between a and b the curve has at least one tangent parallel to the secant joining A and B.
ƒ(b)  ƒ(a) = ƒ′(c). (1) b  a
Proof We picture the graph of ƒ and draw a line through the points A(a, ƒ(a)) and B(b, ƒ(b)). (See Figure 4.14.) The secant line is the graph of the function
g(x) = ƒ(a) +
ƒ(b)  ƒ(a) (x  a) (2) b  a
(pointslope equation). The vertical difference between the graphs of ƒ and g at x is h(x) = ƒ(x)  g(x) Historical Biography JosephLouis Lagrange (1736–1813)
= ƒ(x)  ƒ(a) 
ƒ(b)  ƒ(a) (x  a). (3) b  a
Figure 4.15 shows the graphs of ƒ, g, and h together.
4.2 The Mean Value Theorem
y = f (x)
B(b, f (b))
233
B
y = f(x) h(x)
A(a, f (a))
A
y = g(x)
h(x) = f (x) − g(x)
a
y
0
x
1
Figure 4.16 The function ƒ(x) = 21  x2 satisfies the hypotheses (and conclusion) of the Mean Value Theorem on 3 1, 14 even though ƒ is not differentiable at  1 and 1. y B(2, 4)
3
y = x2
2
1
1
x
2
Figure 4.17 As we find in Example 2, c = 1 is where the tangent is parallel to the secant line. s
Distance (ft)
400
s = f (t) (8, 352)
320 240 160 80 0
x
b
x
Figure 4.15 The secant AB is the graph of the function g(x). The function h(x) = ƒ(x)  g(x) gives the vertical distance between the graphs of ƒ and g at x.
ƒ(b) b ƒ(b) h′(c) = ƒ′(c) b ƒ(b) 0 = ƒ′(c) b ƒ(b)  ƒ(a) ƒ′(c) = , b  a h′(x) = ƒ′(x) 
ƒ(a) a ƒ(a) a ƒ(a) a
Derivative of Eq. (3) . . . . . . with x = c h′(c) = 0 Rearranged
which is what we set out to prove.
(1, 1)
A(0, 0)
a
The function h satisfies the hypotheses of Rolle’s Theorem on 3 a, b4 . It is continuous on 3 a, b4 and differentiable on (a, b) because both ƒ and g are. Also, h(a) = h(b) = 0 because the graphs of ƒ and g both pass through A and B. Therefore h′(c) = 0 at some point c∊(a, b). This is the point we want for Equation (1) in the theorem. To verify Equation (1), we differentiate both sides of Equation (3) with respect to x and then set x = c:
4
x
Figure 4.14 The graph of ƒ and the secant AB over the interval 3a, b4 .
y = "1 − x 2, −1 ≤ x ≤ 1 1
−1
b
At this point, the car’s speed was 30 mph. t
5 Time (sec)
Figure 4.18 Distance versus elapsed time for the car in Example 3.
The hypotheses of the Mean Value Theorem do not require ƒ to be differentiable at either a or b. Onesided continuity at a and b is enough (Figure 4.16).
Example 2 The function ƒ(x) = x2 (Figure 4.17) is continuous for 0 … x … 2 and
differentiable for 0 6 x 6 2. Since ƒ(0) = 0 and ƒ(2) = 4, the Mean Value Theorem says that at some point c in the interval, the derivative ƒ′(x) = 2x must have the value (4  0)>(2  0) = 2. In this case we can identify c by solving the equation 2c = 2 to get c = 1. However, it is not always easy to find c algebraically, even though we know it always exists.
A Physical Interpretation We can think of the number (ƒ(b)  ƒ(a))>(b  a) as the average change in ƒ over 3 a, b4 and ƒ′(c) as an instantaneous change. Then the Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.
Example 3 If a car accelerating from zero takes 8 sec to go 352 ft, its average velocity for the 8sec interval is 352>8 = 44 ft>sec. The Mean Value Theorem says that at some point during the acceleration the speedometer must read exactly 30 mph (44 ft>sec) (Figure 4.18).
234
Chapter 4: Applications of Derivatives
Mathematical Consequences At the beginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollary of the Mean Value Theorem provides the answer that only constant functions have zero derivatives.
Corollary 1 If ƒ′(x) = 0 at each point x of an open interval (a, b), then ƒ(x) = C for all x∊(a, b), where C is a constant.
Proof We want to show that ƒ has a constant value on the interval (a, b). We do so by showing that if x1 and x2 are any two points in (a, b) with x1 6 x2 , then ƒ(x1) = ƒ(x2). Now ƒ satisfies the hypotheses of the Mean Value Theorem on 3 x1, x2 4 : It is differentiable at every point of 3 x1, x2 4 and hence continuous at every point as well. Therefore, ƒ(x2)  ƒ(x1) = ƒ′(c) x2  x1
at some point c between x1 and x2. Since ƒ′ = 0 throughout (a, b), this equation implies successively that
ƒ(x2)  ƒ(x1) = 0, x2  x1
ƒ(x2)  ƒ(x1) = 0,
and
ƒ(x1) = ƒ(x2).
At the beginning of this section, we also asked about the relationship between two functions that have identical derivatives over an interval. The next corollary tells us that their values on the interval have a constant difference.
y
y = x2 + C
Corollary 2 If ƒ′(x) = g′(x) at each point x in an open interval (a, b), then there exists a constant C such that ƒ(x) = g(x) + C for all x∊(a, b). That is, ƒ  g is a constant function on (a, b).
C=2 C=1 C=0
2
C = −1
Proof At each point x∊(a, b) the derivative of the difference function h = ƒ  g is
C = −2
h′(x) = ƒ′(x)  g′(x) = 0.
1 0
x
−1 −2
Figure 4.19 From a geometric point of view, Corollary 2 of the Mean Value Theorem says that the graphs of functions with identical derivatives on an interval can differ only by a vertical shift there. The graphs of the functions with derivative 2x are the parabolas y = x2 + C, shown here for selected values of C.
Thus, h(x) = C on (a, b) by Corollary 1. That is, ƒ(x)  g(x) = C on (a, b), so ƒ(x) = g(x) + C. Corollaries 1 and 2 are also true if the open interval (a, b) fails to be finite. That is, they remain true if the interval is (a, q), ( q, b), or ( q, q). Corollary 2 plays an important role when we discuss antiderivatives in Section 4.8. It tells us, for instance, that since the derivative of ƒ(x) = x2 on ( q, q) is 2x, any other function with derivative 2x on ( q, q) must have the formula x2 + C for some value of C (Figure 4.19).
Example 4 Find the function ƒ(x) whose derivative is sin x and whose graph passes through the point (0, 2). Solution Since the derivative of g(x) = cos x is g′(x) = sin x, we see that ƒ and g have the same derivative. Corollary 2 then says that ƒ(x) = cos x + C for some
4.2 The Mean Value Theorem
235
constant C. Since the graph of ƒ passes through the point (0, 2), the value of C is determined from the condition that ƒ(0) = 2: ƒ(0) = cos (0) + C = 2,
C = 3.
so
The function is ƒ(x) = cos x + 3.
Finding Velocity and Position from Acceleration We can use Corollary 2 to find the velocity and position functions of an object moving along a vertical line. Assume the object or body is falling freely from rest with acceleration 9.8 m>sec2. We assume the position s(t) of the body is measured positive downward from the rest position (so the vertical coordinate line points downward, in the direction of the motion, with the rest position at 0). We know that the velocity y(t) is some function whose derivative is 9.8. We also know that the derivative of g(t) = 9.8t is 9.8. By Corollary 2, y(t) = 9.8t + C for some constant C. Since the body falls from rest, y(0) = 0. Thus 9.8(0) + C = 0,
and
C = 0.
The velocity function must be y(t) = 9.8t. What about the position function s(t)? We know that s(t) is some function whose derivative is 9.8t. We also know that the derivative of ƒ(t) = 4.9t 2 is 9.8t. By Corollary 2, s(t) = 4.9t 2 + C for some constant C. Since s(0) = 0, 4.9(0)2 + C = 0,
and
C = 0.
The position function is s(t) = 4.9t 2 until the body hits the ground. The ability to find functions from their rates of change is one of the very powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.
Proofs of the Laws of Logarithms The algebraic properties of logarithms were stated in Section 1.6. We can prove those properties by applying Corollary 2 of the Mean Value Theorem to each of them. The steps in the proofs are similar to those used in solving problems involving logarithms.
Proof that ln bx = ln b + ln x The argument starts by observing that ln bx and ln x have the same derivative: d b d 1 ln (bx) = = x = ln x. dx bx dx According to Corollary 2 of the Mean Value Theorem, then, the functions must differ by a constant, which means that ln bx = ln x + C for some C. Since this last equation holds for all positive values of x, it must hold for x = 1. Hence, ln (b # 1) = ln 1 + C ln b = 0 + C ln 1 C = ln b.
= 0
236
Chapter 4: Applications of Derivatives
By substituting, we conclude ln bx = ln b + ln x.
Proof that ln x r = r ln x We use the samederivative argument again. For all positive values of x, d 1 d ln xr = r (xr) x dx dx =
Chain Rule
1 r1 rx xr
Derivative Power Rule
d 1 = r#x = (r ln x). dx Since ln xr and r ln x have the same derivative, ln xr = r ln x + C for some constant C. Taking x to be 1 identifies C as zero, and we’re done. You are asked to prove the Quotient Rule for logarithms, b ln a x b = ln b  ln x,
in Exercise 75. The Reciprocal Rule, ln (1>x) = ln x, is a special case of the Quotient Rule, obtained by taking b = 1 and noting that ln 1 = 0.
Laws of Exponents The laws of exponents for the natural exponential ex are consequences of the algebraic properties of ln x. They follow from the inverse relationship between these functions.
Laws of Exponents for ex For all numbers x, x1, and x2, the natural exponential ex obeys the following laws: 1. ex1 # ex2 = ex1 + x2 3.
2. ex =
1 ex
ex1 = ex1  x2 4. (ex1)x2 = ex1x2 = (ex2)x1 ex2
Proof of Law 1 Let
y1 = ex1
and
y2 = ex2. (4)
Then x1 = ln y1 and x2 = ln y2 Take logs of both sides of Eqs. (4). x1 + x2 = ln y1 + ln y2 = ln y1 y2 Product Rule for logarithms x1 + x2 ln y1 y2 Exponentiate. e = e = y1 y2 eln u = u = ex1ex2. The proof of Law 4 is similar. Laws 2 and 3 follow from Law 1 (Exercises 77 and 78).
4.2 The Mean Value Theorem
237
Exercises 4.2 Checking the Mean Value Theorem Find the value or values of c that satisfy the equation
ƒ(b)  ƒ(a) = ƒ′(c) b  a in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–8. 1. ƒ(x) = x2 + 2x  1, 2. ƒ(x) = x2>3,
30, 14
1 3. ƒ(x) = x + x ,
4. ƒ(x) = 2x  1, 1
5. ƒ(x) = sin
x,
7. ƒ(x) = x3  x2,
10. ƒ(x) = x ,
iv) y = x3  33x2 + 216x = x(x  9)(x  24)
3 1, 24
19. Show that if ƒ″ 7 0 throughout an interval 3a, b4 , then ƒ′ has at most one zero in 3a, b4 . What if ƒ″ 6 0 throughout 3a, b4 instead?
3 1, 14
32, 44
30, 14
11. ƒ(x) = 2x(1  x),
2
iii) y = x3  3x2 + 4 = (x + 1)(x  2)2
18. Suppose that ƒ″ is continuous on 3a, b4 and that ƒ has three zeros in the interval. Show that ƒ″ has at least one zero in (a, b). Generalize this result.
20. Show that a cubic polynomial can have at most three real zeros.
Show that the functions in Exercises 21–28 have exactly one zero in the given interval.
3 1, 84
sin x , 12. ƒ(x) = • x 0,
ii) y = x2 + 8x + 15
31, 34
Which of the functions in Exercises 9–14 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. 4>5
i) y = x2  4
nxn  1 + (n  1)an  1xn  2 + g + a1.
x 3,  2 … x … 0 0 6 x … 2 x 2,
9. ƒ(x) = x2>3,
b. Use Rolle’s Theorem to prove that between every two zeros of xn + an  1xn  1 + g + a1 x + a0 there lies a zero of
1 c , 2d 2
6. ƒ(x) = ln (x  1), 8. g(x) = e
30, 14
Roots (Zeros) 17. a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.
1 + 21 + t  3.1, ( 1, 1) 1  t u 25. r(u) = u + sin2 a b  8, ( q, q) 3
x = 0 … x 6 x x … x …
… 1 … 0 2 3
15. The function
ƒ(x) = e
4 + 7, ( q, 0) x2
24. g(t) =
p … x 6 0 2 1 0 … 2 6
22. ƒ(x) = x3 +
32, 14
23. g(t) = 2t + 21 + t  4, (0, q)
30, 14
x  x, 13. ƒ(x) = e 2 2x  3x  3, 2x  3, 1 4. ƒ(x) = e 6x  x2  7,
21. ƒ(x) = x4 + 3x + 1,
x, 0 … x 6 1 0, x = 1
is zero at x = 0 and x = 1 and differentiable on (0, 1), but its derivative on (0, 1) is never zero. How can this be? Doesn’t Rolle’s Theorem say the derivative has to be zero somewhere in (0, 1)? Give reasons for your answer. 16. For what values of a, m, and b does the function 3, ƒ(x) = c  x2 + 3x + a, mx + b,
x = 0 0 6 x 6 1 1 … x … 2
satisfy the hypotheses of the Mean Value Theorem on the interval 30, 24 ?
26. r(u) = 2u  cos2 u + 22, ( q, q)
27. r(u) = sec u 
1 + 5, (0, p>2) u3
28. r(u) = tan u  cot u  u, (0, p>2) Finding Functions from Derivatives 29. Suppose that ƒ(1) = 3 and that ƒ′(x) = 0 for all x. Must ƒ(x) = 3 for all x? Give reasons for your answer.
30. Suppose that ƒ(0) = 5 and that ƒ′(x) = 2 for all x. Must ƒ(x) = 2x + 5 for all x? Give reasons for your answer. 31. Suppose that ƒ′(x) = 2x for all x. Find ƒ(2) if a. ƒ(0) = 0 b. ƒ(1) = 0 c. ƒ(2) = 3. 32. What can be said about functions whose derivatives are constant? Give reasons for your answer. In Exercises 33–38, find all possible functions with the given derivative. 33. a. y′ = x
b. y′ = x2
c. y′ = x3
34. a. y′ = 2x
b. y′ = 2x  1
c. y′ = 3x2 + 2x  1
35. a. y′ = 
1 x2
b. y′ = 1 
1 x2
c. y′ = 5 +
1 x2
238
Chapter 4: Applications of Derivatives
36. a. y′ =
1 2 2x
37. a. y′ = sin 2t 38. a. y′ = sec2 u
b. y′ =
1 2x
c. y′ = 4x 
1 2x
t b. y′ = cos 2
c. y′ = sin 2t + cos
t 2
b. y′ = 2u
c. y′ = 2u  sec2 u
In Exercises 39–42, find the function with the given derivative whose graph passes through the point P. 39. ƒ′(x) = 2x  1, P(0, 0) 40. g′(x) =
1 + 2x, P( 1, 1) x2
3 41. ƒ′(x) = e2x, P a0, b 2
42. r′(t) = sec t tan t  1, P(0, 0) Finding Position from Velocity or Acceleration Exercises 43–46 give the velocity y = ds>dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.
43. y = 9.8t + 5, s(0) = 10 44. y = 32t  2, s(0.5) = 4 45. y = sin pt, s(0) = 0 2t 2 46. y = p cos p , s(p2) = 1 Exercises 47–50 give the acceleration a = d 2s>dt 2, initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t. 47. a = et, y(0) = 20, s(0) = 5 48. a = 9.8, y(0) =  3, s(0) = 0 49. a =  4 sin 2t, y(0) = 2, s(0) =  3 50. a =
9 3t cos p , y(0) = 0, s(0) =  1 p2
Applications 51. Temperature change It took 14 sec for a mercury thermometer to rise from  19°C to 100°C when it was taken from a freezer and placed in boiling water. Show that somewhere along the way the mercury was rising at the rate of 8.5°C>sec.
52. A trucker handed in a ticket at a toll booth showing that in 2 hours she had covered 159 mi on a toll road with speed limit 65 mph. The trucker was cited for speeding. Why? 53. Classical accounts tell us that a 170oar trireme (ancient Greek or Roman warship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme’s speed exceeded 7.5 knots (sea or nautical miles per hour). 54. A marathoner ran the 26.2mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph, assuming the initial and final speeds are zero. 55. Show that at some instant during a 2hour automobile trip the car’s speedometer reading will equal the average speed for the trip. 56. Free fall on the moon On our moon, the acceleration of gravity is 1.6 m>sec2. If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later?
Theory and Examples 57. The geometric mean of a and b The geometric mean of two positive numbers a and b is the number 2ab. Show that the value of c in the conclusion of the Mean Value Theorem for ƒ(x) = 1>x on an interval of positive numbers 3a, b4 is c = 2ab.
58. The arithmetic mean of a and b The arithmetic mean of two numbers a and b is the number (a + b)>2. Show that the value of c in the conclusion of the Mean Value Theorem for ƒ(x) = x2 on any interval 3a, b4 is c = (a + b)>2.
T 59. Graph the function
ƒ(x) = sin x sin (x + 2)  sin2 (x + 1). What does the graph do? Why does the function behave this way? Give reasons for your answers. 60. Rolle’s Theorem a. Construct a polynomial ƒ(x) that has zeros at x =  2,  1, 0, 1, and 2. b. Graph ƒ and its derivative ƒ′ together. How is what you see related to Rolle’s Theorem? c. Do g(x) = sin x and its derivative g′ illustrate the same phenomenon as ƒ and ƒ′? 61. Unique solution Assume that ƒ is continuous on 3a, b4 and differentiable on (a, b). Also assume that ƒ(a) and ƒ(b) have opposite signs and that ƒ′ ≠ 0 between a and b. Show that ƒ(x) = 0 exactly once between a and b. 62. Parallel tangents Assume that ƒ and g are differentiable on 3a, b4 and that ƒ(a) = g(a) and ƒ(b) = g(b). Show that there is at least one point between a and b where the tangents to the graphs of ƒ and g are parallel or the same line. Illustrate with a sketch. 63. Suppose that ƒ′(x) … 1 for 1 … x … 4. Show that ƒ(4) ƒ(1) … 3. 64. Suppose that 0 6 ƒ′(x) 6 1>2 for all xvalues. Show that ƒ(1) 6 ƒ(1) 6 2 + ƒ( 1). 65. Show that 0 cos x  1 0 … 0 x 0 for all xvalues. (Hint: Consider ƒ(t) = cos t on 30, x4 .)
66. Show that for any numbers a and b, the sine inequality 0 sin b sin a 0 … 0 b  a 0 is true.
67. If the graphs of two differentiable functions ƒ(x) and g(x) start at the same point in the plane and the functions have the same rate of change at every point, do the graphs have to be identical? Give reasons for your answer.
68. If 0 ƒ(w)  ƒ(x) 0 … 0 w  x 0 for all values w and x and ƒ is a differentiable function, show that 1 … ƒ′(x) … 1 for all xvalues. 69. Assume that ƒ is differentiable on a … x … b and that ƒ(b) 6 ƒ(a). Show that ƒ′ is negative at some point between a and b.
70. Let ƒ be a function defined on an interval 3a, b4 . What conditions could you place on ƒ to guarantee that min ƒ′ …
ƒ(b)  ƒ(a) … max ƒ′, b  a
where min ƒ′ and max ƒ′ refer to the minimum and maximum values of ƒ′ on 3a, b4 ? Give reasons for your answers.
4.3 Monotonic Functions and the First Derivative Test
T 71. Use the inequalities in Exercise 70 to estimate ƒ(0.1) if ƒ′(x) = 1>(1 + x4 cos x) for 0 … x … 0.1 and ƒ(0) = 1. T 72. Use the inequalities in Exercise 70 to estimate ƒ(0.1) if ƒ′(x) = 1>(1  x4) for 0 … x … 0.1 and ƒ(0) = 2. 73. Let ƒ be differentiable at every value of x and suppose that ƒ(1) = 1, that ƒ′ 6 0 on ( q, 1), and that ƒ′ 7 0 on (1, q). a. Show that ƒ(x) Ú 1 for all x. b. Must ƒ′(1) = 0? Explain. 74. Let ƒ(x) = px2 + qx + r be a quadratic function defined on a closed interval 3a, b4 . Show that there is exactly one point c in (a, b) at which ƒ satisfies the conclusion of the Mean Value Theorem.
239
75. Use the samederivative argument, as was done to prove the Product and Power Rules for logarithms, to prove the Quotient Rule property. 76. Use the samederivative argument to prove the identities a. tan1 x + cot1 x =
p p b. sec1 x + csc1 x = 2 2
77. Starting with the equation ex1ex2 = ex1 + x2, derived in the text, show that ex = 1>ex for any real number x. Then show that ex1 >ex2 = ex1  x2 for any numbers x1 and x2. 78. Show that (ex1)x2 = ex1 x2 = (ex2)x1 for any numbers x1 and x2.
4.3 Monotonic Functions and the First Derivative Test In sketching the graph of a differentiable function, it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function to identify whether local extreme values are present.
Increasing Functions and Decreasing Functions As another corollary to the Mean Value Theorem, we show that functions with positive derivatives are increasing functions and functions with negative derivatives are decreasing functions. A function that is increasing or decreasing on an interval is said to be monotonic on the interval. Corollary 3 Suppose that ƒ is continuous on 3 a, b4 and differentiable on (a, b). If ƒ′(x) 7 0 at each point x∊(a, b), then ƒ is increasing on 3 a, b4 . If ƒ′(x) 6 0 at each point x∊(a, b), then ƒ is decreasing on 3 a, b4 .
Proof Let x1 and x2 be any two points in 3 a, b4 with x1 6 x2. The Mean Value Theorem applied to ƒ on 3 x1, x2 4 says that ƒ(x2)  ƒ(x1) = ƒ′(c)(x2  x1)
for some c between x1 and x2. The sign of the righthand side of this equation is the same as the sign of ƒ ′(c) because x2  x1 is positive. Therefore, ƒ(x2) 7 ƒ(x1) if ƒ′ is positive on (a, b) and ƒ(x2) 6 ƒ(x1) if ƒ′ is negative on (a, b). Corollary 3 tells us that ƒ(x) = 2x is increasing on the interval 3 0, b4 for any b 7 0 because ƒ′(x) = 1> 2x is positive on (0, b). The derivative does not exist at x = 0, but Corollary 3 still applies. The corollary is valid for infinite as well as finite intervals, so ƒ(x) = 2x is increasing on 3 0, q). To find the intervals where a function ƒ is increasing or decreasing, we first find all of the critical points of ƒ. If a 6 b are two critical points for ƒ, and if the derivative ƒ′ is continuous but never zero on the interval (a, b), then by the Intermediate Value Theorem applied to ƒ′, the derivative must be everywhere positive on (a, b), or everywhere negative there. One way we can determine the sign of ƒ′ on (a, b) is simply by evaluating the derivative at a single point c in (a, b). If ƒ′(c) 7 0, then ƒ′(x) 7 0 for all x in (a, b) so ƒ is increasing on 3 a, b4 by Corollary 3; if ƒ′(c) 6 0, then ƒ is decreasing on 3 a, b4 . The next example illustrates how we use this procedure.
240
Chapter 4: Applications of Derivatives
EXAMPLE 1
Find the critical points of ƒ(x) = x3  12x  5 and identify the open intervals on which ƒ is increasing and on which ƒ is decreasing.
y y = x3 − 12x − 5 20
Solution The function ƒ is everywhere continuous and differentiable. The first derivative
(−2, 11) 10
−4 −3 −2 −1 0
1
2
3
x
4
is zero at x = 2 and x = 2. These critical points subdivide the domain of ƒ to create nonoverlapping open intervals ( q, 2), (2, 2), and (2, q) on which ƒ′ is either positive or negative. We determine the sign of ƒ′ by evaluating ƒ′ at a convenient point in each subinterval. The behavior of ƒ is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the graph of ƒ is given in Figure 4.20.
−10 −20
ƒ′(x) = 3x2  12 = 3(x2  4) = 3(x + 2)(x  2)
(2, −21)
FIGURE 4.20 The function ƒ(x) = x3  12x  5 is monotonic on three separate intervals (Example 1).
Interval
 q 6 x 6 2
2 6 x 6 2
2 6 x 6 q
ƒ′(3) = 15
ƒ′(0) = 12 
ƒ′(3) = 15 +
decreasing
increasing
ƒ′ evaluated
+
Sign of ƒ′
increasing
Behavior of ƒ
−3
−2
−1
0
1
2
x
3
We used “strict” lessthan inequalities to identify the intervals in the summary table for Example 1, since open intervals were specified. Corollary 3 says that we could use … inequalities as well. That is, the function ƒ in the example is increasing on  q 6 x … 2, decreasing on 2 … x … 2, and increasing on 2 … x 6 q. We do not talk about whether a function is increasing or decreasing at a single point.
HISTORICAL BIOGRAPHY Edmund Halley (1656–1742)
First Derivative Test for Local Extrema In Figure 4.21, at the points where ƒ has a minimum value, ƒ′ 6 0 immediately to the left and ƒ′ 7 0 immediately to the right. (If the point is an endpoint, there is only one side to consider.) Thus, the function is decreasing on the left of the minimum value and it is increasing on its right. Similarly, at the points where ƒ has a maximum value, ƒ′ 7 0 immediately to the left and ƒ′ 6 0 immediately to the right. Thus, the function is increasing on the left of the maximum value and decreasing on its right. In summary, at a local extreme point, the sign of ƒ′(x) changes. Absolute max f ′ undefined Local max f′ = 0 No extremum f′ = 0 f′ > 0
y = f(x)
No extremum f′ = 0
f′ > 0
f′ < 0
f′ < 0
f′ < 0 Local min
Local min f′ = 0
f′ > 0 Absolute min a
c1
c2
c3
c4
c5
b
x
FIGURE 4.21 The critical points of a function locate where it is increasing and where it is decreasing. The first derivative changes sign at a critical point where a local extremum occurs.
These observations lead to a test for the presence and nature of local extreme values of differentiable functions.
241
4.3 Monotonic Functions and the First Derivative Test
First Derivative Test for Local Extrema Suppose that c is a critical point of a continuous function ƒ, and that ƒ is differentiable at every point in some interval containing c except possibly at c itself. Moving across this interval from left to right, 1. if ƒ′ changes from negative to positive at c, then ƒ has a local minimum at c; 2. if ƒ′ changes from positive to negative at c, then ƒ has a local maximum at c; 3. if ƒ′ does not change sign at c (that is, ƒ′ is positive on both sides of c or negative on both sides), then ƒ has no local extremum at c. The test for local extrema at endpoints is similar, but there is only one side to consider in determining whether ƒ is increasing or decreasing, based on the sign of ƒ′.
Proof of the First Derivative Test Part (1). Since the sign of ƒ′ changes from negative to positive at c, there are numbers a and b such that a 6 c 6 b, ƒ′ 6 0 on (a, c), and ƒ′ 7 0 on (c, b). If x∊(a, c), then ƒ(c) 6 ƒ(x) because ƒ′ 6 0 implies that ƒ is decreasing on 3 a, c4 . If x∊(c, b), then ƒ(c) 6 ƒ(x) because ƒ′ 7 0 implies that ƒ is increasing on 3 c, b4 . Therefore, ƒ(x) Ú ƒ(c) for every x∊(a, b). By definition, ƒ has a local minimum at c. Parts (2) and (3) are proved similarly.
Example 2 Find the critical points of ƒ(x) = x1>3(x  4) = x4>3  4x1>3. Identify the open intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values. Solution The function ƒ is continuous at all x since it is the product of two continuous functions, x1>3 and (x  4). The first derivative d 4>3 4 4 (x  4x1>3) = x1>3  x2>3 3 3 dx 4(x  1) 4 2>3 = x (x  1) = 3 3x2>3 is zero at x = 1 and undefined at x = 0. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where ƒ might have an extreme value. The critical points partition the xaxis into open intervals on which ƒ′ is either positive or negative. The sign pattern of ƒ′ reveals the behavior of ƒ between and at the critical points, as summarized in the following table. ƒ′(x) =
Interval Sign of ƒ′
y 4 y = x13(x − 4) 1 1
2
3
4
x
−2 −3
0 6 x 6 1 
x 7 1 +
decreasing
decreasing
increasing
−1
2
−1 0 −1
Behavior of ƒ
x 6 0 
(1, −3)
Figure 4.22 The function ƒ(x) = x1>3(x  4) decreases when x 6 1 and increases when x 7 1 (Example 2).
0
1
x
2
Corollary 3 to the Mean Value Theorem implies that ƒ decreases on ( q, 0), decreases on (0, 1), and increases on (1, q). The First Derivative Test for Local Extrema tells us that ƒ does not have an extreme value at x = 0 (ƒ′ does not change sign) and that ƒ has a local minimum at x = 1 (ƒ′ changes from negative to positive). The value of the local minimum is ƒ(1) = 11>3(1  4) = 3. This is also an absolute minimum since ƒ is decreasing on ( q, 1) and increasing on (1, q). Figure 4.22 shows this value in relation to the function’s graph. Note that limx S 0 ƒ′(x) =  q, so the graph of ƒ has a vertical tangent at the origin.
242
Chapter 4: Applications of Derivatives
Example 3 Find the critical points of ƒ(x) = (x2  3)ex. Identify the open intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values. Solution The function ƒ is continuous and differentiable for all real numbers, so the critical points occur only at the zeros of ƒ′. Using the Derivative Product Rule, we find the derivative ƒ′(x) = (x2  3) #
d x d 2 e + (x  3) # ex dx dx = (x2  3) # ex + (2x) # ex = (x2 + 2x  3)ex.
Since ex is never zero, the first derivative is zero if and only if
y
x2 + 2x  3 = 0 (x + 3)(x  1) = 0.
y = (x 2 − 3)e x
The zeros x = 3 and x = 1 partition the xaxis into open intervals as follows.
4 3 2 1 −5 −4 −3 −2 −1 −1
1
2
3
x
Interval Sign of ƒ′ Behavior of ƒ
−2
x 6 3 +
3 6 x 6 1 
1 6 x +
increasing
decreasing
increasing
−4
−3
−3
−2
0
−1
1
2
x
3
−4
We can see from the table that there is a local maximum (about 0.299) at x = 3 and a local minimum (about 5.437) at x = 1. The local minimum value is also an absolute minimum because ƒ(x) 7 0 for 0 x 0 7 23. There is no absolute maximum. The function increases on ( q, 3) and (1, q) and decreases on (3, 1). Figure 4.23 shows the graph.
−5 −6
Figure 4.23 The graph of ƒ(x) = (x2  3)ex (Example 3).
Exercises 4.3 Analyzing Functions from Derivatives Answer the following questions about the functions whose derivatives are given in Exercises 1–14:
a. What are the critical points of ƒ? b. On what open intervals is ƒ increasing or decreasing? c. At what points, if any, does ƒ assume local maximum and minimum values? 1. ƒ′(x) = x(x  1)
2. ƒ′(x) = (x  1)(x + 2)
3. ƒ′(x) = (x  1)2(x + 2)
4. ƒ′(x) = (x  1)2(x + 2)2
5. ƒ′(x) = (x  1)ex 6. ƒ′(x) = (x  7)(x + x2(x  1) 7. ƒ′(x) = , x + 2 (x  2)(x + 8. ƒ′(x) = (x + 1)(x 9. ƒ′(x) = 1 
14. ƒ′(x) = (sin x + cos x)(sin x  cos x), 0 … x … 2p Identifying Extrema In Exercises 15–44:
a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.
1
4) , x ≠ 1, 3 3) 10. ƒ′(x) = 3 
6 2x
−3 −2 −1 −1
, x≠0
−2
y
16.
y 2
x ≠ 2
12. ƒ′(x) = x1>2(x  3)
13. ƒ′(x) = (sin x  1)(2 cos x + 1), 0 … x … 2p
15.
1)(x + 5)
4 , x ≠ 0 x2
11. ƒ′(x) = x1>3(x + 2)
2
y = f (x) 1
2
1 3
x
−3 −2 −1 −1 −2
y = f (x) 1
2
3
x
4.3 Monotonic Functions and the First Derivative Test
17.
18.
y
243
57. ƒ(x) = sin 2x, 0 … x … p
y
58. ƒ(x) = sin x  cos x, 0 … x … 2p 2
y = f (x)
1
1
−3 −2 −1 −1
2
2
y = f (x) 3
x
1 1
−3 −2 −1 −1
−2
2
3
x
−2
59. ƒ(x) = 23 cos x + sin x, p 60. ƒ(x) = 2x + tan x, 2 x x 61. ƒ(x) =  2 sin , 0 … 2 2 62. ƒ(x) =  2 cos x  cos2 x,
0 … x … 2p p 6 x 6 2 x … 2p p … x … p
2
19. g(t) =  t 2  3t + 3
20. g(t) = 3t 2 + 9t + 5
21. h(x) =  x3 + 2x2
22. h(x) = 2x3  18x
2
3
23. ƒ(u) = 3u  4u
24. ƒ(u) = 6u  u 3
25. ƒ(r) = 3r 3 + 16r
26. h(r) = (r + 7)3
27. ƒ(x) = x4  8x2 + 16
28. g(x) = x4  4x3 + 4x2
29. H(t) =
3 4 t  t6 2
30. K(t) = 15t 3  t 5
31. ƒ(x) = x  6 2x  1 2
33. g(x) = x 28  x
35. ƒ(x) =
x2  3 , x ≠ 2 x  2
37. ƒ(x) = x1>3(x + 8) 1>3
2
39. h(x) = x (x  4) 2x
41. ƒ(x) = e
x
+ e
43. ƒ(x) = x ln x
32. g(x) = 4 2x  x2 + 3 34. g(x) = x2 25  x
x3 3x + 1 38. g(x) = x2>3(x + 5)
36. ƒ(x) =
2
2>3
2
40. k(x) = x (x  4) 42. ƒ(x) = e
2x
2
44. ƒ(x) = x ln x
In Exercises 45–56: a. Identify the function’s local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? T c. Support your findings with a graphing calculator or computer grapher. 45. ƒ(x) = 2x  x2,  q 6 x … 2 46. ƒ(x) = (x + 1)2,  q 6 x … 0 47. g(x) = x2  4x + 4, 1 … x 6 q 48. g(x) =  x2  6x  9,  4 … x 6 q 49. ƒ(t) = 12t  t 3,  3 … t 6 q 50. ƒ(t) = t 3  3t 2,  q 6 t … 3 x3  2x2 + 4x, 0 … x 6 q 3 52. k(x) = x3 + 3x2 + 3x + 1,  q 6 x … 0 51. h(x) =
53. ƒ(x) = 225  x2,  5 … x … 5
54. ƒ(x) = 2x2  2x  3, 3 … x 6 q
x  2 , 0 … x 6 1 x2  1 x2 , 2 6 x … 1 56. g(x) = 4  x2 55. g(x) =
In Exercises 57–64: a. Find the local extrema of each function on the given interval, and say where they occur. T b. Graph the function and its derivative together. Comment on the behavior of ƒ in relation to the signs and values of ƒ′.
63. ƒ(x) = csc x  2 cot x, 0 6 x 6 p p p 6 x 6 64. ƒ(x) = sec2 x  2 tan x, 2 2 Theory and Examples Show that the functions in Exercises 65 and 66 have local extreme values at the given values of u, and say which kind of local extreme the function has. u 65. h(u) = 3 cos , 0 … u … 2p, at u = 0 and u = 2p 2 u 66. h(u) = 5 sin , 0 … u … p, at u = 0 and u = p 2 67. Sketch the graph of a differentiable function y = ƒ(x) through the point (1, 1) if ƒ′(1) = 0 and
a. ƒ′(x) 7 0 for x 6 1 and ƒ′(x) 6 0 for x 7 1; b. ƒ′(x) 6 0 for x 6 1 and ƒ′(x) 7 0 for x 7 1; c. ƒ′(x) 7 0 for x ≠ 1; d. ƒ′(x) 6 0 for x ≠ 1. 68. Sketch the graph of a differentiable function y = ƒ(x) that has a. a local minimum at (1, 1) and a local maximum at (3, 3); b. a local maximum at (1, 1) and a local minimum at (3, 3); c. local maxima at (1, 1) and (3, 3); d. local minima at (1, 1) and (3, 3). 69. Sketch the graph of a continuous function y = g(x) such that a. g(2) = 2, 0 6 g′ 6 1 for x 6 2, g′(x) S 1 as x S 2,  1 6 g′ 6 0 for x 7 2, and g′(x) S 1+ as x S 2+; b. g(2) = 2, g′ 6 0 for x 6 2, g′(x) S  q as x S 2, g′ 7 0 for x 7 2, and g′(x) S q as x S 2+. 70. Sketch the graph of a continuous function y = h(x) such that a. h(0) = 0, 2 … h(x) … 2 for all x, h′(x) S q as x S 0, and h′(x) S q as x S 0+; b. h(0) = 0, 2 … h(x) … 0 for all x, h′(x) S q as x S 0, and h′(x) S  q as x S 0+. 71. Discuss the extremevalue behavior of the function ƒ(x) = x sin (1>x), x ≠ 0. How many critical points does this function have? Where are they located on the xaxis? Does ƒ have an absolute minimum? An absolute maximum? (See Exercise 49 in Section 2.3.) 72. Find the open intervals on which the function ƒ(x) = ax2 + bx + c, a ≠ 0, is increasing and decreasing. Describe the reasoning behind your answer. 73. Determine the values of constants a and b so that ƒ(x) = ax2 + bx has an absolute maximum at the point (1, 2). 74. Determine the values of constants a, b, c, and d so that ƒ(x) = ax3 + bx2 + cx + d has a local maximum at the point (0, 0) and a local minimum at the point (1, 1).
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Chapter 4: Applications of Derivatives
75. Locate and identify the absolute extreme values of
79. Find the absolute maximum value of ƒ(x) = x2 ln (1>x) and say where it is assumed.
a. ln (cos x) on 3 p>4, p>34,
80. a. Prove that ex Ú 1 + x if x Ú 0.
b. cos (ln x) on 31>2, 24.
b. Use the result in part (a) to show that
76. a. Prove that ƒ(x) = x  ln x is increasing for x 7 1. b. Using part (a), show that ln x 6 x if x 7 1. 77. Find the absolute maximum and minimum values of ƒ(x) = ex  2x on 30, 14 .
78. Where does the periodic function ƒ(x) = 2esin (x>2) take on its extreme values and what are these values? y y = 2e sin (x2)
ex Ú 1 + x +
81. Show that increasing functions and decreasing functions are onetoone. That is, show that for any x1 and x2 in I, x2 ≠ x1 implies ƒ(x2) ≠ ƒ(x1). Use the results of Exercise 81 to show that the functions in Exercises 82–86 have inverses over their domains. Find a formula for dƒ 1 >dx using Theorem 3, Section 3.8. 82. ƒ(x) = (1>3)x + (5>6) 3
x
0
1 2 x. 2
84. ƒ(x) = 1  8x
83. ƒ(x) = 27x3
85. ƒ(x) = (1  x)3
86. ƒ(x) = x5>3
4.4 Concavity and Curve Sketching y
CA
VE
UP
y = x3
CO N W
0
f ′ increases x
CO NC AV E
DO
f ′ decreases
N
Figure 4.24 The graph of ƒ(x) = x3 is concave down on ( q, 0) and concave up on (0, q) (Example 1a).
We have seen how the first derivative tells us where a function is increasing, where it is decreasing, and whether a local maximum or local minimum occurs at a critical point. In this section we see that the second derivative gives us information about how the graph of a differentiable function bends or turns. With this knowledge about the first and second derivatives, coupled with our previous understanding of symmetry and asymptotic behavior studied in Sections 1.1 and 2.6, we can now draw an accurate graph of a function. By organizing all of these ideas into a coherent procedure, we give a method for sketching graphs and revealing visually the key features of functions. Identifying and knowing the locations of these features is of major importance in mathematics and its applications to science and engineering, especially in the graphical analysis and interpretation of data.
Concavity As you can see in Figure 4.24, the curve y = x3 rises as x increases, but the portions defined on the intervals ( q, 0) and (0, q) turn in different ways. As we approach the origin from the left along the curve, the curve turns to our right and falls below its tangents. The slopes of the tangents are decreasing on the interval ( q, 0). As we move away from the origin along the curve to the right, the curve turns to our left and rises above its tangents. The slopes of the tangents are increasing on the interval (0, q). This turning or bending behavior defines the concavity of the curve.
Definition The graph of a differentiable function y = ƒ(x) is (a) concave up on an open interval I if ƒ′ is increasing on I; (b) concave down on an open interval I if ƒ′ is decreasing on I.
If y = ƒ(x) has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to the first derivative function. We conclude that ƒ′ increases if ƒ″ 7 0 on I, and decreases if ƒ″ 6 0.
4.4 Concavity and Curve Sketching
y
The Second Derivative Test for Concavity Let y = ƒ(x) be twicedifferentiable on an interval I.
4
y = x2
1. If ƒ″ 7 0 on I, the graph of ƒ over I is concave up. 2. If ƒ″ 6 0 on I, the graph of ƒ over I is concave down.
CA CON
UP
3
U VE
CO NC AV E
2 P
y″ > 0 −2
−1
245
1
If y = ƒ(x) is twicedifferentiable, we will use the notations ƒ″ and y″ interchangeably when denoting the second derivative.
y″ > 0
0
1
x
2
EXAMPLE 1
FIGURE 4.25 The graph of ƒ(x) = x2 is concave up on every interval (Example 1b).
(a) The curve y = x3 (Figure 4.24) is concave down on ( q, 0) where y″ = 6x 6 0 and
concave up on (0, q) where y″ = 6x 7 0. (b) The curve y = x2 (Figure 4.25) is concave up on ( q, q) because its second derivative y″ = 2 is always positive.
EXAMPLE 2
Determine the concavity of y = 3 + sin x on 3 0, 2p4 .
Solution The first derivative of y = 3 + sin x is y′ = cos x, and the second derivative is y″ = sin x. The graph of y = 3 + sin x is concave down on (0, p), where y″ = sin x is negative. It is concave up on (p, 2p), where y″ = sin x is positive (Figure 4.26).
Points of Inflection
y 4 3
y = 3 + sin x (p, 3)
2 1 0 −1
p
2p
x
The curve y = 3 + sin x in Example 2 changes concavity at the point (p, 3). Since the first derivative y′ = cos x exists for all x, we see that the curve has a tangent line of slope 1 at the point (p, 3). This point is called a point of inflection of the curve. Notice from Figure 4.26 that the graph crosses its tangent line at this point and that the second derivative y″ = sin x has value 0 when x = p. In general, we have the following definition.
y″ = −sin x
FIGURE 4.26 Using the sign of y″ to determine the concavity of y (Example 2).
DEFINITION A point (c, ƒ(c)) where the graph of a function has a tangent line and where the concavity changes is a point of inflection.
We observed that the second derivative of ƒ(x) = 3 + sin x is equal to zero at the inflection point (p, 3). Generally, if the second derivative exists at a point of inflection (c, ƒ(c)), then ƒ″(c) = 0. This follows immediately from the Intermediate Value Theorem whenever ƒ″ is continuous over an interval containing x = c because the second derivative changes sign moving across this interval. Even if the continuity assumption is dropped, it is still true that ƒ″(c) = 0, provided the second derivative exists (although a more advanced argument is required in this noncontinuous case). Since a tangent line must exist at the point of inflection, either the first derivative ƒ′(c) exists (is finite) or the graph has a vertical tangent at the point. At a vertical tangent neither the first nor second derivative exists. In summary, we conclude the following result. At a point of inflection (c, ƒ(c)), either ƒ″(c) = 0 or ƒ″(c) fails to exist.
The next example illustrates a function having a point of inflection where the first derivative exists, but the second derivative fails to exist.
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Chapter 4: Applications of Derivatives
Example 3 The graph of ƒ(x) = x5>3 has a horizontal tangent at the origin because
y y=
2
ƒ′(x) = (5>3)x2>3 = 0 when x = 0. However, the second derivative
x53
ƒ″(x) =
1 0
−1
1 Point of inflection
−1
x
d 5 2>3 10 1>3 a x b = x 9 dx 3
fails to exist at x = 0. Nevertheless, ƒ″(x) 6 0 for x 6 0 and ƒ″(x) 7 0 for x 7 0, so the second derivative changes sign at x = 0 and there is a point of inflection at the origin. The graph is shown in Figure 4.27.
−2
Figure 4.27 The graph of ƒ(x) = x5>3 has a horizontal tangent at the origin where the concavity changes, although ƒ″ does not exist at x = 0 (Example 3).
Here is an example showing that an inflection point need not occur even though both derivatives exist and ƒ″ = 0.
Example 4 The curve y = x4 has no inflection point at x = 0 (Figure 4.28). Even
though the second derivative y″ = 12x2 is zero there, it does not change sign.
As our final illustration, we show a situation in which a point of inflection occurs at a vertical tangent to the curve where neither the first nor the second derivative exists.
y y = x4 2
Example 5 The graph of y = x1>3 has a point of inflection at the origin because the
1
second derivative is positive for x 6 0 and negative for x 7 0:
y″ = 0 −1
0
1
x
y″ =
Figure 4.28 The graph of y = x4 has no inflection point at the origin, even though y″ = 0 there (Example 4). y Point of inflection
y = x13 0
x
Figure 4.29 A point of inflection where y′ and y″ fail to exist (Example 5).
d 2 1>3 d 1 2>3 2 1 x 2 = dx a x b =  x5>3. 2 3 9 dx
However, both y′ = x2>3 >3 and y″ fail to exist at x = 0, and there is a vertical tangent there. See Figure 4.29. Caution Example 4 in Section 4.1 (Figure 4.9) shows that the function ƒ(x) = x2>3 does not have a second derivative at x = 0 and does not have a point of inflection there (there is no change in concavity at x = 0). Combined with the behavior of the function in Example 5 above, we see that when the second derivative does not exist at x = c, an inflection point may or may not occur there. So we need to be careful about interpreting functional behavior whenever first or second derivatives fail to exist at a point. At such points the graph can have vertical tangents, corners, cusps, or various discontinuities. To study the motion of an object moving along a line as a function of time, we often are interested in knowing when the object’s acceleration, given by the second derivative, is positive or negative. The points of inflection on the graph of the object’s position function reveal where the acceleration changes sign.
Example 6 A particle is moving along a horizontal coordinate line (positive to the right) with position function s(t) = 2t 3  14t 2 + 22t  5,
t Ú 0.
Find the velocity and acceleration, and describe the motion of the particle. Solution The velocity is y(t) = s′(t) = 6t 2  28t + 22 = 2(t  1)(3t  11), and the acceleration is a(t) = y′(t) = s″(t) = 12t  28 = 4(3t  7). When the function s(t) is increasing, the particle is moving to the right; when s(t) is decreasing, the particle is moving to the left. Notice that the first derivative (y = s′) is zero at the critical points t = 1 and t = 11>3.
4.4 Concavity and Curve Sketching
247
1 6 t 6 11>3 11>3 6 t Interval 0 6 t 6 1 Sign of Y = s′ +  + Behavior of s increasing decreasing increasing Particle motion right left right The particle is moving to the right in the time intervals 3 0, 1) and (11>3, q), and moving to the left in (1, 11>3). It is momentarily stationary (at rest) at t = 1 and t = 11>3. The acceleration a(t) = s″(t) = 4(3t  7) is zero when t = 7>3. Interval Sign of a = s″ Graph of s
0 6 t 6 7>3  concave down
7>3 6 t + concave up
The particle starts out moving to the right while slowing down, and then reverses and begins moving to the left at t = 1 under the influence of the leftward acceleration over the time interval 3 0, 7>3). The acceleration then changes direction at t = 7>3 but the particle continues moving leftward, while slowing down under the rightward acceleration. At t = 11>3 the particle reverses direction again: moving to the right in the same direction as the acceleration, so it is speeding up.
Second Derivative Test for Local Extrema Instead of looking for sign changes in ƒ′ at critical points, we can sometimes use the following test to determine the presence and nature of local extrema.
THEOREM 5—Second Derivative Test for Local Extrema Suppose ƒ″ is continuous on an open interval that contains x = c. 1. If ƒ′(c) = 0 and ƒ″(c) 6 0, then ƒ has a local maximum at x = c. 2. If ƒ′(c) = 0 and ƒ″(c) 7 0, then ƒ has a local minimum at x = c. 3. If ƒ′(c) = 0 and ƒ″(c) = 0, then the test fails. The function ƒ may have a local maximum, a local minimum, or neither.
f ′ = 0, f ″ < 0 1 local max
f ′ = 0, f ″ > 0 1 local min
Proof Part (1). If ƒ″(c) 6 0, then ƒ″(x) 6 0 on some open interval I containing the point c, since ƒ″ is continuous. Therefore, ƒ′ is decreasing on I. Since ƒ′(c) = 0, the sign of ƒ′ changes from positive to negative at c so ƒ has a local maximum at c by the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions y = x4, y = x4, and y = x3. For each function, the first and second derivatives are zero at x = 0. Yet the function y = x4 has a local minimum there, y = x4 has a local maximum, and y = x3 is increasing in any open interval containing x = 0 (having neither a maximum nor a minimum there). Thus the test fails. This test requires us to know ƒ″ only at c itself and not in an interval about c. This makes the test easy to apply. That’s the good news. The bad news is that the test is inconclusive if ƒ″ = 0 or if ƒ″ does not exist at x = c. When this happens, use the First Derivative Test for local extreme values.
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Chapter 4: Applications of Derivatives
Together ƒ′ and ƒ″ tell us the shape of the function’s graph—that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is turning or bending as defined by its concavity. We use this information to sketch a graph of the function that captures its key features.
Example 7 Sketch a graph of the function ƒ(x) = x4  4x3 + 10 using the following steps. (a) Identify where the extrema of ƒ occur. (b) Find the intervals on which ƒ is increasing and the intervals on which ƒ is decreasing. (c) Find where the graph of ƒ is concave up and where it is concave down. (d) Sketch the general shape of the graph for ƒ. (e) Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. Then sketch the curve. Solution The function ƒ is continuous since ƒ′(x) = 4x3  12x2 exists. The domain of ƒ is ( q, q), and the domain of ƒ′ is also ( q, q). Thus, the critical points of ƒ occur only at the zeros of ƒ′. Since ƒ′(x) = 4x3  12x2 = 4x2(x  3), the first derivative is zero at x = 0 and x = 3. We use these critical points to define intervals where ƒ is increasing or decreasing.
3 6 x Interval x 6 0 0 6 x 6 3 Sign of ƒ′   + Behavior of ƒ decreasing decreasing increasing
(a) Using the First Derivative Test for local extrema and the table above, we see that there is no extremum at x = 0 and a local minimum at x = 3. (b) Using the table above, we see that ƒ is decreasing on ( q, 04 and 3 0, 34 , and increasing on 3 3, q). (c) ƒ″(x) = 12x2  24x = 12x(x  2) is zero at x = 0 and x = 2. We use these points to define intervals where ƒ is concave up or concave down.
Interval Sign of ƒ″ Behavior of ƒ
x 6 0 + concave up
0 6 x 6 2  concave down
2 6 x + concave up
We see that ƒ is concave up on the intervals ( q, 0) and (2, q), and concave down on (0, 2). (d) Summarizing the information in the last two tables, we obtain the following. x * 0
0 * x * 2
2 * x * 3
3 * x
decreasing decreasing decreasing increasing concave up concave down concave up concave up
4.4 Concavity and Curve Sketching
The general shape of the curve is shown in the accompanying figure.
y y = x 4 − 4x 3 + 10 20 15 (0, 10)
decr
decr
decr
incr
conc up
conc down
conc up
conc up
Inflection 10 point 5 −1
0 −5 −10
1 Inflection point
2
3
4
(2, −6)
−15 −20
249
(3, −17) Local minimum
FIGURE 4.30 The graph of ƒ(x) = x4  4x3 + 10 (Example 7).
x
0
2
3
infl point
infl point
local min
General shape
(e) Plot the curve’s intercepts (if possible) and the points where y′ and y″ are zero. Indicate any local extreme values and inflection points. Use the general shape as a guide to sketch the curve. (Plot additional points as needed.) Figure 4.30 shows the graph of ƒ. The steps in Example 7 give a procedure for graphing the key features of a function. Asymptotes were defined and discussed in Section 2.6. We can find them for rational functions, and the methods in the next section give tools to help find them for more general functions. Procedure for Graphing y = ƒ(x) 1. Identify the domain of ƒ and any symmetries the curve may have. 2. Find the derivatives y′ and y″. 3. Find the critical points of ƒ, if any, and identify the function’s behavior at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if any occur, and determine the concavity of the curve. 6. Identify any asymptotes that may exist. 7. Plot key points, such as the intercepts and the points found in Steps 3–5, and sketch the curve together with any asymptotes that exist.
EXAMPLE 8
Sketch the graph of ƒ(x) =
(x + 1)2 . 1 + x2
Solution 1. The domain of ƒ is ( q, q) and there are no symmetries about either axis or the origin (Section 1.1). 2. Find ƒ′ and ƒ″. (x + 1)2 1 + x2 (1 + x2) # 2(x + 1)  (x + 1)2 # 2x ƒ′(x) = (1 + x2)2 2 2(1  x ) = (1 + x2)2 ƒ(x) =
ƒ″(x) = =
[email protected] at x =  1, [email protected] (y = 1) at x = 0
Critical points: x =  1, x = 1
(1 + x2)2 # 2( 2x)  2(1  x2)3 2(1 + x2) # 2x4 (1 + x2)4 4x(x2  3) (1 + x2)3
After some algebra
3. Behavior at critical points. The critical points occur only at x = {1 where ƒ′(x) = 0 (Step 2) since ƒ′ exists everywhere over the domain of ƒ. At x = 1, ƒ″(1) = 1 7 0, yielding a relative minimum by the Second Derivative Test. At x = 1, f″(1) = 1 6 0, yielding a relative maximum by the Second Derivative test.
250
Chapter 4: Applications of Derivatives
4. Increasing and decreasing. We see that on the interval ( q, 1) the derivative ƒ′(x) 6 0, and the curve is decreasing. On the interval (1, 1), ƒ′(x) 7 0 and the curve is increasing; it is decreasing on (1, q) where ƒ′(x) 6 0 again. 5. Inflection points. Notice that the denominator of the second derivative (Step 2) is always positive. The second derivative ƒ″ is zero when x =  23, 0, and 23. The second derivative changes sign at each of these points: negative on 1  q,  23 2, positive on 1  23, 0 2, negative on 1 0, 23 2, and positive again on 1 23, q 2. Thus each point is a point of inflection. The curve is concave down on the interval 1  q,  23 2, concave up on 1  23, 0 2, concave down on 1 0, 23 2, and concave up again on 1 23, q 2. 6. Asymptotes. Expanding the numerator of ƒ(x) and then dividing both numerator and denominator by x2 gives (x + 1)2 x2 + 2x + 1 = Expanding numerator 1 + x2 1 + x2 1 + (2>x) + (1>x2) . Dividing by x2 = (1>x2) + 1
ƒ(x) = y 2
(1, 2)
Point of inflection where x = " 3 y=1
1 Horizontal asymptote −1 Point of inflection where x = − " 3
1
Figure 4.31 The graph of y = (Example 8).
x
(x + 1)2 1 + x2
We see that ƒ(x) S 1+ as x S q and that ƒ(x) S 1 as x S  q. Thus, the line y = 1 is a horizontal asymptote. Since ƒ decreases on ( q, 1) and then increases on (1, 1), we know that ƒ(1) = 0 is a local minimum. Although ƒ decreases on (1, q), it never crosses the horizontal asymptote y = 1 on that interval (it approaches the asymptote from above). So the graph never becomes negative, and ƒ(1) = 0 is an absolute minimum as well. Likewise, ƒ(1) = 2 is an absolute maximum because the graph never crosses the asymptote y = 1 on the interval ( q, 1), approaching it from below. Therefore, there are no vertical asymptotes (the range of ƒ is 0 … y … 2). 7. The graph of ƒ is sketched in Figure 4.31. Notice how the graph is concave down as it approaches the horizontal asymptote y = 1 as x S  q, and concave up in its approach to y = 1 as x S q.
Example 9 Sketch the graph of ƒ(x) =
x2 + 4 . 2x
Solution 1. The domain of ƒ is all nonzero real numbers. There are no intercepts because neither x nor ƒ(x) can be zero. Since ƒ(x) = ƒ(x), we note that ƒ is an odd function, so the graph of ƒ is symmetric about the origin. 2. We calculate the derivatives of the function, but first rewrite it in order to simplify our computations: ƒ(x) =
x2 + 4 x 2 = + x Function simplified for differentiation 2x 2
x2  4 1 2 Combine fractions to solve easily ƒ′(x) =  2 = 2 x 2x2 4 ƒ″(x) = 3 Exists throughout the entire domain of ƒ x ƒ′(x) =
0.
3. The critical points occur at x = {2 where ƒ′(x) = 0. Since ƒ″(2) 6 0 and ƒ″(2) 7 0, we see from the Second Derivative Test that a relative maximum occurs at x = 2 with ƒ(2) = 2, and a relative minimum occurs at x = 2 with ƒ(2) = 2.
4.4 Concavity and Curve Sketching
y 4 2 −4
−2 (−2, −2)
4. On the interval ( q, 2) the derivative ƒ′ is positive because x2  4 7 0 so the graph is increasing; on the interval (2, 0) the derivative is negative and the graph is decreasing. Similarly, the graph is decreasing on the interval (0, 2) and increasing on (2, q). 5. There are no points of inflection because ƒ″(x) 6 0 whenever x 6 0, ƒ″(x) 7 0 whenever x 7 0, and ƒ″ exists everywhere and is never zero throughout the domain of ƒ. The graph is concave down on the interval ( q, 0) and concave up on the interval (0, q). 6. From the rewritten formula for ƒ(x), we see that
2 y= x +4 2x
(2, 2) y= x 2
0
2
4
x
−2 −4
Figure 4.32 The graph of y = (Example 9).
251
lim a
x2 + 4 2x
xS0 +
x 2 + x b = + q and 2
lim a
xS0 
x 2 + x b =  q, 2
so the yaxis is a vertical asymptote. Also, as x S q or as x S  q, the graph of ƒ(x) approaches the line y = x>2. Thus y = x>2 is an oblique asymptote. 7. The graph of ƒ is sketched in Figure 4.32.
Example 10 Sketch the graph of ƒ(x) = e2>x. Solution The domain of ƒ is ( q, 0)h(0, q) and there are no symmetries about either axis or the origin. The derivatives of ƒ are ƒ′(x) = e2>x a
and
y 5
ƒ″(x) = y = e 2x
4 3 Inflection 2 point 1 −2
−1
y=1 0
1
2
3
x
Figure 4.33 The graph of y = e2>x has a point of inflection at ( 1, e2). The line y = 1 is a horizontal asymptote and x = 0 is a vertical asymptote (Example 10).
2e2>x 2 b =  2 2 x x
x2(2e2>x)(2>x2)  2e2>x(2x) x4
=
4e2>x(1 + x) . x4
Both derivatives exist everywhere over the domain of ƒ. Moreover, since e2>x and x2 are both positive for all x ≠ 0, we see that ƒ′ 6 0 everywhere over the domain and the graph is everywhere decreasing. Examining the second derivative, we see that ƒ″(x) = 0 at x = 1. Since e2>x 7 0 and x4 7 0, we have ƒ″ 6 0 for x 6 1 and ƒ″ 7 0 for x 7 1, x ≠ 0. Therefore, the point (1, e2) is a point of inflection. The curve is concave down on the interval ( q, 1) and concave up over (1, 0) h (0, q). From Example 7, Section 2.6, we see that limx S 0 ƒ(x) = 0. As x S 0+, we see that 2>x S q, so limx S 0+ ƒ(x) = q and the yaxis is a vertical asymptote. Also, as x S  q or x S q, 2>x S 0 and so limx S  q ƒ(x) = limx Sq ƒ(x) = e0 = 1. Therefore, y = 1 is a horizontal asymptote. There are no absolute extrema, since ƒ never takes on the value 0 and has no absolute maximum. The graph of ƒ is sketched in Figure 4.33.
Graphical Behavior of Functions from Derivatives As we saw in Examples 7–10, we can learn much about a twicedifferentiable function y = ƒ(x) by examining its first derivative. We can find where the function’s graph rises and falls and where any local extrema are located. We can differentiate y′ to learn how the graph bends as it passes over the intervals of rise and fall. We can determine the shape of the function’s graph. Information we cannot get from the derivative is how to place the graph in the xyplane. But, as we discovered in Section 4.2, the only additional information we need to position the graph is the value of ƒ at one point. Information about the asymptotes is found using limits (Section 2.6). The following
252
Chapter 4: Applications of Derivatives
figure summarizes how the first derivative and second derivative affect the shape of a graph.
y = f (x)
y = f (x)
Differentiable 1 smooth, connected; graph may rise and fall
y′ > 0 1 rises from left to right; may be wavy
or
y = f (x)
y′ < 0 1 falls from left to right; may be wavy
−
or +
y″ > 0 1 concave up throughout; no waves; graph may rise or fall
+
−
or
−
y″ < 0 1 concave down throughout; no waves; graph may rise or fall
y″ changes sign at an inflection point
y′ = 0 and y″ < 0 at a point; graph has local maximum
y′ = 0 and y″ > 0 at a point; graph has local minimum
+
y′ changes sign 1 graph has local maximum or local minimum
Exercises 4.4 Analyzing Functions from Graphs Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the intervals on which the functions are concave up and concave down.
1. y = x 3 − x 2 − 2x + 1 3 y
2
2.
3
4 y = x − 2x2 + 4 4 y
0
2p 3
x
4.
x
x 0
y = 9 x13(x 2 − 7) 14 y
0
x
x
0
8. y = 2 cos x −
y
y = 3 (x 2 − 1)23 4 y
0
2p 3
7. y = sin 0 x 0 , −2p ≤ x ≤ 2p 0
3.
6. p p y = x + sin 2x, − 2p ≤ x ≤ 2p y = tan x − 4x, − < x < 3 2 3 2 y y
−
x
0
5.
−p
3p " 2 x, −p ≤ x ≤ 2 y
0
3p 2
x
NOT TO SCALE
Graphing Functions In Exercises 9–58, identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. x
9. y = x2  4x + 3
10. y = 6  2x  x2
11. y = x3  3x + 3
12. y = x(6  2x)2
13. y =  2x3 + 6x2  3
14. y = 1  9x  6x2  x3
15. y = (x  2)3 + 1 16. y = 1  (x + 1)3 17. y = x4  2x2 = x2(x2  2) 18. y =  x4 + 6x2  4 = x2(6  x2)  4 19. y = 4x3  x4 = x3(4  x) 20. y = x4 + 2x3 = x3(x + 2) 21. y = x5  5x4 = x4(x  5) 22. y = xa
65. y′ = (8x  5x2)(4  x)2 66. y′ = (x2  2x)(x  5)2 p p 67. y′ = sec2 x,  6 x 6 2 2 p p 68. y′ = tan x,  6 x 6 2 2 u u 69. y′ = cot , 0 6 u 6 2p 70. y′ = csc2 , 0 6 u 6 2p 2 2 p p 71. y′ = tan2 u  1,  6 u 6 2 2 72. y′ = 1  cot2 u, 0 6 u 6 p 73. y′ = cos t, 0 … t … 2p
4 x  5b 2
74. y′ = sin t, 0 … t … 2p 75. y′ = (x + 1)2>3
76. y′ = (x  2)1>3
24. y = x  sin x, 0 … x … 2p
77. y′ = x2>3(x  1)
78. y′ = x4>5(x + 1)
25. y = 23x  2 cos x, 0 … x … 2p p p 4 6 x 6 26. y = x  tan x, 3 2 2
79. y′ = 2 0 x 0 = e
23. y = x + sin x, 0 … x … 2p
80. y′ = e
27. y = sin x cos x, 0 … x … p 28. y = cos x + 23 sin x, 0 … x … 2p
29. y = x1>5 31. y =
30. y = x2>5
x
32. y =
2
2x + 1
33. y = 2x  3x2>3 35. y = x2>3 a
5  xb 2
37. y = x 28  x2
39. y = 216  x2 2
x  3 x  2 8x 43. y = 2 x + 4 45. y = 0 x2  1 0 41. y =
21  x2
2x + 1
34. y = 5x2>5  2x
2 x, 2x,
48. y = 2 0 x  4 0
51. y = ln (3  x2)
52. y = x (ln x)2
53. y = e  2e
 3x
55. y = ln (cos x) 1 1 + ex
2
y = f ′(x) x
x P
y = f ″(x)
y = f ″(x)
83. y P
y = f ′(x) x
0
84.
y y = f ′(x)
x
54. y = xe
ln x 2x ex 58. y = 1 + ex 56. y =
Sketching the General Shape, Knowing y ′ Each of Exercises 59–80 gives the first derivative of a continuous function y = ƒ(x). Find y″ and then use Steps 2–4 of the graphing procedure on page 249 to sketch the general shape of the graph of ƒ.
59. y′ = 2 + x  x2
y
y = f ″(x)
ex 50. y = x
x
82. P
x 6 0 x Ú 0
49. y = xe1>x x
y y = f ′(x)
38. y = (2  x2)3>2 2 40. y = x2 + x
5 x4 + 5 46. y = 0 x2  2 x 0
x2, x … 0 x 2, x 7 0
81.
36. y = x2>3(x  5)
3 3 42. y = 2 x + 1
2x, x … 0 2x, x 7 0
Sketching y from Graphs of y′ and y″ Each of Exercises 81–84 shows the graphs of the first and second derivatives of a function y = ƒ(x). Copy the picture and add to it a sketch of the approximate graph of ƒ, given that the graph passes through the point P.
44. y =
47. y = 2 0 x 0 = e
57. y =
253
4.4 Concavity and Curve Sketching
60. y′ = x2  x  6
61. y′ = x(x  3)
62. y′ = x2(2  x)
63. y′ = x(x2  12)
64. y′ = (x  1)2(2x + 3)
x
0 y = f ″(x) P
Graphing Rational Functions Graph the rational functions in Exercises 85–102 using all the steps in the graphing procedure on page 249. x2  49 2x2 + x  1 85. y = 86. y = 2 2 x  1 x + 5x  14
x4 + 1 x2 1 89. y = 2 x  1 87. y =
x2  4 2x x2 90. y = 2 x  1 88. y =
Chapter 4: Applications of Derivatives
93. y = 95. y = 97. y = 99. y = 101. y = 102. y =
Motion Along a Line The graphs in Exercises 107 and 108 show the position s = ƒ(t) of an object moving up and down on a coordinate line. (a) When is the object moving away from the origin? Toward the origin? At approximately what times is the (b) velocity equal to zero? (c) Acceleration equal to zero? (d) When is the acceleration positive? Negative?
x2  4 x2  2 x2  4 x + 1 x2  x + 1 x  1 x3 + x  2 x  x2 x  1 x2(x  2)
s = f (t)
5
0
Theory and Examples 103. The accompanying figure shows a portion of the graph of a twicedifferentiable function y = ƒ(x). At each of the five labeled points, classify y′ and y″ as positive, negative, or zero. y
R
5
ƒ( 2) = 8,
ƒ′(2) = ƒ′(2) = 0,
ƒ(0) = 4,
ƒ′(x) 6 0 for
ƒ(2) = 0,
ƒ″(x) 6 0 for x 6 0,
0 x 0 7 2,
ƒ″(x) 7 0 for x 7 0.
0 x 0 6 2,
y Derivatives 0, 0, 0, 0, 0, 0, 0,
y″ y″ y″ y″ y″ y″ y″
t
c c = f (x)
105. Sketch the graph of a twicedifferentiable function y = ƒ(x) with the following properties. Label coordinates where possible.
6 = 7 7 7 = 6
15
Cost
104. Sketch a smooth connected curve y = ƒ(x) with
y′ x 6 2 2 1 y′ 2 6 x 6 4 y′ 4 4 y′ 4 6 x 6 6 y′ 6 7 y′ x 7 6 y′
10 Time (sec)
109. Marginal cost The accompanying graph shows the hypothetical cost c = ƒ(x) of manufacturing x items. At approximately what production level does the marginal cost change from decreasing to increasing?
x
x
t
s = f (t)
0
T
0
15
s
108.
Q
ƒ′(x) 7 0 for
10 Time (sec)
S
y = f (x) P
s
107. Displacement
x2  2 92. y = x2  1 x2 94. y = x + 1 x2  x + 1 96. y = x  1 x3  3x2 + 3x  1 98. y = x2 + x  2 x 100. y = x2  1 8 (Agnesi>s witch) x2 + 4 4x (Newton>s serpentine) x2 + 4
91. y = 
Displacement
254
7 7 7 = 6 6 6
0 0 0 0 0 0 0
106. Sketch the graph of a twicedifferentiable function y = ƒ(x) that passes through the points ( 2, 2), ( 1, 1), (0, 0), (1, 1), and (2, 2) and whose first two derivatives have the following sign patterns. + + y′: 2 0 2 + y″: 1 1
20 40 60 80 100 120 Thousands of units produced
x
110. The accompanying graph shows the monthly revenue of the Widget Corporation for the past 12 years. During approximately what time intervals was the marginal revenue increasing? Decreasing? y y = r(t)
0
5
10
t
111. Suppose the derivative of the function y = ƒ(x) is y′ = (x  1)2(x  2). At what points, if any, does the graph of ƒ have a local minimum, local maximum, or point of inflection? (Hint: Draw the sign pattern for y′.)
4.5 Indeterminate Forms and L’Hôpital’s Rule
112. Suppose the derivative of the function y = ƒ(x) is
255
120. Suppose that the second derivative of the function y = ƒ(x) is
2
y′ = (x  1) (x  2)(x  4).
y″ = x2(x  2)3(x + 3).
At what points, if any, does the graph of ƒ have a local minimum, local maximum, or point of inflection?
For what xvalues does the graph of ƒ have an inflection point?
113. For x 7 0, sketch a curve y = ƒ(x) that has ƒ(1) = 0 and ƒ′(x) = 1>x. Can anything be said about the concavity of such a curve? Give reasons for your answer. 114. Can anything be said about the graph of a function y = ƒ(x) that has a continuous second derivative that is never zero? Give reasons for your answer. 115. If b, c, and d are constants, for what value of b will the curve y = x3 + bx2 + cx + d have a point of inflection at x = 1? Give reasons for your answer. 116. Parabolas a. Find the coordinates of the vertex of the parabola y = ax2 + bx + c, a ≠ 0. b. When is the parabola concave up? Concave down? Give reasons for your answers. 117. Quadratic curves What can you say about the inflection points of a quadratic curve y = ax2 + bx + c, a ≠ 0? Give reasons for your answer. 118. Cubic curves What can you say about the inflection points of a cubic curve y = ax3 + bx2 + cx + d, a ≠ 0? Give reasons for your answer. 119. Suppose that the second derivative of the function y = ƒ(x) is y″ = (x + 1)(x  2). For what xvalues does the graph of ƒ have an inflection point?
121. Find the values of constants a, b, and c so that the graph of y = ax3 + bx2 + cx has a local maximum at x = 3, local minimum at x = 1, and inflection point at (1, 11). 122. Find the values of constants a, b, and c so that the graph of y = (x2 + a)>(bx + c) has a local minimum at x = 3 and a local maximum at (1, 2). Computer Explorations In Exercises 123–126, find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function’s first and second derivatives. How are the values at which these graphs intersect the xaxis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function?
123. y = x5  5x4  240
124. y = x3  12x2
125. y =
4 5 x + 16x2  25 5
126. y =
x3 x4  4x2 + 12x + 20 4 3
127. Graph ƒ(x) = 2x4  4x2 + 1 and its first two derivatives together. Comment on the behavior of ƒ in relation to the signs and values of ƒ′ and ƒ″. 128. Graph ƒ(x) = x cos x and its second derivative together for 0 … x … 2p. Comment on the behavior of the graph of ƒ in relation to the signs and values of ƒ″.
4.5 Indeterminate Forms and L’Hôpital’s Rule HISTORICAL BIOGRAPHY Guillaume François Antoine de l’Hôpital (1661–1704) Johann Bernoulli (1667–1748)
John (Johann) Bernoulli discovered a rule using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or + q. The rule is known today as l’Hôpital’s Rule, after Guillaume de l’Hôpital. He was a French nobleman who wrote the first introductory differential calculus text, where the rule first appeared in print. Limits involving transcendental functions often require some use of the rule for their calculation.
Indeterminate Form 0 , 0
If we want to know how the function F(x) =
x  sin x x3
behaves near x = 0 (where it is undefined), we can examine the limit of F(x) as x S 0. We cannot apply the Quotient Rule for limits (Theorem 1 of Chapter 2) because the limit of the denominator is 0. Moreover, in this case, both the numerator and denominator approach 0, and 0>0 is undefined. Such limits may or may not exist in general, but the limit does exist for the function F(x) under discussion by applying l’Hôpital’s Rule, as we will see in Example 1d.
256
Chapter 4: Applications of Derivatives
If the continuous functions ƒ(x) and g (x) are both zero at x = a, then lim
xSa
ƒ(x) g(x)
cannot be found by substituting x = a. The substitution produces 0>0, a meaningless expression, which we cannot evaluate. We use 0>0 as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as q > q, q # 0, q  q, 00, and 1q, which cannot be evaluated in a consistent way; these are called indeterminate forms as well. Sometimes, but not always, limits that lead to indeterminate forms may be found by cancelation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find limx S 0 (sin x)>x. But we have had success with the limit ƒ′(a) = lim
xSa
ƒ(x)  ƒ(a) x  a ,
from which we calculate derivatives and which produces the indeterminant form 0>0 when we attempt to substitute x = a. L’Hôpital’s Rule enables us to draw on our success with derivatives to evaluate limits that otherwise lead to indeterminate forms.
THEOREM 6—L’Hôpital’s Rule Suppose that ƒ(a) = g(a) = 0, that ƒ and g are differentiable on an open interval I containing a, and that g′(x) ≠ 0 on I if x ≠ a. Then lim
xSa
ƒ(x) ƒ′(x) = lim , g(x) x S a g′(x)
assuming that the limit on the right side of this equation exists.
We give a proof of Theorem 6 at the end of this section. Caution To apply l’Hôpital’s Rule to ƒ>g, divide the derivative of ƒ by the derivative of g. Do not fall into the trap of taking the derivative of ƒ>g. The quotient to use is ƒ′>g′, not (ƒ>g)′.
Example 1 The following limits involve 0>0 indeterminate forms, so we apply l’Hôpital’s Rule. In some cases, it must be applied repeatedly. (a) lim
3x  sin x 3  cos x 3  cos x 2 = lim = 2 = x 1 1 xS0 x=0
(b) lim
21 + x  1
(c) lim
21 + x  1  x>2
xS0
xS0
xS0
x
1 2 21 + x 1 = lim = 2 xS0 1
x2
0 ; apply l’Hôpital’s Rule. 0
(1>2)(1 + x)1>2  1>2 2x xS0
0 Still ; apply l’Hôpital’s Rule again. 0
(1>4)(1 + x)3>2 1 =  8 2 xS0
0 Not ; limit is found. 0
= lim
= lim
4.5 Indeterminate Forms and L’Hôpital’s Rule
(d) lim
xS0
x  sin x x3
0 ; apply l’Hôpital’s Rule. 0
1  cos x 3x2 sin x = lim x S 0 6x cos x 1 = lim = 6 xS0 6
0 Still ; apply l’Hôpital’s Rule again. 0
= lim
xS0
257
0 Still ; apply l’Hôpital’s Rule again. 0 0 Not ; limit is found. 0
Here is a summary of the procedure we followed in Example 1. Using L’Hôpital’s Rule To find lim
xSa
ƒ(x) g(x)
by l’Hôpital’s Rule, we continue to differentiate ƒ and g, so long as we still get the form 0>0 at x = a. But as soon as one or the other of these derivatives is different from zero at x = a we stop differentiating. L’Hôpital’s Rule does not apply when either the numerator or denominator has a finite nonzero limit.
Example 2 Be careful to apply l’Hôpital’s Rule correctly: 1  cos x 00 x + x2 sin x = lim Not 00 x S 0 1 + 2x
lim
xS0
It is tempting to try to apply l’Hôpital’s Rule again, which would result in lim
xS0
cos x 1 = , 2 2
but this is not the correct limit. l’Hôpital’s Rule can be applied only to limits that give indeterminate forms, and limx S 0 (sin x)>(1 + 2x) does not give an indeterminate form. Instead, this limit is 0>1 = 0, and the correct answer for the original limit is 0. L’Hôpital’s Rule applies to onesided limits as well.
Example 3 In this example the onesided limits are different. (a) lim+ xS0
Recall that q and + q mean the same thing.
(b) limxS0
sin x x2
0 0
= lim+ xS0
cos x q = 2x
sin x x2
Positive for x 7 0 0 0
= limxS0
cos x =  q 2x
Indeterminate Forms
Negative for x 6 0
H , H, H # 0, H − H
Sometimes when we try to evaluate a limit as x S a by substituting x = a we get an indeterminant form like q > q, q # 0, or q  q, instead of 0>0. We first consider the form q > q.
258
Chapter 4: Applications of Derivatives
More advanced treatments of calculus prove that l’Hôpital’s Rule applies to the indeterminate form q > q , as well as to 0>0. If ƒ(x) S { q and g(x) S { q as x S a, then lim
xSa
ƒ(x) ƒ′(x) = lim g(x) x S a g′(x)
provided the limit on the right exists. In the notation x S a, a may be either finite or infinite. Moreover, x S a may be replaced by the onesided limits x S a+ or x S a.
Example 4 Find the limits of these q > q forms: (a)
sec x x S p>2 1 + tan x lim
(b) lim
x Sq
ln x 2 2x
(c) lim
x Sq
ex . x2
Solution (a) The numerator and denominator are discontinuous at x = p>2, so we investigate the onesided limits there. To apply l’Hôpital’s Rule, we can choose I to be any open interval with x = p>2 as an endpoint. sec x q lim q from the left so we apply l’Hôpital’s Rule. x S (p>2) 1 + tan x =
lim
x S (p>2)
sec x tan x = sec2 x
lim sin x = 1
x S (p>2)
The righthand limit is 1 also, with ( q)>( q) as the indeterminate form. Therefore, the twosided limit is equal to 1. (b) lim
x Sq
(c) lim
x Sq
1>x 1>x ln x 1 = lim = lim = 0 Sq Sq x x 1> 2x 2 2x 1> 2x 2x
2x 1 = x =
2x
ex ex ex q = lim = lim = 2 x Sq 2x x Sq 2 x
Next we turn our attention to the indeterminate forms q # 0 and q  q. Sometimes these forms can be handled by using algebra to convert them to a 0>0 or q > q form. Here again we do not mean to suggest that q # 0 or q  q is a number. They are only notations for functional behaviors when considering limits. Here are examples of how we might work with these indeterminate forms.
Example 5 Find the limits of these q # 0 forms: 1
(a) lim ax sin x b x Sq
(b) lim+ 2x ln x xS0
Solution
sin h 1 1 = 1 a. lim ax sin x b = lim+ a sin hb = lim+ h x Sq hS0 h hS0 b. lim+ 2x ln x = lim+ xS0
xS0
ln x 1> 2x
1>x
= lim+
= lim+1 2 2x 2 = 0
xS0
xS0
1>2x3>2
q # 0; let h = 1>x. q # 0 converted to q > q l’Hôpital’s Rule applied
4.5 Indeterminate Forms and L’Hôpital’s Rule
259
Example 6 Find the limit of this q  q form: lim a
xS0
1 1  xb. sin x
Solution If x S 0+, then sin x S 0+ and
1 1  x S q  q. sin x Similarly, if x S 0, then sin x S 0 and 1 1  x S  q  ( q) =  q + q. sin x Neither form reveals what happens in the limit. To find out, we first combine the fractions: 1 1 x  sin x  x = . Common denominator is x sin x. sin x x sin x Then we apply l’Hôpital’s Rule to the result: lim a
xS0
x  sin x 1 1  x b = lim sin x x S 0 x sin x
00
= lim
1  cos x sin x + x cos x
= lim
sin x 0 = = 0. 2 cos x  x sin x 2
xS0
xS0
Still 00
Indeterminate Powers Limits that lead to the indeterminate forms 1q, 00, and q0 can sometimes be handled by first taking the logarithm of the function. We use l’Hôpital’s Rule to find the limit of the logarithm expression and then exponentiate the result to find the original function limit. This procedure is justified by the continuity of the exponential function and Theorem 10 in Section 2.5, and it is formulated as follows. (The formula is also valid for onesided limits.) If limx S a ln ƒ(x) = L, then lim ƒ(x) = lim eln ƒ(x) = eL.
xSa
xSa
Here a may be either finite or infinite.
Example 7 Apply l’Hôpital’s Rule to show that limx S 0+ (1 + x)1>x = e. Solution The limit leads to the indeterminate form 1q . We let ƒ(x) = (1 + x)1>x and find limx S 0+ ln ƒ(x). Since 1 ln ƒ(x) = ln (1 + x)1>x = x ln (1 + x), l’Hôpital’s Rule now applies to give lim+ ln ƒ(x) = lim+
xS0
xS0
ln (1 + x) 00 x
1 1 + x = lim+ 1 xS0 =
l’Hôpital’s Rule applied
1 = 1. 1
Therefore, lim+ (1 + x)1>x = lim+ ƒ(x) = lim+ eln ƒ(x) = e1 = e. xS0
xS0
xS0
260
Chapter 4: Applications of Derivatives
Example 8 Find limx Sq x1>x. Solution The limit leads to the indeterminate form q0 . We let ƒ(x) = x1>x and find limx Sq ln ƒ(x). Since ln x ln ƒ(x) = ln x1>x = x , l’Hôpital’s Rule gives ln x lim ln ƒ(x) = lim x x Sq
x Sq
= lim
x Sq
=
1>x 1
q
q l’Hôpital’s Rule applied
0 = 0. 1
Therefore lim x1>x = lim ƒ(x) = lim eln ƒ(x) = e0 = 1. x Sq
x Sq
x Sq
Proof of L’Hôpital’s Rule y y = f ′(a)(x − a) f (x) y = g′(a)(x − a)
ƒ(x) = ƒ′(a)(x  a) + P1(x  a) and g(x) = g′(a)(x  a) + P2(x  a)
g(x) 0
a
Before we prove l’Hôpital’s Rule, we consider a special case to provide some geometric insight for its reasonableness. Consider the two functions ƒ(x) and g(x) having continuous derivatives and satisfying ƒ(a) = g(a) = 0, g′(a) ≠ 0. The graphs of ƒ(x) and g(x), together with their linearizations y = ƒ′(a)(x  a) and y = g′(a)(x  a), are shown in Figure 4.34. We know that near x = a, the linearizations provide good approximations to the functions. In fact,
x
where P1 S 0 and P2 S 0 as x S a. So, as Figure 4.34 suggests, lim
xSa
Figure 4.34 The two functions in l’Hôpital’s Rule, graphed with their linear approximations at x = a.
ƒ′(a)(x  a) + P1(x  a) ƒ(x) = lim g(x) x S a g′(a)(x  a) + P2(x  a) = lim
ƒ′(a) + P1 ƒ′(a) = g′(a) + P2 g′(a)
g′(a) ≠ 0
= lim
ƒ′(x) , g′(x)
Continuous derivatives
xSa
xSa
Historical Biography AugustinLouis Cauchy (1789–1857)
as asserted by l’Hôpital’s Rule. We now proceed to a proof of the rule based on the more general assumptions stated in Theorem 6, which do not require that g′(a) ≠ 0 and that the two functions have continuous derivatives. The proof of l’Hôpital’s Rule is based on Cauchy’s Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Cauchy’s Theorem first and then show how it leads to l’Hôpital’s Rule. THEOREM 7—Cauchy’s Mean Value Theorem Suppose functions ƒ and g are continuous on 3 a, b4 and differentiable throughout (a, b) and also suppose g′(x) ≠ 0 throughout (a, b). Then there exists a number c in (a, b) at which
When g(x) = x, Theorem 7 is the Mean Value Theorem.
ƒ′(c) ƒ(b)  ƒ(a) . = g′(c) g(b)  g(a)
Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that g(a) ≠ g(b). For if g(b) did equal g(a), then the Mean Value Theorem would give g′(c) =
g(b)  g(a) = 0 b  a
4.5 Indeterminate Forms and L’Hôpital’s Rule
261
for some c between a and b, which cannot happen because g′(x) ≠ 0 in (a, b). We next apply the Mean Value Theorem to the function F(x) = ƒ(x)  ƒ(a) 
ƒ(b)  ƒ(a) 3 g(x)  g(a) 4 . g(b)  g(a)
This function is continuous and differentiable where ƒ and g are, and F(b) = F(a) = 0. Therefore, there is a number c between a and b for which F′(c) = 0. When expressed in terms of ƒ and g, this equation becomes F′(c) = ƒ′(c) so that y
ƒ′(c) ƒ(b)  ƒ(a) . = g′(c) g(b)  g(a)
slope =
f ′(c) g′(c) B (g(b), f (b))
P
slope =
A 0
f (b) − f (a) g(b) − g(a)
(g(a), f (a))
Cauchy’s Mean Value Theorem has a geometric interpretation for a general winding curve C in the plane joining the two points A = (g(a), ƒ(a)) and B = (g(b), ƒ(b)). In Chapter 11 you will learn how the curve C can be formulated so that there is at least one point P on the curve for which the tangent to the curve at P is parallel to the secant line joining the points A and B. The slope of that tangent line turns out to be the quotient ƒ′>g′ evaluated at the number c in the interval (a, b), which is the lefthand side of the equation in Theorem 7. Because the slope of the secant line joining A and B is ƒ(b)  ƒ(a) , g(b)  g(a)
x
Figure 4.35 There is at least one point P on the curve C for which the slope of the tangent to the curve at P is the same as the slope of the secant line joining the points A(g(a), ƒ(a)) and B(g(b), ƒ(b)).
ƒ(b)  ƒ(a) 3 g′(c) 4 = 0 g(b)  g(a)
the equation in Cauchy’s Mean Value Theorem says that the slope of the tangent line equals the slope of the secant line. This geometric interpretation is shown in Figure 4.35. Notice from the figure that it is possible for more than one point on the curve C to have a tangent line that is parallel to the secant line joining A and B.
Proof of l’Hôpital’s Rule We first establish the limit equation for the case x S a+. The method needs almost no change to apply to x S a, and the combination of these two cases establishes the result. Suppose that x lies to the right of a. Then g′(x) ≠ 0, and we can apply Cauchy’s Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that ƒ′(c) ƒ(x)  ƒ(a) = . g′(c) g(x)  g(a) But ƒ(a) = g(a) = 0, so ƒ′(c) ƒ(x) = . g′(c) g(x) As x approaches a, c approaches a because it always lies between a and x. Therefore, lim+
xSa
ƒ(x) ƒ′(c) ƒ′(x) = lim+ = lim+ , g(x) c S a g′(c) x S a g′(x)
which establishes l’Hôpital’s Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval 3 x, a4 , x 6 a.
262
Chapter 4: Applications of Derivatives
Exercises 4.5 Finding Limits in Two Ways In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2.
x + 2 x S 2 x 2  4
sin 5x x xS0
1. lim
2. lim
2
3. lim
x Sq
5x  3x 7x2 + 1
x  1  x  3
x S 1 4x 3
5x  2x x Sq 7x 3 + 3
11. lim
x Sq
55. lim+ x xS0
1>ln x
8x2 15. lim x S 0 cos x  1
sin x  x 16. lim xS0 x3
61. lim a
1  sin u 19. lim u S p>2 1 + cos 2u
23. lim
tS0
25.
lim ax 
24. lim
tS0 1
p b sec x 2
3sin u  1 27. lim u uS0 x2x xS0 2  1
29. lim
26.
lim
32. lim
33. lim+
ln (x2 + 2x) ln x
34. lim+
35. lim
yS0
25y + 25  5
y
xS0
x Sq
xS0
(ln x)2 ln (sin x)
1 1 41. lim+ a b x  1 ln x xS1
36. lim
yS0
69.
lim
2x + 1
sec x tan x
x S (p>2)x
x
2  3 + 4x
xS0
1 x 60. lim+ a1 + x b S x 0
62. lim a x Sq
x2 + 1 1>x b x + 2
64. lim+ x (ln x)2 xS0
66. lim+ sin x # ln x xS0
68. lim+ xS0
2x
2sin x
cot x 70. lim+ csc x xS0
72. lim
2x + 4x  2x
x x S q 5
2
ex x x Sq xe
x e1>x 75. Which one is correct, and which one is wrong? Give reasons for your answers.
ln (ex  1) ln x
2ay + a2  a
y
a. lim
xS3
, a 7 0
xS0
3x + 1 1 b x sin x
42. lim+ (csc x  cot x + cos x) xS0
p  xb 2
29x + 1
73. lim
38. lim+ (ln x  ln sin x)
xS0
x Sq
x x Sq 3
log2 x (x + 3)
40. lim+ a
65. lim+ x tan a
71. lim
x Sq log 3
37. lim (ln 2x  ln (x + 1)) 39. lim+
p  xb tan x 2
3x  1 x xS0 2  1
ln (x + 1) log2 x
xS0
67. lim
(1>2)u  1 28. lim u uS0
31. lim
xS0
a
63. lim+ x2 ln x
58. lim (ex + x)1>x
Theory and Applications L’Hôpital’s Rule does not help with the limits in Exercises 67–74. Try it—you just keep on cycling. Find the limits some other way.
t sin t  cos t
x S (p>2)
x + 2 x b x  1
xS0
30. lim
x
x Sq
3u + p + (p>3))
lim
u S p>3 sin (u
22. lim
t(1  cos t) t  sin t
x S (p>2)
x Sq
ln (csc x) x S p>2 (x  (p>2))2
x2 x S 0 ln (sec x)
xS q
1>(2 ln x)
xS0
x  1 20. lim x S 1 ln x  sin px
21. lim
56. lim x1>ln x
x Sq
59. lim+ xx
18.
xSe
57. lim (1 + 2x)
sin 5t 14. lim t S 0 2t
2u  p  u)
54. lim+ (ln x)1>(x  e)
53. lim (ln x)
x  8x x Sq 12x 2 + 5x
u S p>2 cos (2p
xS1
1>x
sin t 2 13. lim t tS0
17. lim
sin 3x  3x + x2 sin x sin 2x xS0
50. lim
52. lim+ x1>(x  1)
xS1
2
12. lim
(ex  1)2 x S 0 x sin x
48. lim
51. lim+ x1>(1  x)
3t 3 + 3 10. lim 3 t S 1 4t  t + 3
3
x Sq
Indeterminate Powers and Products Find the limits in Exercises 51–66.
x2  25 8. lim xS  5 x + 5
t 3  4t + 15 9. lim 2 t S 3 t  t  12
eh  (1 + h) h2
46. lim x2ex
u  sin u cos u tan u  u uS0
Applying l’Hôpital’s Rule Use l’Hôpital’s rule to find the limits in Exercises 7–50.
x  2 7. lim 2 xS2 x  4
hS0
49. lim
2x2 + 3x 6. lim 3 x Sq x + x + 1
1  cos x 5. lim xS0 x2
44. lim
u S 0 eu
x  sin x 47. lim x tan x xS0
3
4. lim
cos u  1  u  1 et + t 2 45. lim t t Sq e  t 43. lim
74. lim+ xS0
x  3 x  3 0 1 1 = b. lim 2 = lim = = 0 6 xS3 x  3 x2  3 x S 3 2x 6
76. Which one is correct, and which one is wrong? Give reasons for your answers. a. lim
x2  2x 2x  2 = lim  sin x x S 0 2x  cos x 2 2 = = 1 = lim 2 + 0 x S 0 2 + sin x
x S 0 x2
x2  2x 2x  2 2 = = 2 = lim 0  1 x S 0 x 2  sin x x S 0 2x  cos x
b. lim
4.5 Indeterminate Forms and L’Hôpital’s Rule
77. Only one of these calculations is correct. Which one? Why are the others wrong? Give reasons for your answers. a. lim x ln x = 0 # ( q) = 0
a. Use l’Hôpital’s Rule to show that
x S 0+
b. lim+ x ln x = 0 # ( q) =  q
T b. Graph
xS0
ln x q = q = 1 c. lim+ x ln x = lim+ xS0 x S 0 (1>x) d. lim+ x ln x = lim+ xS0
xS0
= lim+
xS0
(1>x) ( 1>x2)
= lim+ (x) = 0 xS0
a. ƒ(x) = x,
g(x) = x2,
(a, b) = ( 2, 0)
b. ƒ(x) = x,
g(x) = x2,
(a, b) arbitrary
g(x) = x2,
c. ƒ(x) = x >3  4x,
(a, b) = (0, 3)
79. Continuous extension Find a value of c that makes the function 9x  3 sin 3x , x ≠ 0 5x3 ƒ(x) = c c, x = 0 continuous at x = 0. Explain why your value of c works. 80. For what values of a and b is lim a
xS0
T 81. H − H Form
tan 2x a sin bx + 2 + x b = 0? x3 x
a. Estimate the value of lim 1 x  2x2 + x 2
x Sq
by graphing ƒ(x) = x  2x2 + x over a suitably large interval of xvalues. b. Now confirm your estimate by finding the limit with l’Hôpital’s Rule. As the first step, multiply ƒ(x) by the fraction 1 x + 2x2 + x 2 > 1 x + 2x2 + x 2 and simplify the new numerator.
82. Find lim
x Sq
1 2x2
+ 1  2x 2.
T 83. 0 , 0 Form Estimate the value of
2x2  (3x + 1) 2x + 2 x  1 xS1 lim
by graphing. Then confirm your estimate with l’Hôpital’s Rule. 84. This exercise explores the difference between the limit lim a1 +
x Sq
and the limit
1 x b x2
1 x lim a1 + x b = e.
x Sq
1 x 1 x b and g(x) = a1 + x b 2 x
together for x Ú 0. How does the behavior of ƒ compare with that of g? Estimate the value of limx Sq ƒ(x).
78. Find all values of c that satisfy the conclusion of Cauchy’s Mean Value Theorem for the given functions and interval.
3
1 x lim a1 + x b = e. x Sq
ƒ(x) = a1 +
ln x (1>x)
263
c. Confirm your estimate of limx Sq ƒ(x) by calculating it with l’Hôpital’s Rule. 85. Show that r k lim a1 + b = er. k
k Sq
86. Given that x 7 0, find the maximum value, if any, of a. x1>x 2
b. x1>x
n
c. x1>x (n a positive integer) n
d. Show that limx Sq x1>x = 1 for every positive integer n. 87. Use limits to find horizontal asymptotes for each function. 3x + e2x 1 a. y = x tan a x b b. y = 2x + e3x 2
88. Find ƒ′(0) for ƒ(x) = e
e1/x , 0,
x ≠ 0 x = 0.
T 89. The continuous extension of (sin x)x to 30, p4
a. Graph ƒ(x) = (sin x)x on the interval 0 … x … p. What value would you assign to ƒ to make it continuous at x = 0?
b. Verify your conclusion in part (a) by finding limx S 0+ ƒ(x) with l’Hôpital’s Rule. c. Returning to the graph, estimate the maximum value of ƒ on 30, p4 . About where is max ƒ taken on?
d. Sharpen your estimate in part (c) by graphing ƒ′ in the same window to see where its graph crosses the xaxis. To simplify your work, you might want to delete the exponential factor from the expression for ƒ′ and graph just the factor that has a zero.
T 90. The function (sin x)tan x (Continuation of Exercise 89.) a. Graph ƒ(x) = (sin x)tan x on the interval  7 … x … 7. How do you account for the gaps in the graph? How wide are the gaps? b. Now graph ƒ on the interval 0 … x … p. The function is not defined at x = p>2, but the graph has no break at this point. What is going on? What value does the graph appear to give for ƒ at x = p>2? (Hint: Use l’Hôpital’s Rule to find lim ƒ as x S (p>2) and x S (p>2)+.) c. Continuing with the graphs in part (b), find max ƒ and min ƒ as accurately as you can and estimate the values of x at which they are taken on.
264
Chapter 4: Applications of Derivatives
4.6 Applied Optimization What are the dimensions of a rectangle with fixed perimeter having maximum area? What are the dimensions for the least expensive cylindrical can of a given volume? How many items should be produced for the most profitable production run? Each of these questions asks for the best, or optimal, value of a given function. In this section we use derivatives to solve a variety of optimization problems in mathematics, physics, economics, and business.
x
Solving Applied Optimization Problems 1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized? 2. Draw a picture. Label any part that may be important to the problem. 3. Introduce variables. List every relation in the picture and in the problem as an equation or algebraic expression, and identify the unknown variable. 4. Write an equation for the unknown quantity. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. This may require considerable manipulation. 5. Test the critical points and endpoints in the domain of the unknown. Use what you know about the shape of the function’s graph. Use the first and second derivatives to identify and classify the function’s critical points.
12
x x
x
12 (a)
x
12 − 2x 12 12 − 2x x
EXAMPLE 1 An opentop box is to be made by cutting small congruent squares from the corners of a 12in.by12in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible?
x
(b)
FIGURE 4.36 An open box made by cutting the corners from a square sheet of tin. What size corners maximize the box’s volume (Example 1)?
Solution We start with a picture (Figure 4.36). In the figure, the corner squares are x in. on a side. The volume of the box is a function of this variable: V(x) = x(12  2x)2 = 144x  48x2 + 4x3.
Since the sides of the sheet of tin are only 12 in. long, x … 6 and the domain of V is the interval 0 … x … 6. A graph of V (Figure 4.37) suggests a minimum value of 0 at x = 0 and x = 6 and a maximum near x = 2. To learn more, we examine the first derivative of V with respect to x:
Maximum y
Volume
y = x(12 − 2x)2, 0≤x≤6
0
dV = 144  96x + 12x2 = 12(12  8x + x2) = 12(2  x)(6  x). dx
min
min 2
6
V = hlw
x
NOT TO SCALE
FIGURE 4.37 The volume of the box in Figure 4.36 graphed as a function of x.
Of the two zeros, x = 2 and x = 6, only x = 2 lies in the interior of the function’s domain and makes the criticalpoint list. The values of V at this one critical point and two endpoints are Critical point value: V(2) = 128 Endpoint values:
V(0) = 0,
V(6) = 0.
The maximum volume is 128 in3. The cutout squares should be 2 in. on a side.
4.6 Applied Optimization
265
EXAMPLE 2 You have been asked to design a oneliter can shaped like a right circular cylinder (Figure 4.38). What dimensions will use the least material?
2r
Solution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is h
pr 2h = 1000. Surface area of can:
FIGURE 4.38 This oneliter can uses the least material when h = 2r (Example 2).
1 liter = 1000 cm3
2 A = 2pr + 2prh ()* ()*
circular cylindrical ends wall
How can we interpret the phrase “least material”? For a first approximation we can ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible while satisfying the constraint pr 2h = 1000 cm3. To express the surface area as a function of one variable, we solve for one of the variables in pr 2h = 1000 and substitute that expression into the surface area formula. Solving for h is easier: h =
1000 . pr 2
Thus, A = 2pr 2 + 2prh = 2pr 2 + 2pr a = 2pr 2 +
1000 b pr 2
2000 r .
Our goal is to find a value of r 7 0 that minimizes the value of A. Figure 4.39 suggests that such a value exists.
A Tall and thin can Short and wide can —— , r > 0 A = 2pr2 + 2000 r
Tall and thin
min
0 Short and wide
3
500 p
r
FIGURE 4.39 The graph of A = 2pr 2 + 2000>r is concave up.
Notice from the graph that for small r (a tall, thin cylindrical container), the term 2000>r dominates (see Section 2.6) and A is large. For large r (a short, wide cylindrical container), the term 2pr 2 dominates and A again is large.
266
Chapter 4: Applications of Derivatives
Since A is differentiable on r 7 0, an interval with no endpoints, it can have a minimum value only where its first derivative is zero. dA 2000 = 4pr dr r2 2000 0 = 4pr r2
Set dA>dr
4pr 3 = 2000 r = 3 What happens at r = 2 500>p? The second derivative
= 0.
Multiply by r 2.
3 500 ≈ 5.42 Solve for r. A p
d 2A 4000 = 4p + 3 r dr 2 is positive throughout the domain of A. The graph is therefore everywhere concave up and 3 the value of A at r = 2 500>p is an absolute minimum. The corresponding value of h (after a little algebra) is h =
1000 500 = 2 3 p = 2r. pr 2 A
The oneliter can that uses the least material has height equal to twice the radius, here with r ≈ 5.42 cm and h ≈ 10.84 cm.
Examples from Mathematics and Physics
Example 3 A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?
y x 2 + y2 = 4 Qx, "4 − x 2 R 2 −2 −x
0
x 2
Figure 4.40 The rectangle inscribed in the semicircle in Example 3.
x
Solution Let 1 x, 24  x2 2 be the coordinates of the corner of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.40). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower righthand corner: Length: 2x,
Height: 24  x2,
Area: 2x 24  x2.
Notice that the values of x are to be found in the interval 0 … x … 2, where the selected corner of the rectangle lies. Our goal is to find the absolute maximum value of the function on the domain 3 0, 24 . The derivative
A(x) = 2x 24  x2
dA 2x2 = + 2 24  x2 dx 24  x2
is not defined when x = 2 and is equal to zero when 2x2 + 2 24  x2 24  x2 2x2 + 2(4  x2) 8  4x2 x2
= 0 = 0 = 0 = 2
x = { 22.
4.6 Applied Optimization
267
Of the two zeros, x = 22 and x =  22, only x = 22 lies in the interior of A’s domain and makes the criticalpoint list. The values of A at the endpoints and at this one critical point are Critical point value: A1 22 2 = 2 22 24  2 = 4 Endpoint values: A(0) = 0, A(2) = 0.
The area has a maximum value of 4 when the rectangle is 24  x2 = 22 units high and 2x = 2 22 units long.
Example 4 The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat’s principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Describe the path that a ray of light will follow in going from a point A in a medium where the speed of light is c1 to a point B in a second medium where its speed is c2.
Historical Biography Willebrord Snell van Royen (1580–1626)
y A a
u1
Angle of incidence u1
Medium 1 P
0
x Medium 2
Solution Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xyplane and that the line separating the two media is the xaxis (Figure 4.41). In a uniform medium, where the speed of light remains constant, “shortest time” means “shortest path,” and the ray of light will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance traveled equals rate times time, so
b
u2 d−x
d Angle of refraction
x
B
Time =
Figure 4.41 A light ray refracted (deflected from its path) as it passes from one medium to a denser medium (Example 4).
distance rate .
From Figure 4.41, the time required for light to travel from A to P is 2a2 + x2 AP . t1 = c = c1 1
From P to B, the time is 2b2 + (d  x)2 PB . t2 = c = c2 2
The time from A to B is the sum of these: t = t1 + t2 =
2a2 + x2
c1
+
2b2 + (d  x)2
c2
.
This equation expresses t as a differentiable function of x whose domain is 3 0, d4 . We want to find the absolute minimum value of t on this closed interval. We find the derivative
dtdx negative
0
dtdx zero
− − − − − +++++++++ x0
dt x d  x = dx c1 2a2 + x2 c2 2b2 + (d  x)2
and observe that it is continuous. In terms of the angles u1 and u2 in Figure 4.41,
dtdx positive x d
Figure 4.42 The sign pattern of dt>dx in Example 4.
sin u1 sin u2 dt = c  c . 1 2 dx The function t has a negative derivative at x = 0 and a positive derivative at x = d. Since dt>dx is continuous over the interval 3 0, d4 , by the Intermediate Value Theorem for continuous functions (Section 2.5), there is a point x0 ∊ 3 0, d4 where dt>dx = 0 (Figure 4.42).
Chapter 4: Applications of Derivatives
There is only one such point because dt>dx is an increasing function of x (Exercise 62). At this unique point we then have sin u1 sin u2 c1 = c2 . This equation is Snell’s Law or the Law of Refraction, and is an important principle in the theory of optics. It describes the path the ray of light follows.
Examples from Economics Suppose that r(x) = the revenue from selling x items c(x) = the cost of producing the x items p(x) = r(x)  c(x) = the profit from producing and selling x items. Although x is usually an integer in many applications, we can learn about the behavior of these functions by defining them for all nonzero real numbers and by assuming they are differentiable functions. Economists use the terms marginal revenue, marginal cost, and marginal profit to name the derivatives r′(x), c′(x), and p′(x) of the revenue, cost, and profit functions. Let’s consider the relationship of the profit p to these derivatives. If r(x) and c(x) are differentiable for x in some interval of production possibilities, and if p(x) = r(x)  c(x) has a maximum value there, it occurs at a critical point of p(x) or at an endpoint of the interval. If it occurs at a critical point, then p′(x) = r′(x) c′(x) = 0 and we see that r′(x) = c′(x). In economic terms, this last equation means that
At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.43).
y Cost c(x)
Dollars
268
Revenue r(x) Breakeven point B
0
Maximum profit, c′(x) = r ′(x)
Local maximum for loss (minimum profit), c′(x) = r′(x) x Items produced
FIGURE 4.43 The graph of a typical cost function starts concave down and later turns concave up. It crosses the revenue curve at the breakeven point B. To the left of B, the company operates at a loss. To the right, the company operates at a profit, with the maximum profit occurring where c′(x) = r′(x). Farther to the right, cost exceeds revenue (perhaps because of a combination of rising labor and material costs and market saturation) and production levels become unprofitable again.
4.6 Applied Optimization
269
Example 5 Suppose that r(x) = 9x and c(x) = x3  6x2 + 15x, where x represents millions of MP3 players produced. Is there a production level that maximizes profit? If so, what is it?
y c(x) = x 3 − 6x2 + 15x
Solution Notice that r′(x) = 9 and c′(x) = 3x2  12x + 15. 3x2  12x + 15 = 9 Set c′(x) 3x2  12x + 6 = 0
r(x) = 9x
The two solutions of the quadratic equation are Maximum for profit
x1 =
Local maximum for loss 0 2 − "2
2
= r′(x).
2 + "2
x2 =
x
12  272 = 2  22 ≈ 0.586 6
and
12 + 272 = 2 + 22 ≈ 3.414. 6
The possible production levels for maximum profit are x ≈ 0.586 million MP3 players or x ≈ 3.414 million. The second derivative of p(x) = r(x)  c(x) is p″(x) = c″(x) since r″(x) is everywhere zero. Thus, p″(x) = 6(2  x), which is negative at x = 2 + 22 and positive at x = 2  22. By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graphs of r(x) and c(x) are shown in Figure 4.44.
NOT TO SCALE
Figure 4.44 The cost and revenue curves for Example 5.
Example 6 A cabinetmaker uses mahogany wood to produce 5 desks each day. Each delivery of one container of wood is $5000, whereas the storage of that material is $10 per day per unit stored, where a unit is the amount of material needed by her to produce 1 desk. How much material should be ordered each time, and how often should the material be delivered, to minimize her average daily cost in the production cycle between deliveries? Solution If she asks for a delivery every x days, then she must order 5x units to have enough material for that delivery cycle. The average amount in storage is approximately onehalf of the delivery amount, or 5x>2. Thus, the cost of delivery and storage for each cycle is approximately Cost per cycle = delivery costs + storage costs y
5000 c(x) = x + 25x y = 25x
Cost per cycle = 5000 ()*
delivery cost
Cost
5000 y= x
Figure 4.45 The average daily cost c(x) is the sum of a hyperbola and a linear function (Example 6).
a
#
5x b 2
()*
average amount stored
x
()* number of days stored
#
10
()*
storage cost per day
We compute the average daily cost c(x) by dividing the cost per cycle by the number of days x in the cycle (see Figure 4.45). c(x) =
min x value Cycle length
+
x
5000 x + 25x,
x 7 0.
As x S 0 and as x S q , the average daily cost becomes large. So we expect a minimum to exist, but where? Our goal is to determine the number of days x between deliveries that provides the absolute minimum cost. We find the critical points by determining where the derivative is equal to zero: c′(x) = 
500 + 25 = 0 x2
x = { 2200 ≈ {14.14.
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Chapter 4: Applications of Derivatives
Of the two critical points, only 2200 lies in the domain of c(x). The critical point value of the average daily cost is c1 2200 2 =
5000 2200
+ 25 2200 = 500 22 ≈ $707.11.
We note that c(x) is defined over the open interval (0, q) with c″(x) = 10000>x3 7 0. Thus, an absolute minimum exists at x = 2200 ≈ 14.14 days. The cabinetmaker should schedule a delivery of 5(14) = 70 units of the mahogany wood every 14 days.
Exercises 4.6 Mathematical Applications Whenever you are maximizing or minimizing a function of a single variable, we urge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your answer.
1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in2, and what are its dimensions? 2. Show that among all rectangles with an 8m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. a. Express the ycoordinate of P in terms of x. (Hint: Write an equation for the line AB.) b. Express the area of the rectangle in terms of x. c. What is the largest area the rectangle can have, and what are its dimensions? y B
singlestrand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? 8. The shortest fence A 216 m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 9. Designing a tank Your iron works has contracted to design and build a 500 ft3, squarebased, opentop, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. a. What dimensions do you tell the shop to use? b. Briefly describe how you took weight into account. 10. Catching rainwater A 1125 ft3 opentop rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy. a. If the total cost is c = 5(x2 + 4xy) + 10xy,
P(x, ?)
what values of x and y will minimize it? A −1
0
x
1
x
4. A rectangle has its base on the xaxis and its upper two vertices on the parabola y = 12  x2. What is the largest area the rectangle can have, and what are its dimensions? 5. You are planning to make an open rectangular box from an 8in.by15in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this way, and what is its volume? 6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when a = b. 7. The best fencing plan A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a
b. Give a possible scenario for the cost function in part (a). 11. Designing a poster You are designing a rectangular poster to contain 50 in2 of printing with a 4in. margin at the top and bottom and a 2in. margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.
3
3
y x
4.6 Applied Optimization
13. Two sides of a triangle have lengths a and b, and the angle between them is u. What value of u will maximize the triangle’s area? (Hint: A = (1>2)ab sin u.) 14. Designing a can What are the dimensions of the lightest opentop right circular cylindrical can that will hold a volume of 1000 cm3? Compare the result here with the result in Example 2.
x
x
x
x
24″
24″ x
3
15. Designing a can You are designing a 1000 cm right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be
271
x x
36″
x 18″
The sheet is then unfolded.
A = 8r 2 + 2prh rather than the A = 2pr 2 + 2prh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now? T 16. Designing a box with a lid A piece of cardboard measures 10 in. by 15 in. Two equal squares are removed from the corners of a 10in. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with lid.
NOT TO SCALE
x
x
x
x
10″
Base
Lid
x
x x
x 15″
Base
24″
36″
18. A rectangle is to be inscribed under the arch of the curve y = 4 cos (0.5x) from x = p to x = p. What are the dimensions of the rectangle with largest area, and what is the largest area? 19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume? 20. a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume? Girth = distance around here
a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. T 17. Designing a suitcase A 24in.by36in. sheet of cardboard is folded in half to form a 24in.by18in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. e. Find a value of x that yields a volume of 1120 in3. f. Write a paragraph describing the issues that arise in part (b).
Length
Square end
T b. Graph the volume of a 108in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). 21. (Continuation of Exercise 20.) a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by w and the girth is 2h + 2w. What dimensions will give the box its largest volume now?
272
Chapter 4: Applications of Derivatives Girth
D
C
R
"L2 − x 2
h
Crease
L
Q (originally at A) x
w h
x A
T b. Graph the volume as a function of h and compare what you see with your answer in part (a). 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.
P
B
26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?
y x x
Circumference = x
y
y
23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction. 24. The trough in the figure is to be made to the dimensions shown. Only the angle u can be varied. What value of u will maximize the trough’s volume?
(a)
(b)
27. Constructing cones A right triangle whose hypotenuse is 23 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.
h r
1′ u
"3
y x 28. Find the point on the line a + = 1 that is closest to the origin. b
u 1′ 1′
20′
29. Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible. 30. Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible.
25. Paper folding A rectangular sheet of 8.5in.by11in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. 2
3
a. Show that L = 2x >(2x  8.5). 2
b. What value of x minimizes L ?
c. What is the minimum value of L?
31. A wire b m long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a minimum, what is the length of each part? 32. Answer Exercise 31 if one piece is bent into a square and the other into a circle. 33. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.
w
5 h 3
4
4.6 Applied Optimization
34. Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 3. (See accompanying figure.)
y02 R = g sin 2a,
w
where g is the downward acceleration due to gravity. Find the angle a for which the range R is the largest possible.
h
35. What value of a makes ƒ(x) = x2 + (a>x) have
r=3
a. a local minimum at x = 2? b. a point of inflection at x = 1? 3
273
2
36. What values of a and b make ƒ(x) = x + ax + bx have a. a local maximum at x = 1 and a local minimum at x = 3? b. a local minimum at x = 4 and a point of inflection at x = 1? Physical Applications 37. Vertical motion The height above ground of an object moving vertically is given by
s =  16t 2 + 96t + 112,
T 43. Strength of a beam The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. (See the accompanying figure.) a. Find the dimensions of the strongest beam that can be cut from a 12in.diameter cylindrical log. b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k = 1. Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k = 1. Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it.
with s in feet and t in seconds. Find a. the object’s velocity when t = 0; b. its maximum height and when it occurs;
12″
c. its velocity when s = 0.
d
38. Quickest route Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time? 39. Shortest beam The 8ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.
Beam
Building
8′ wall 27′
40. Motion on a line The positions of two particles on the saxis are s1 = sin t and s2 = sin (t + p>3), with s1 and s2 in meters and t in seconds. a. At what time(s) in the interval 0 … t … 2p do the particles meet? b. What is the farthest apart that the particles ever get? c. When in the interval 0 … t … 2p is the distance between the particles changing the fastest? 41. The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two lights, one having an intensity eight times that of the other, are 6 m apart. How far from the stronger light is the total illumination least? 42. Projectile motion The range R of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity y0 at an angle a with the horizontal, then in Chapter 13 we find that
w
T 44. Stiffness of a beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth. a. Find the dimensions of the stiffest beam that can be cut from a 12in.diameter cylindrical log. b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k = 1. Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k = 1. Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it. 45. Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is s = 10 cos pt. a. What is the cart’s maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart’s speed then?
0
10
s
46. Two masses hanging side by side from springs have positions s1 = 2 sin t and s2 = sin 2t, respectively. a. At what times in the interval 0 6 t do the masses pass each other? (Hint: sin 2t = 2 sin t cos t.)
274
Chapter 4: Applications of Derivatives
b. When in the interval 0 … t … 2p is the vertical distance between the masses the greatest? What is this distance? (Hint: cos 2t = 2 cos2 t  1.)
s1
m1
0 s2
m2
churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate y = dx>dt of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, y may be considered to be a function of x alone, and y = kx(a  x) = kax  kx2,
s
47. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day. a. Start counting time with t = 0 at noon and express the distance s between the ships as a function of t. b. How rapidly was the distance between the ships changing at noon? One hour later? c. The visibility that day was 5 nautical miles. Did the ships ever sight each other? T d. Graph s and ds>dt together as functions of t for 1 … t … 3, using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c). e. The graph of ds>dt looks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that ds>dt approaches a limiting value as t S q. What is this value? What is its relation to the ships’ individual speeds? 48. Fermat’s principle in optics Light from a source A is reflected by a plane mirror to a receiver at point B, as shown in the accompanying figure. Show that for the light to obey Fermat’s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely geometric argument, which you may prefer.) Normal Light receiver Light source A
Angle of incidence u1
Angle of reflection u2
B
Plane mirror
49. Tin pest When metallic tin is kept below 13.2°C, it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their
where x = the amount of product a = the amount of substance at the beginning k = a positive constant. At what value of x does the rate y have a maximum? What is the maximum value of y? 50. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax3 + bx2 + cx + d, where y( L) = H and y(0) = 0. a. What is dy>dx at x = 0? b. What is dy>dx at x = L? c. Use the values for dy>dx at x = 0 and x =  L together with y(0) = 0 and y( L) = H to show that x 3 x 2 y(x) = H c 2a b + 3a b d . L L
y
Landing path
H = Cruising altitude
Airport L
x
Business and Economics 51. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by a n = x  c + b(100  x), where a and b are positive constants. What selling price will bring a maximum profit?
52. You operate a tour service that offers the following rates: $200 per person if 50 people (the minimum number to book the tour) go on the tour. For each additional person, up to a maximum of 80 people total, the rate per person is reduced by $2. It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people does it take to maximize your profit?
4.6 Applied Optimization
53. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km A(q) = q + cm + , 2 where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of placing an order (the same, no matter how often you order), c is the cost of one item (a constant), m is the number of items sold each week (a constant), and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to find the quantity that will minimize A(q). What is it? (The formula you get for the answer is called the Wilson lot size formula.) b. Shipping costs sometimes depend on order size. When they do, it is more realistic to replace k by k + bq, the sum of k and a constant multiple of q. What is the most economical quantity to order now? 54. Production level Prove that the production level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost. 55. Show that if r(x) = 6x and c(x) = x3  6x2 + 15x are your revenue and cost functions, then the best you can do is break even (have revenue equal cost). 56. Production level Suppose that c(x) = x3  20x2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.
Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity y can be modeled by the equation
where r0 is the rest radius of the trachea in centimeters and c is a positive constant whose value depends in part on the length of the trachea. Show that y is greatest when r = (2>3)r0; that is, when the trachea is about 33% contracted. The remarkable fact is that Xray photographs confirm that the trachea contracts about this much during a cough. T b. Take r0 to be 0.5 and c to be 1 and graph y over the interval 0 … r … 0.5. Compare what you see with the claim that y is at a maximum when r = (2>3)r0. Theory and Examples 61. An inequality for positive integers Show that if a, b, c, and d are positive integers, then
(a2 + 1)(b2 + 1)(c2 + 1)(d 2 + 1) Ú 16. abcd 62. The derivative dt>dx in Example 4 a. Show that ƒ(x) =
x 2a2 + x2
is an increasing function of x. b. Show that
58. The 800room Mega Motel chain is filled to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?
is a decreasing function of x.
and C is a constant.
R = M2 a
C M  b 2 3
60. How we cough a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough.
r0 … r … r0 , 2
y = c(r0  r)r 2 cm>sec,
57. You are to construct an open rectangular box with a square base and a volume of 48 ft3. If material for the bottom costs $6>ft2 and material for the sides costs $4>ft2, what dimensions will result in the least expensive box? What is the minimum cost?
Biology 59. Sensitivity to medicine (Continuation of Exercise 72, Section 3.3.) Find the amount of medicine to which the body is most sensitive by finding the value of M that maximizes the derivative dR>dM, where
275
g(x) =
d  x 2b + (d  x)2 2
c. Show that dt x d  x = dx c1 2a2 + x2 c2 2b2 + (d  x)2
is an increasing function of x.
63. Let ƒ(x) and g(x) be the differentiable functions graphed here. Point c is the point where the vertical distance between the curves is the greatest. Is there anything special about the tangents to the two curves at c? Give reasons for your answer.
y = f (x) y = g(x)
a
c
b
x
276
Chapter 4: Applications of Derivatives
64. You have been asked to determine whether the function ƒ(x) = 3 + 4 cos x + cos 2x is ever negative.
T b. Graph the distance function D(x) and y = 2x together and reconcile what you see with your answer in part (a).
a. Explain why you need to consider values of x only in the interval 30, 2p4.
y
b. Is ƒ ever negative? Explain.
(x, " x)
65. a. The function y = cot x  22 csc x has an absolute maximum value on the interval 0 6 x 6 p. Find it. T b. Graph the function and compare what you see with your answer in part (a). 66. a. The function y = tan x + 3 cot x has an absolute minimum value on the interval 0 6 x 6 p>2. Find it. T b. Graph the function and compare what you see with your answer in part (a). 67. a. How close does the curve y = 2x come to the point (3>2, 0)? (Hint: If you minimize the square of the distance, you can avoid square roots.)
0
y = "x x
3 a , 0b 2
68. a. How close does the semicircle y = 216  x2 come to the point 1 1, 23 2?
T b. Graph the distance function and y = 216  x2 together and reconcile what you see with your answer in part (a).
4.7 Newton’s Method In this section we study a numerical method, called Newton’s method or the Newton– Raphson method, which is a technique to approximate the solution to an equation ƒ(x) = 0. Essentially it uses tangent lines of the graph of y = ƒ(x) near the points where ƒ is zero to estimate the solution. (A value of x where ƒ is zero is a root of the function ƒ and a solution of the equation ƒ(x) = 0.)
Procedure for Newton’s Method The goal of Newton’s method for estimating a solution of an equation ƒ(x) = 0 is to produce a sequence of approximations that approach the solution. We pick the first number x0 of the sequence. Then, under favorable circumstances, the method does the rest by moving step by step toward a point where the graph of ƒ crosses the xaxis (Figure 4.46). At each step the method approximates a zero of ƒ with a zero of one of its linearizations. Here is how it works. The initial estimate, x0, may be found by graphing or just plain guessing. The method then uses the tangent to the curve y = ƒ(x) at (x0, ƒ(x0)) to approximate the curve, calling the point x1 where the tangent meets the xaxis (Figure 4.46). The number x1 is usually a better approximation to the solution than is x0. The point x2 where the tangent to the curve at (x1, ƒ(x1)) crosses the xaxis is the next approximation in the sequence. We continue on, using each approximation to generate the next, until we are close enough to the root to stop. We can derive a formula for generating the successive approximations in the following way. Given the approximation xn, the pointslope equation for the tangent to the curve at (xn, ƒ(xn)) is
y y = f (x) (x0, f (x0))
(x1, f (x1)) (x2, f(x2 ))
y = ƒ(xn) + ƒ′(xn)(x  xn).
Root sought 0
x3 Fourth
x2 Third
x1 Second
x0 First
APPROXIMATIONS
Figure 4.46 Newton’s method starts with an initial guess x0 and (under favorable circumstances) improves the guess one step at a time.
x
We can find where it crosses the xaxis by setting y = 0 (Figure 4.47): 0 = ƒ(xn) + ƒ′(xn)(x  xn) ƒ(xn) = x  xn ƒ′(xn) x = xn 
ƒ(xn) ƒ′(xn)
If ƒ′(xn)
≠ 0
This value of x is the next approximation xn + 1. Here is a summary of Newton’s method.
4.7 Newton’s Method
y y = f (x) Point: (xn, f (xn )) Slope: f ′(xn ) Tangent line equation: y − f (xn ) = f ′(xn )(x − xn ) (xn, f (xn))
Tangent line (graph of linearization of f at xn )
277
Newton’s Method 1. Guess a first approximation to a solution of the equation ƒ(x) = 0. A graph of y = ƒ(x) may help. 2. Use the first approximation to get a second, the second to get a third, and so on, using the formula xn + 1 = xn 
ƒ(xn) , ƒ′(xn)
if ƒ′(xn) ≠ 0.(1)
Root sought 0
xn xn+1 = xn −
x
f (xn ) f '(xn )
Figure 4.47 The geometry of the successive steps of Newton’s method. From xn we go up to the curve and follow the tangent line down to find xn + 1.
Applying Newton’s Method Applications of Newton’s method generally involve many numerical computations, making them well suited for computers or calculators. Nevertheless, even when the calculations are done by hand (which may be very tedious), they give a powerful way to find solutions of equations. In our first example, we find decimal approximations to 22 by estimating the positive root of the equation ƒ(x) = x2  2 = 0.
Example 1 Find the positive root of the equation ƒ(x) = x2  2 = 0. Solution With ƒ(x) = x2  2 and ƒ′(x) = 2x, Equation (1) becomes xn 2  2 2xn xn 1 = xn + x n 2
xn + 1 = xn 
=
xn 1 + x . n 2
The equation xn + 1 =
xn 1 + x n 2
enables us to go from each approximation to the next with just a few keystrokes. With the starting value x0 = 1, we get the results in the first column of the following table. (To five decimal places, 22 = 1.41421.) Error x0 x1 x2 x3
= = = =
Number of correct digits
1 0.41421 1 1.5 0.08579 1 1.41667 0.00246 3 1.41422 0.00001 5
Newton’s method is the method used by most software applications to calculate roots because it converges so fast (more about this later). If the arithmetic in the table in Example 1 had been carried to 13 decimal places instead of 5, then going one step further would have given 22 correctly to more than 10 decimal places.
278
Chapter 4: Applications of Derivatives
Example 2 Find the xcoordinate of the point where the curve y = x3  x crosses
y
the horizontal line y = 1.
20
y = x3 − x − 1
Solution The curve crosses the line when x3  x = 1 or x3  x  1 = 0. When does ƒ(x) = x3  x  1 equal zero? Since ƒ(1) = 1 and ƒ(2) = 5, we know by the Intermediate Value Theorem there is a root in the interval (1, 2) (Figure 4.48). We apply Newton’s method to ƒ with the starting value x0 = 1. The results are displayed in Table 4.1 and Figure 4.49. At n = 5, we come to the result x6 = x5 = 1.3247 17957. When xn + 1 = xn, Equation (1) shows that ƒ(xn) = 0. We have found a solution of ƒ(x) = 0 to nine decimals.
15 10 5
1
0
−1
2
x
3
Figure 4.48 The graph of ƒ(x) = x3  x  1 crosses the xaxis once; this is the root we want to find (Example 2).
(1.5, 0.875) Root sought
x1
x2
1
with x0 = 1
n
xn
xn+1 = xn −
ƒ′(xn)
ƒ(xn)
1 2 0 1 1 1.5 0.875 5.75 2 1.3478 26087 0.1006 82173 4.4499 05482 3 1.3252 00399 0.0020 58362 4.2684 68292 4 1.3247 18174 0.0000 00924 4.2646 34722 5 1.3247 17957 [email protected] 4.2646 32999
y = x3 − x − 1
x0
Table 4.1 The result of applying Newton’s method to ƒ(x) = x 3  x  1 ƒ(xn) ƒ′(xn)
1.5 1.3478 26087 1.3252 00399 1.3247 18174 1.3247 17957 1.3247 17957
x
1.5 1.3478 (1, −1)
Figure 4.49 The first three xvalues in Table 4.1 (four decimal places).
In Figure 4.50 we have indicated that the process in Example 2 might have started at the point B0(3, 23) on the curve, with x0 = 3. Point B0 is quite far from the xaxis, but the tangent at B0 crosses the xaxis at about (2.12, 0), so x1 is still an improvement over x0. If we use Equation (1) repeatedly as before, with ƒ(x) = x3  x  1 and ƒ′(x) = 3x2  1, we obtain the nineplace solution x7 = x6 = 1.3247 17957 in seven steps.
Convergence of the Approximations In Chapter 10 we define precisely the idea of convergence for the approximations xn in Newton’s method. Intuitively, we mean that as the number n of approximations increases without bound, the values xn get arbitrarily close to the desired root r. (This notion is similar to the idea of the limit of a function g(t) as t approaches infinity, as defined in Section 2.6.) In practice, Newton’s method usually gives convergence with impressive speed, but this is not guaranteed. One way to test convergence is to begin by graphing the function to estimate a good starting value for x0. You can test that you are getting closer to a zero of the function by evaluating 0 ƒ(xn) 0 , and check that the approximations are converging by evaluating 0 xn  xn + 1 0 . Newton’s method does not always converge. For instance, if
y 25 B0(3, 23) 20 y = x3 − x − 1 15
10 B1(2.12, 6.35) 5 −1" 3
−1
0
ƒ(x) = e
Root sought 1" 3
x2 x1 1
1.6 2.12
x0 3
Figure 4.50 Any starting value x0 to the right of x = 1> 23 will lead to the root in Example 2.
x
 2r  x, 2x  r,
x 6 r x Ú r,
the graph will be like the one in Figure 4.51. If we begin with x0 = r  h, we get x1 = r + h, and successive approximations go back and forth between these two values. No amount of iteration brings us closer to the root than our first guess. If Newton’s method does converge, it converges to a root. Be careful, however. There are situations in which the method appears to converge but no root is there. Fortunately, such situations are rare.
4.7 Newton’s Method
When Newton’s method converges to a root, it may not be the root you have in mind. Figure 4.52 shows two ways this can happen.
y y = f (x)
r 0
x0
x1
279
x
y = f (x) Starting point Root sought
Figure 4.51 Newton’s method fails to converge. You go from x0 to x1 and back to x0, never getting any closer to r.
x0
Root found
y = f (x) x1
x
x1
x2
Root found
x
x0 Root sought
Starting point
Figure 4.52 If you start too far away, Newton’s method may miss the root you want.
Exercises 4.7 Root Finding 1. Use Newton’s method to estimate the solutions of the equation x2 + x  1 = 0. Start with x0 =  1 for the lefthand solution and with x0 = 1 for the solution on the right. Then, in each case, find x2.
2. Use Newton’s method to estimate the one real solution of x3 + 3x + 1 = 0. Start with x0 = 0 and then find x2. 3. Use Newton’s method to estimate the two zeros of the function ƒ(x) = x4 + x  3. Start with x0 = 1 for the lefthand zero and with x0 = 1 for the zero on the right. Then, in each case, find x2. 4. Use Newton’s method to estimate the two zeros of the function ƒ(x) = 2x  x2 + 1. Start with x0 = 0 for the lefthand zero and with x0 = 2 for the zero on the right. Then, in each case, find x2. 5. Use Newton’s method to find the positive fourth root of 2 by solving the equation x4  2 = 0. Start with x0 = 1 and find x2. 6. Use Newton’s method to find the negative fourth root of 2 by solving the equation x4  2 = 0. Start with x0 = 1 and find x2. 7. Guessing a root Suppose that your first guess is lucky, in the sense that x0 is a root of ƒ(x) = 0. Assuming that ƒ′(x0) is defined and not 0, what happens to x1 and later approximations? 8. Estimating pi You plan to estimate p>2 to five decimal places by using Newton’s method to solve the equation cos x = 0. Does it matter what your starting value is? Give reasons for your answer. Theory and Examples 9. Oscillation Show that if h 7 0, applying Newton’s method to
ƒ(x) = e
2x,
2 x,
x Ú 0 x 6 0
leads to x1 = h if x0 = h and to x1 = h if x0 = h. Draw a picture that shows what is going on.
10. Approximations that get worse and worse Apply Newton’s method to ƒ(x) = x1>3 with x0 = 1 and calculate x1, x2, x3, and x4. Find a formula for 0 xn 0 . What happens to 0 xn 0 as n S q? Draw a picture that shows what is going on.
11. Explain why the following four statements ask for the same information: i) Find the roots of ƒ(x) = x3  3x  1. ii) Find the xcoordinates of the intersections of the curve y = x3 with the line y = 3x + 1. iii) Find the xcoordinates of the points where the curve y = x3  3x crosses the horizontal line y = 1. iv) F ind the values of x where the derivative of g(x) = (1>4)x4  (3>2)x2  x + 5 equals zero. 12. Locating a planet To calculate a planet’s space coordinates, we have to solve equations like x = 1 + 0.5 sin x. Graphing the function ƒ(x) = x  1  0.5 sin x suggests that the function has a root near x = 1.5. Use one application of Newton’s method to improve this estimate. That is, start with x0 = 1.5 and find x1. (The value of the root is 1.49870 to five decimal places.) Remember to use radians. T 13. Intersecting curves The curve y = tan x crosses the line y = 2x between x = 0 and x = p>2. Use Newton’s method to find where. T 14. Real solutions of a quartic Use Newton’s method to find the two real solutions of the equation x4  2x3  x2  2x + 2 = 0. ow many solutions does the equation sin 3x = 0.99  x2 T 15. a. H have? b. Use Newton’s method to find them. 16. Intersection of curves a. Does cos 3x ever equal x? Give reasons for your answer. b. Use Newton’s method to find where. 17. Find the four real zeros of the function ƒ(x) = 2x4  4x2 + 1.
280
Chapter 4: Applications of Derivatives
T 18. Estimating pi Estimate p to as many decimal places as your calculator will display by using Newton’s method to solve the equation tan x = 0 with x0 = 3. 19. Intersection of curves At what value(s) of x does cos x = 2x? 20. Intersection of curves At what value(s) of x does cos x =  x? 21. The graphs of y = x2(x + 1) and y = 1>x (x 7 0) intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places. y
the submarine travels on the parabolic path y = x2 and that the buoy is located at the point (2, 1>2). a. Show that the value of x that minimizes the distance between the submarine and the buoy is a solution of the equation x = 1>(x2 + 1). b. Solve the equation x = 1>(x2 + 1) with Newton’s method. y y = x2
y = x 2(x + 1)
3 2
ar, 1rb
1 0
−1
Submarine track in two dimensions
1 CPA y = 1x
1
2
1
0
2
x
1 Sonobuoy a2, − 2b
x
22. The graphs of y = 2x and y = 3  x2 intersect at one point x = r. Use Newton’s method to estimate the value of r to four decimal places. 2
23. Intersection of curves At what value(s) of x does ex = x2  x + 1?
T 29. Curves that are nearly flat at the root Some curves are so flat that, in practice, Newton’s method stops too far from the root to give a useful estimate. Try Newton’s method on ƒ(x) = (x  1)40 with a starting value of x0 = 2 to see how close your machine comes to the root x = 1. See the accompanying graph.
24. Intersection of curves At what value(s) of x does ln (1  x2) = x  1?
y
25. Use the Intermediate Value Theorem from Section 2.5 to show that ƒ(x) = x3 + 2x  4 has a root between x = 1 and x = 2. Then find the root to five decimal places. 26. Factoring a quartic Find the approximate values of r1 through r4 in the factorization 8x4  14x3  9x2 + 11x  1 = 8(x  r1)(x  r2)(x  r3)(x  r4). y
y = 8x 4 − 14x 3 − 9x 2 + 11x − 1
−2 −4 −6 −8 −10 −12
Slope = −40
Slope = 40
1
2 −1
y = (x − 1) 40
1
2
T 27. Converging to different zeros Use Newton’s method to find the zeros of ƒ(x) = 4x4  4x2 using the given starting values. a. x0 =  2 and x0 = 0.8, lying in 1  q,  22>2 2 b. x0 =  0.5 and x0 = 0.25, lying in 1  221>7, 221>7 2 c. x0 = 0.8 and x0 = 2, lying in 1 22>2, q 2 d. x0 =  221>7 and x0 = 221>7
(2, 1)
x
28. The sonobuoy problem In submarine location problems, it is often necessary to find a submarine’s closest point of approach (CPA) to a sonobuoy (sound detector) in the water. Suppose that
Nearly flat 0
1
2
x
30. The accompanying figure shows a circle of radius r with a chord of length 2 and an arc s of length 3. Use Newton’s method to solve for r and u (radians) to four decimal places. Assume 0 6 u 6 p.
s=3
r u r
2
4.8 Antiderivatives
281
4.8 Antiderivatives We have studied how to find the derivative of a function and how to use it to solve a wide range of problems. However, many other problems require that we recover a function from its known derivative (from its known rate of change). For instance, the laws of physics tell us the acceleration of an object falling from an initial height, and we can use this to compute its velocity and its height at any time. More generally, starting with a function ƒ, we want to find a function F whose derivative is ƒ. If such a function F exists, it is called an antiderivative of ƒ. We will see in the next chapter that antiderivatives are the link connecting the two major elements of calculus: derivatives and definite integrals.
Finding Antiderivatives Definition A function F is an antiderivative of ƒ on an interval I if F′(x) = ƒ(x) for all x in I.
The process of recovering a function F(x) from its derivative ƒ(x) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function ƒ, G to represent an antiderivative of g, and so forth.
Example 1 Find an antiderivative for each of the following functions. 1
(a) ƒ(x) = 2x (b) g(x) = cos x (c) h(x) = x + 2e2x
Solution We need to think backward here: What function do we know has a derivative equal to the given function? (a) F(x) = x2 (b) G(x) = sin x (c) H(x) = ln 0 x 0 + e2x
Each answer can be checked by differentiating. The derivative of F(x) = x2 is 2x. The derivative of G(x) = sin x is cos x, and the derivative of H(x) = ln 0 x 0 + e2x is (1>x) + 2e2x.
The function F(x) = x2 is not the only function whose derivative is 2x. The function x + 1 has the same derivative. So does x2 + C for any constant C. Are there others? Corollary 2 of the Mean Value Theorem in Section 4.2 gives the answer: Any two antiderivatives of a function differ by a constant. So the functions x2 + C, where C is an arbitrary constant, form all the antiderivatives of ƒ(x) = 2x. More generally, we have the following result. 2
Theorem 8 If F is an antiderivative of ƒ on an interval I, then the most general antiderivative of ƒ on I is F(x) + C where C is an arbitrary constant.
Thus the most general antiderivative of ƒ on I is a family of functions F(x) + C whose graphs are vertical translations of one another. We can select a particular antiderivative from this family by assigning a specific value to C. Here is an example showing how such an assignment might be made.
282
Chapter 4: Applications of Derivatives
Example 2 Find an antiderivative of ƒ(x) = 3x2 that satisfies F(1) = 1.
y C=2 C=1 C=0 C = −1 C = −2
y = x3 + C 2 1
F(x) = x3 + C gives all the antiderivatives of ƒ(x). The condition F(1) = 1 determines a specific value for C. Substituting x = 1 into F(x) = x3 + C gives
x
0 −1
Solution Since the derivative of x3 is 3x2 , the general antiderivative
F(1) = (1)3 + C = 1 + C.
(1, −1)
Since F(1) = 1, solving 1 + C = 1 for C gives C = 2. So
−2
F(x) = x3  2
Figure 4.53 The curves y = x3 + C fill the coordinate plane without overlapping. In Example 2, we identify the curve y = x3  2 as the one that passes through the given point (1,  1).
is the antiderivative satisfying F(1) = 1. Notice that this assignment for C selects the particular curve from the family of curves y = x3 + C that passes through the point (1, 1) in the plane (Figure 4.53). By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrary constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number of important functions. The rules in Table 4.2 are easily verified by differentiating the general antiderivative formula to obtain the function to its left. For example, the derivative of (tan kx)>k + C is sec2 kx, whatever the value of the constants C or k ≠ 0, and this establishes Formula 4 for the most general antiderivative of sec2 kx.
Example 3 Find the general antiderivative of each of the following functions. 1
(a) ƒ(x) = x5
(b) g(x) =
x (d) i(x) = cos 2
(e) j(x) = e3x
2x
(c) h(x) = sin 2x
(f) k(x) = 2x
Table 4.2 Antiderivative formulas, k a nonzero constant
Function
General antiderivative
Function
General antiderivative
1. xn
1 xn + 1 + C, n ≠ 1 n + 1
8. ekx
1 kx e + C k
2. sin kx
1  cos kx + C k
1 9. x
3. cos kx
1 sin kx + C k
10.
ln 0 x 0 + C, x ≠ 0
4. sec2 kx
1 tan kx + C k
11.
5. csc2 kx
1  cot kx + C k
6. sec kx tan kx
1 sec kx + C k
7. csc kx cot kx
1  csc kx + C k
12.
1 21  k 2x2
1 1 sin kx + C k
1 1 + k 2x2
1 tan1 kx + C k
1 x 2k 2x2  1
sec1 kx + C, kx 7 1
13. akx
a
1 b akx + C, a 7 0, a ≠ 1 k ln a
4.8 Antiderivatives
283
Solution In each case, we can use one of the formulas listed in Table 4.2. x6 + C 6 ( b) g(x) = x1>2, so
Formula 1 with n = 5
x1>2 + C = 2 2x + C 1>2 cos 2x (c) H(x) = + C 2 sin (x>2) x ( d) I(x) = + C = 2 sin + C 2 1>2
Formula 1 with n =  1>2
1 (e) J(x) =  e3x + C 3
Formula 8 with k =  3
(a) F(x) =
G(x) =
(f) K(x) = a
Formula 2 with k = 2 Formula 3 with k = 1>2
1 b 2x + C ln 2
Formula 13 with a = 2, k = 1
Other derivative rules also lead to corresponding antiderivative rules. We can add and subtract antiderivatives and multiply them by constants.
Table 4.3 Antiderivative linearity rules
Function
General antiderivative
1. Constant Multiple Rule: 2. Negative Rule: 3. Sum or Difference Rule:
kƒ(x) ƒ(x) ƒ(x) { g(x)
kF(x) + C, k a constant F(x) + C F(x) { G(x) + C
The formulas in Table 4.3 are easily proved by differentiating the antiderivatives and verifying that the result agrees with the original function. Formula 2 is the special case k = 1 in Formula 1.
Example 4 Find the general antiderivative of ƒ(x) =
3 2x
+ sin 2x.
Solution We have that ƒ(x) = 3g(x) + h(x) for the functions g and h in Example 3. Since G(x) = 2 2x is an antiderivative of g(x) from Example 3b, it follows from the Constant Multiple Rule for antiderivatives that 3G(x) = 3 # 2 2x = 6 2x is an antiderivative of 3g(x) = 3> 2x. Likewise, from Example 3c we know that H(x) = (1>2) cos 2x is an antiderivative of h(x) = sin 2x. From the Sum Rule for antiderivatives, we then get that F(x) = 3G(x) + H(x) + C 1 = 6 2x  cos 2x + C 2 is the general antiderivative formula for ƒ(x), where C is an arbitrary constant.
Initial Value Problems and Differential Equations Antiderivatives play several important roles in mathematics and its applications. Methods and techniques for finding them are a major part of calculus, and we take up that study in
284
Chapter 4: Applications of Derivatives
Chapter 8. Finding an antiderivative for a function ƒ(x) is the same problem as finding a function y(x) that satisfies the equation dy = ƒ(x). dx This is called a differential equation, since it is an equation involving an unknown function y that is being differentiated. To solve it, we need a function y(x) that satisfies the equation. This function is found by taking the antiderivative of ƒ(x). We can fix the arbitrary constant arising in the antidifferentiation process by specifying an initial condition y(x0) = y0. This condition means the function y(x) has the value y0 when x = x0. The combination of a differential equation and an initial condition is called an initial value problem. Such problems play important roles in all branches of science. The most general antiderivative F(x) + C (such as x3 + C in Example 2) of the function ƒ(x) gives the general solution y = F(x) + C of the differential equation dy>dx = ƒ(x). The general solution gives all the solutions of the equation (there are infinitely many, one for each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by finding the particular solution that satisfies the initial condition y(x0) = y0. In Example 2, the function y = x3  2 is the particular solution of the differential equation dy>dx = 3x2 satisfying the initial condition y(1) = 1.
Antiderivatives and Motion We have seen that the derivative of the position function of an object gives its velocity, and the derivative of its velocity function gives its acceleration. If we know an object’s acceleration, then by finding an antiderivative we can recover the velocity, and from an antiderivative of the velocity we can recover its position function. This procedure was used as an application of Corollary 2 in Section 4.2. Now that we have a terminology and conceptual framework in terms of antiderivatives, we revisit the problem from the point of view of differential equations.
Example 5 A hotair balloon ascending at the rate of 12 ft>sec is at a height 80 ft above the ground when a package is dropped. How long does it take the package to reach the ground? s y(0) = 12
Solution Let y(t) denote the velocity of the package at time t, and let s(t) denote its height above the ground. The acceleration of gravity near the surface of the earth is 32 ft>sec2. Assuming no other forces act on the dropped package, we have Negative because gravity acts in the dy = 32. direction of decreasing s dt
This leads to the following initial value problem (Figure 4.54):
dy = −32 dt
Differential equation:
Initial condition:
dy = 32 dt y(0) = 12. Balloon initially rising
This is our mathematical model for the package’s motion. We solve the initial value problem to obtain the velocity of the package.
s(t)
1. Solve the differential equation: The general formula for an antiderivative of 32 is 0
ground
Figure 4.54 A package dropped from a rising hotair balloon (Example 5).
y = 32t + C.
Having found the general solution of the differential equation, we use the initial condition to find the particular solution that solves our problem.
4.8 Antiderivatives
285
2. Evaluate C: 12 = 32(0) + C Initial condition y(0)
= 12
C = 12.
The solution of the initial value problem is y = 32t + 12.
Since velocity is the derivative of height, and the height of the package is 80 ft at time t = 0 when it is dropped, we now have a second initial value problem: Differential equation: Initial condition:
ds = 32t + 12 Set y dt s(0) = 80.
= ds>dt in the previous equation.
We solve this initial value problem to find the height as a function of t. 1. Solve the differential equation: Finding the general antiderivative of 32t + 12 gives s = 16t 2 + 12t + C. 2. Evaluate C: 80 = 16(0)2 + 12(0) + C Initial condition s(0) C = 80.
= 80
The package’s height above ground at time t is s = 16t 2 + 12t + 80.
Use the solution: To find how long it takes the package to reach the ground, we set s equal to 0 and solve for t: 16t 2 + 12t + 80 = 0 4t 2 + 3t + 20 = 0 3 { 2329 Quadratic formula 8 t ≈ 1.89, t ≈ 2.64. t =
The package hits the ground about 2.64 sec after it is dropped from the balloon. (The negative root has no physical meaning.)
Indefinite Integrals A special symbol is used to denote the collection of all antiderivatives of a function ƒ.
Definition The collection of all antiderivatives of ƒ is called the indefinite integral of ƒ with respect to x, and is denoted by ƒ(x) dx. L The symbol 1 is an integral sign. The function ƒ is the integrand of the integral, and x is the variable of integration.
After the integral sign in the notation we just defined, the integrand function is always followed by a differential to indicate the variable of integration. We will have more to say
286
Chapter 4: Applications of Derivatives
about why this is important in Chapter 5. Using this notation, we restate the solutions of Example 1, as follows: L
2x dx = x2 + C,
L
cos x dx = sin x + C,
L This notation is related to the main application of antiderivatives, which will be explored in Chapter 5. Antiderivatives play a key role in computing limits of certain infinite sums, an unexpected and wonderfully useful role that is described in a central result of Chapter 5, called the Fundamental Theorem of Calculus. 1 a x + 2e2x b dx = ln 0 x 0 + e2x + C.
Example 6 Evaluate L
(x2  2x + 5) dx.
Solution If we recognize that (x3 >3)  x2 + 5x is an antiderivative of x2  2x + 5, we can evaluate the integral as antiderivative
$++%++& x3 (x2  2x + 5) dx =  x2 + 5x + " C. 3 L arbitrary constant
If we do not recognize the antiderivative right away, we can generate it termbyterm with the Sum, Difference, and Constant Multiple Rules: L
(x2  2x + 5) dx =
L
x2 dx 
=
L
x2 dx  2 x dx + 5 1 dx L L
= a =
L
2x dx +
L
5 dx
x3 x2 + C1 b  2a + C2 b + 5(x + C3) 3 2
x3 + C1  x2  2C2 + 5x + 5C3. 3
This formula is more complicated than it needs to be. If we combine C1, 2C2, and 5C3 into a single arbitrary constant C = C1  2C2 + 5C3, the formula simplifies to x3  x2 + 5x + C 3 and still gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate termbyterm. Write L
(x2  2x + 5) dx = =
L
x2 dx 
L
2x dx +
L
5 dx
x3  x2 + 5x + C. 3
Find the simplest antiderivative you can for each part and add the arbitrary constant of integration at the end.
4.8 Antiderivatives
287
Exercises 4.8 Finding Antiderivatives In Exercises 1–24, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
25.
L
x2 c. x2  2x + 1 1. a. 2x b.
27.
L
29.
L
x7 c. x7  6x + 8 2. a. 6x b. 4
3. a.  3x
4
b. x
c. x
3
4
+ 2x + 3
2 9. a. x1>3 3
x + x2 2 5 b. 2 x 1 b. 3 2x 1 b. 2 2x 1 b. 3 3 2x 1 b. x2>3 3
1 10. a. x1>2 2
1 b.  x3>2 2
3 c.  x5>2 2
1 11. a. x
7 b. x
5 c. 1  x
4. a. 2x3 1 5. a. 2 x 2 6. a.  3 x 3 7. a. 2x 2 4 3 8. a. 2 x 3
12. a.
1 3x
b.
b.
2 5x
c. x3 + x  1 31.
5 c. 2  2 x 1 c. x3  3 x
33.
41.
L
t 2t + 2t dt t2
43.
L
(2 cos t) dt
45.
L
7 sin
47.
23. a.
2 21  x2
1 x 24. a. x  a b 2
b.
b. x2 + 2x
L L
x5>4 dx a
2x
2
+
2 2x
b dx
1 1 b dy a L 7 y5>4
L
( 5 sin t) dt
u du 3
46.
L
3 cos 5u du
L
(3 csc2 x) dx
48.
L
49.
L
csc u cot u du 2
50.
2 sec u tan u du L5
51.
L
(e3x + 5ex) dx
52.
L
(2ex  3e2x) dx
53.
L
(ex + 4x) dx
54.
L
(1.3)x dx
c. ex>5
55.
(4 sec x tan x  2 sec2 x) dx
5 x c. a b 3
L
56.
1 (csc2 x  csc x cot x) dx L2
57.
L
(sin 2x  csc2 x) dx
58.
L
(2 cos 2x  3 sin 3x) dx
59.
L
1 + cos 4t dt 2
60.
L
1  cos 6t dt 2
61.
L
63.
L
1 ax 
px + p cos x 2 3x c. sec2 2 2
c. 1  8 csc 2x px px cot 2 2 px px tan c. sec 2 2
c. x 22  1
1 2(x + 1)
1 2 a  3 + 2xb dx L 5 x
44.
4 1 3x x2
c. ex>2
2
(1  x2  3x5) dx
4 + 2t dt t3
b. ex
b. xp
38.
L
L
19. a. e3x
22. a. x 23
2 b dy y1>4
t2 + 4t 3 b dt L 2 a
42.
b. 4 sec 3x tan 3x
b. 2x
a8y 
(5  6x) dx
x3(x + 1) dx
18. a. sec x tan x
21. a. 3x
3 + 2 x 2 dx
L
L
c.  p csc
36.
40.
b.  csc 5x cot 5x
b. e
1 2x
2x(1  x3) dx
17. a. csc x cot x
34.
L
c. cos
20. a. e
x1>3 dx
39.
p px cos 2 2 x 2 b. sec2 3 3 3 3x b.  csc2 2 2
4x>3
L
32.
1 c.  x4>3 3
2x
14. a. p cos px
2x
1 1 a 2  x2  b dx 3 L x
30.
L
1
3
c. sin px  3 sin 3x
16. a. csc x
(2x3  5x + 7) dx
28.
37.
3 c. 2 x +
2x
b. 3 sin x
2
t b dt 2
L
1
13. a.  p sin px
15. a. sec2 x
a3t 2 +
26.
35.
c. 2x +
c. 1 +
b.
(x + 1) dx
c.
1 1 + 4x2
c. px  x1
Finding Indefinite Integrals In Exercises 25–70, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
5 b dx x2 + 1
3x 23 dx
62. 64.
a
sec2 x b dx 3
2 1 a  1>4 b dy y L 21  y2 L
x 22  1 dx
288
Chapter 4: Applications of Derivatives
L (Hint: 1 + tan2 u = sec2 u) (1 + tan2 u) du
65.
cot2 x dx L (Hint: 1 + cot2 x = csc2 x) 67.
L
66.
L
68.
(2 + tan2 u) du
(1  cot2 x) dx
csc u cos u (tan u + sec u) du 70. du 69. L L csc u  sin u Checking Antiderivative Formulas Verify the formulas in Exercises 71–82 by differentiation.
71.
L
(7x  2)3 dx =
(7x  2)4 + C 28 (3x + 5) 3
1
L
(3x + 5)2 dx = 
73.
L
sec2 (5x  1) dx =
74.
L
75.
1 1 + C dx = 2 x + 1 L (x + 1)
76.
x 1 + C dx = 2 x + 1 (x + 1) L
72.
77. 78.
csc2 a
+ C
1 tan (5x  1) + C 5
x  1 x  1 b dx = 3 cot a b + C 3 3
1 dx = ln 0 x + 1 0 + C, x ≠  1 Lx + 1 L
84. Right, or wrong? Say which for each formula and give a brief reason for each answer. a. b.
L
tan u sec2 u du =
sec3 u + C 3
L
tan u sec2 u du =
1 2 tan u + C 2
1 tan u sec2 u du = sec2 u + C 2 L 85. Right, or wrong? Say which for each formula and give a brief reason for each answer. (2x + 1)3 a. + C (2x + 1)2 dx = 3 L c.
b.
L
3(2x + 1)2 dx = (2x + 1)3 + C
c.
L
6(2x + 1)2 dx = (2x + 1)3 + C
86. Right, or wrong? Say which for each formula and give a brief reason for each answer. a.
L
22x + 1 dx = 2x2 + x + C
b.
L
22x + 1 dx = 2x2 + x + C
1 22x + 1 dx = 1 22x + 1 23 + C 3 L 87. Right, or wrong? Give a brief reason why. c.
15(x + 3)2 x + 3 3 dx = a b + C 4 x  2 L (x  2)
xex dx = xex  ex + C
88. Right, or wrong? Give a brief reason why.
x dx 1 79. = a tan1 a a b + C 2 2 La + x 80. 81. 82.
x dx = sin1 a a b + C 2 L 2a  x 2
tan1 x tan1 x 1 dx = ln x  ln (1 + x2) + C x 2 2 L x L
x cos (x2)  sin (x2) sin (x2) dx = + C x 2 x L Initial Value Problems 89. Which of the following graphs shows the solution of the initial value problem
dy = 2x, y = 4 when x = 1? dx
(sin1 x)2 dx = x(sin1 x)2  2x + 2 21  x2 sin1 x + C
y
y
y
83. Right, or wrong? Say which for each formula and give a brief reason for each answer. a. b. c.
L
x sin x dx =
x2 sin x + C 2
L
x sin x dx =  x cos x + C
L
x sin x dx =  x cos x + sin x + C
4
(1, 4)
3
4
(1, 4)
4
3 2
3
2 1
1
1
−1 0 (a)
1
x
−1 0 (b)
Give reasons for your answer.
(1, 4)
2
1
x
−1 0 (c)
1
x
289
4.8 Antiderivatives
90. Which of the following graphs shows the solution of the initial value problem dy =  x, y = 1 when x =  1? dx y
y
(−1, 1) 0
x
(a)
112. y(4) =  cos x + 8 sin 2x ; y‴(0) = 0, y″(0) = y′(0) = 1, y(0) = 3
y
(−1, 1)
(−1, 1) 0
x
(b)
Give reasons for your answer. Solve the initial value problems in Exercises 91–112. dy = 2x  7, y(2) = 0 91. dx dy = 10  x, y(0) = 1 92. dx dy 1 = + x, x 7 0; y(2) = 1 93. dx x2 dy = 9x2  4x + 5, y( 1) = 0 94. dx dy = 3x2>3, y( 1) =  5 95. dx dy 1 = , y(4) = 0 96. dx 2 2x
97.
ds = 1 + cos t, s(0) = 4 dt
98.
ds = cos t + sin t, s(p) = 1 dt
99.
dr =  p sin pu, r(0) = 0 du
0 (c)
d3 u 1 = 0; u″(0) = 2, u′(0) =  , u(0) = 22 2 dt 3 111. y(4) =  sin t + cos t ; y‴(0) = 7, y″(0) = y′(0) =  1, y(0) = 0 110.
x
113. Find the curve y = ƒ(x) in the xyplane that passes through the point (9, 4) and whose slope at each point is 3 2x. 114. a. Find a curve y = ƒ(x) with the following properties: d 2y i) 2 = 6x dx ii) Its graph passes through the point (0, 1) and has a horizontal tangent there. b. How many curves like this are there? How do you know? Solution (Integral) Curves Exercises 115–118 show solution curves of differential equations. In each exercise, find an equation for the curve through the labeled point.
115.
2
x
1
0
x
2
−1 −1
117.
118. dy = sin x − cos x dx y
y 6
dy = 1 + psin px dx 2" x
1 0 (−p, −1)
103.
d 2y 106. 2 = 0; y′(0) = 2, y(0) = 0 dx
1
1 (−1, 1) −1
p dy = 8t + csc2 t, y a b = 7 102. 2 dt
d 2y = 2  6x; y′(0) = 4, y(0) = 1 dx2
(1, 0.5) 0
dy 1 = sec t tan t, y(0) = 1 101. 2 dt
105.
y dy = x − 1 dx 2
1
dr = cos pu, r(0) = 1 100. du
dy 3 = , t 7 1, y(2) = 0 dt t 2t 2  1 dy 8 = + sec2 t, y(0) = 1 104. dt 1 + t2
116. dy = 1 − 4 x13 3 dx
y
2
x
4
2
0
(1, 2)
1
2
3
x
−2
Applications 119. Finding displacement from an antiderivative of velocity
a. Suppose that the velocity of a body moving along the saxis is ds = y = 9.8t  3. dt
107.
d 2r 2 dr 2 = 3; = 1, r(1) = 1 dt t = 1 dt 2 t
108.
d 2s 3t ds 2 = ; = 3, s(4) = 4 8 dt t = 4 dt 2
109.
d 3y = 6; y″(0) = 8, y′(0) = 0, y(0) = 5 dx3
i) Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0. ii) Find the body’s displacement from t = 1 to t = 3 given that s =  2 when t = 0. iii) Now find the body’s displacement from t = 1 to t = 3 given that s = s0 when t = 0.
290
Chapter 4: Applications of Derivatives
b. Suppose that the position s of a body moving along a coordinate line is a differentiable function of time t. Is it true that once you know an antiderivative of the velocity function ds>dt you can find the body’s displacement from t = a to t = b even if you do not know the body’s exact position at either of those times? Give reasons for your answer. 120. Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 m>sec2. How fast will the rocket be going 1 min later? 121. Stopping a car in time You are driving along a highway at a steady 60 mph (88 ft>sec) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 242 ft? To find out, carry out the following steps.
1. Solve the initial value problem Differential equation: Initial conditions:
d 2s = k dt 2
(k constant)
ds = 88 and s = 0 when t = 0. dt
Measuring time and distance from when the brakes are applied
2. Find the value of t that makes ds>dt = 0. (The answer will
where y0 and s0 are the body’s velocity and position at time t = 0. Derive this equation by solving the initial value problem Differential equation: Initial conditions:
1 s =  gt 2 + y0 t + s0 ,(2) 2
where s is the body’s height above the surface. The equation has a minus sign because the acceleration acts downward, in the direction of decreasing s. The velocity y0 is positive if the object is rising at time t = 0 and negative if the object is falling. Instead of using the result of Exercise 125, you can derive Equation (2) directly by solving an appropriate initial value problem. What initial value problem? Solve it to be sure you have the right one, explaining the solution steps as you go along. 127. Suppose that ƒ(x) =
3. Find the value of k that makes s = 242 for the value of t you 122. Stopping a motorcycle The State of Illinois Cycle Rider Safety Program requires motorcycle riders to be able to brake from 30 mph (44 ft>sec) to 0 in 45 ft. What constant deceleration does it take to do that? 123. Motion along a coordinate line A particle moves on a coordinate line with acceleration a = d 2s>dt 2 = 15 2t  1 3> 2t 2, subject to the conditions that ds>dt = 4 and s = 0 when t = 1. Find a. the velocity y = ds>dt in terms of t b. the position s in terms of t. T 124. The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 ft. How long did it take the hammer and feather to fall 4 ft on the moon? To find out, solve the following initial value problem for s as a function of t. Then find the value of t that makes s equal to 0. d 2s Differential equation: =  5.2 ft>sec2 dt 2 Initial conditions:
ds = 0 and s = 4 when t = 0 dt
125. Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is a s = t 2 + y0 t + s0 ,(1) 2
ds = y0 and s = s0 when t = 0. dt
126. Free fall near the surface of a planet For free fall near the surface of a planet where the acceleration due to gravity has a constant magnitude of g [email protected]>sec2, Equation (1) in Exercise 125 takes the form
involve k.)
found in Step 2.
d 2s = a dt 2
Find:
d d 1 1  2x 2 and g(x) = dx (x + 2). dx
a.
L
ƒ(x) dx
b.
L
c.
L
[ƒ(x)] dx
d.
L
e.
L
3ƒ(x) + g(x) 4 dx
f.
L
g(x) dx 3g(x) 4 dx 3ƒ(x)  g(x) 4 dx
128. Uniqueness of solutions If differentiable functions y = F(x) and y = g(x) both solve the initial value problem dy = ƒ(x), dx
y(x0) = y0,
on an interval I, must F(x) = G(x) for every x in I? Give reasons for your answer. Computer Explorations Use a CAS to solve the initial value problems in Exercises 129–132. Plot the solution curves.
129. y′ = cos2 x + sin x, y(p) = 1 1 130. y′ = x + x, y(1) =  1 1 , y(0) = 2 24  x2 2 132. y″ = x + 2x, y(1) = 0, y′(1) = 0
131. y′ =
Chapter 4 Practice Exercises
Chapter 4
291
Questions to Guide Your Review
1. What can be said about the extreme values of a function that is continuous on a closed interval? 2. What does it mean for a function to have a local extreme value on its domain? An absolute extreme value? How are local and absolute extreme values related, if at all? Give examples.
14. What is a cusp? Give examples. 15. List the steps you would take to graph a rational function. Illustrate with an example. 16. Outline a general strategy for solving maxmin problems. Give examples.
3. How do you find the absolute extrema of a continuous function on a closed interval? Give examples.
17. Describe l’Hôpital’s Rule. How do you know when to use the rule and when to stop? Give an example.
4. What are the hypotheses and conclusion of Rolle’s Theorem? Are the hypotheses really necessary? Explain.
18. How can you sometimes handle limits that lead to indeterminate forms q > q, q # 0, and q  q ? Give examples.
5. What are the hypotheses and conclusion of the Mean Value Theorem? What physical interpretations might the theorem have? 6. State the Mean Value Theorem’s three corollaries.
19. How can you sometimes handle limits that lead to indeterminate forms 1q, 00, and qq ? Give examples.
7. How can you sometimes identify a function ƒ(x) by knowing ƒ′ and knowing the value of ƒ at a point x = x0? Give an example.
20. Describe Newton’s method for solving equations. Give an example. What is the theory behind the method? What are some of the things to watch out for when you use the method?
8. What is the First Derivative Test for Local Extreme Values? Give examples of how it is applied.
21. Can a function have more than one antiderivative? If so, how are the antiderivatives related? Explain.
9. How do you test a twicedifferentiable function to determine where its graph is concave up or concave down? Give examples.
22. What is an indefinite integral? How do you evaluate one? What general formulas do you know for finding indefinite integrals?
10. What is an inflection point? Give an example. What physical significance do inflection points sometimes have?
23. How can you sometimes solve a differential equation of the form dy>dx = ƒ(x)?
11. What is the Second Derivative Test for Local Extreme Values? Give examples of how it is applied.
24. What is an initial value problem? How do you solve one? Give an example.
12. What do the derivatives of a function tell you about the shape of its graph?
25. If you know the acceleration of a body moving along a coordinate line as a function of time, what more do you need to know to find the body’s position function? Give an example.
13. List the steps you would take to graph a polynomial function. Illustrate with an example.
Chapter 4
Practice Exercises
Extreme Values 1. Does ƒ(x) = x3 + 2x + tan x have any local maximum or minimum values? Give reasons for your answer.
6. Does ƒ(x) = 2ex >(1 + x2) have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of ƒ.
2. Does g(x) = csc x + 2 cot x have any local maximum values? Give reasons for your answer.
In Exercises 7 and 8, find the absolute maximum and absolute minimum values of ƒ over the interval.
3. Does ƒ(x) = (7 + x)(11  3x)1>3 have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of ƒ. 4. Find values of a and b such that the function ƒ(x) =
ax + b x2  1
has a local extreme value of 1 at x = 3. Is this extreme value a local maximum, or a local minimum? Give reasons for your answer. 5. Does g(x) = ex  x have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of g.
7. ƒ(x) = x  2 ln x, 1 … x … 3 8. ƒ(x) = (4>x) + ln x2, 1 … x … 4 9. The greatest integer function ƒ(x) = : x ; , defined for all values of x, assumes a local maximum value of 0 at each point of 30, 1). Could any of these local maximum values also be local minimum values of ƒ? Give reasons for your answer. 10. a. Give an example of a differentiable function ƒ whose first derivative is zero at some point c even though ƒ has neither a local maximum nor a local minimum at c. b. How is this consistent with Theorem 2 in Section 4.1? Give reasons for your answer.
292
Chapter 4: Applications of Derivatives
11. The function y = 1>x does not take on either a maximum or a minimum on the interval 0 6 x 6 1 even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for continuous functions? Why?
that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal>min. (An acrefoot is 43,560 ft3, the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.)
12. What are the maximum and minimum values of the function y = 0 x 0 on the interval  1 … x 6 1? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why?
20. The formula F(x) = 3x + C gives a different function for each value of C. All of these functions, however, have the same derivative with respect to x, namely F′(x) = 3. Are these the only differentiable functions whose derivative is 3? Could there be any others? Give reasons for your answers.
T 13. A graph that is large enough to show a function’s global behavior may fail to reveal important local features. The graph of ƒ(x) = (x8 >8)  (x6 >2)  x5 + 5x3 is a case in point. a. Graph ƒ over the interval  2.5 … x … 2.5. Where does the graph appear to have local extreme values or points of inflection?
b. Now factor ƒ′(x) and show that ƒ has a local maximum at 3 x =2 5 ≈ 1.70998 and local minima at x = { 23 ≈ {1.73205. c. Zoom in on the graph to find a viewing window that shows 3 the presence of the extreme values at x = 2 5 and x = 23.
The moral here is that without calculus the existence of two of the three extreme values would probably have gone unnoticed. On any normal graph of the function, the values would lie close enough together to fall within the dimensions of a single pixel on the screen. (Source: Uses of Technology in the Mathematics Curriculum, by Benny Evans and Jerry Johnson, Oklahoma State University, published in 1990 under a grant from the National Science Foundation, USE8950044.)
21. Show that
even though
d x d 1 a b = ab x + 1 dx x + 1 dx x 1 ≠ . x + 1 x + 1
Doesn’t this contradict Corollary 2 of the Mean Value Theorem? Give reasons for your answer. 22. Calculate the first derivatives of ƒ(x) = x2 >(x2 + 1) and g(x) =  1>(x2 + 1). What can you conclude about the graphs of these functions? Analyzing Graphs In Exercises 23 and 24, use the graph to answer the questions.
23. Identify any global extreme values of ƒ and the values of x at which they occur. y y = f (x)
T 14. (Continuation of Exercise 13.) a. Graph ƒ(x) = (x8 >8)  (2>5)x5  5x  (5>x2) + 11 over the interval  2 … x … 2. Where does the graph appear to have local extreme values or points of inflection?
(1, 1)
c. Zoom in to find a viewing window that shows the presence of 7 3 the extreme values at x = 2 5 and x = 2 2.
The Mean Value Theorem 15. a. Show that g(t) = sin2 t  3t decreases on every interval in its domain.
b. How many solutions does the equation sin2 t  3t = 5 have? Give reasons for your answer. 16. a. Show that y = tan u increases on every open interval in its domain. b. If the conclusion in part (a) is really correct, how do you explain the fact that tan p = 0 is less than tan (p>4) = 1? 17. a. S how that the equation x4 + 2x2  2 = 0 has exactly one solution on 30, 14 . T b. Find the solution to as many decimal places as you can.
18. a. S how that ƒ(x) = x>(x + 1) increases on every open interval in its domain. b. Show that ƒ(x) = x3 + 2x has no local maximum or minimum values.
19. Water in a reservoir As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acreft in 24 hours. Show
x
0
7
b. Show that ƒ has a local maximum value at x = 25 ≈ 1.2585 3 and a local minimum value at x = 2 2 ≈ 1.2599.
1 a2, 2b
24. Estimate the open intervals on which the function y = ƒ(x) is a. increasing. b. decreasing. c. Use the given graph of ƒ′ to indicate where any local extreme values of the function occur, and whether each extreme is a relative maximum or minimum. y (2, 3) y = f ′(x) (−3, 1) x −1 −2
Each of the graphs in Exercises 25 and 26 is the graph of the position function s = ƒ(t) of an object moving on a coordinate line (t represents time). At approximately what times (if any) is each object’s (a) velocity equal to zero? (b) Acceleration equal to zero? During approximately what time intervals does the object move (c) forward? (d) Backward?
Chapter 4 Practice Exercises
25. s
69. lim (csc x  cot x)
s = f (t) 0
xS0
6
3
9
12 14
s
26.
t
2
4
6
8
72. lim a x Sq
t
28. y = x3  3x2 + 3
77. lim
31. y = x3(8  x)
32. y = x2(2x2  9)
33. y = x  3x2>3
34. y = x1>3(x  4)
29. y =  x3 + 6x2  9x + 3
30. y = (1>8)(x3 + 3x2  9x  27)
35. y = x 23  x
37. y = (x  3) e
38. y = xe
39. y = ln (x2  4x + 3)
40. y = ln (sin x)
1 41. y = sin a x b
1 42. y = tan a x b
36. y = x 24  x
2
x2
1
Each of Exercises 43– 48 gives the first derivative of a function y = ƒ(x). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph. 43. y′ = 16  x2
44. y′ = x2  x  6
45. y′ = 6x(x + 1)(x  2)
46. y′ = x2(6  4x)
47. y′ = x4  2x2
48. y′ = 4x2  x4
In Exercises 49–52, graph each function. Then use the function’s first derivative to explain what you see. 49. y = x2>3 + (x  1)1>3
50. y = x2>3 + (x  1)2>3
51. y = x1>3 + (x  1)1>3
52. y = x2>3  (x  1)1>3
Sketch the graphs of the rational functions in Exercises 53–60. 53. y =
x + 1 x  3
54. y =
2x x + 5
55. y =
x2 + 1 x
56. y =
x2  x + 1 x
57. y =
x3 + 2 2x
58. y =
59. y =
x2  4 x2  3
x4  1 x2 x2 60. y = 2 x  4
Using L’Hôpital’s Rule Use l’Hôpital’s Rule to find the limits in Exercises 61–72. x2 + 3x  4 xa  1 61. lim 62. lim b S S x 1 x 1 x 1x  1 tan x tan x 63. lim x 64. lim xSp x S 0 x + sin x sin2 x sin mx 65. lim 66. lim x S 0 tan(x 2) x S 0 sin nx x S p>2
2sin x  1 x xS0 e  1
2sin x  1 x xS0 e  1
27. y = x2  (x3 >6)
67. lim  sec 7x cos 3x
x3 x3 b  2 x  1 x + 1 2
Find the limits in Exercises 73–84. 10x  1 3u  1 73. lim 74. lim x u xS0 uS0 75. lim
1
1 1  2b x4 x
x Sq
Graphs and Graphing Graph the curves in Exercises 27– 42.
2 x
xS0
71. lim 1 2x2 + x + 1  2x2  x 2
s = f (t)
0
70. lim a
293
68. lim+ 2x sec x xS0
76. lim
5  5 cos x  x  1 t  ln (1 + 2t) 79. lim+ tS0 t2 et 1 8 1. lim+ a t  t b tS0 x xS0 e
b kx 83. lim a1 + x b x Sq
4  4ex xex sin2 (px) 80. lim x  4 xS4 e + 3  x
78. lim
xS0
82. lim+ e1>y ln y yS0
7 2 84. lim a1 + x + 2 b x Sq x
Optimization 85. The sum of two nonnegative numbers is 36. Find the numbers if
a. the difference of their square roots is to be as large as possible. b. the sum of their square roots is to be as large as possible. 86. The sum of two nonnegative numbers is 20. Find the numbers a. if the product of one number and the square root of the other is to be as large as possible. b. if one number plus the square root of the other is to be as large as possible. 87. An isosceles triangle has its vertex at the origin and its base parallel to the xaxis with the vertices above the axis on the curve y = 27  x2. Find the largest area the triangle can have. 88. A customer has asked you to design an opentop rectangular stainless steel vat. It is to have a square base and a volume of 32 ft3, to be welded from quarterinch plate, and to weigh no more than necessary. What dimensions do you recommend? 89. Find the height and radius of the largest right circular cylinder that can be put in a sphere of radius 23.
90. The figure here shows two right circular cones, one upside down inside the other. The two bases are parallel, and the vertex of the smaller cone lies at the center of the larger cone’s base. What values of r and h will give the smaller cone the largest possible volume?
12′ r
h 6′
294
Chapter 4: Applications of Derivatives
91. Manufacturing tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 … x … 4 and y =
40  10x . 5  x
Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make? 92. Particle motion The positions of two particles on the saxis are s1 = cos t and s2 = cos (t + p>4). a. What is the farthest apart the particles ever get? b. When do the particles collide? T 93. Opentop box An opentop rectangular box is constructed from a 10in.by16in. piece of cardboard by cutting squares of equal side length from the corners and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume. Support your answers graphically. 94. The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here? Round your answer down to the nearest foot. y
(8, 6)
6 0
x
8
96. Let ƒ(x) = x4  x3. Show that the equation ƒ(x) = 75 has a solution in the interval 33, 44 and use Newton’s method to find it.
Finding Indefinite Integrals Find the indefinite integrals (most general antiderivatives) in Exercises 97–120. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
L
99.
L
101.
dr 2 L (r + 5)
103.
L
105.
L
(x3 + 5x  7) dx a3 2t +
4 b dt t2
3u 2u 2 + 1 du x3(1 + x4)1>4 dx
L
sec2
s ds 10
109. csc 22u cot 22u du L 111.
L
sin2
112.
L
cos2
113.
L
115. 117. 119.
L
csc2 ps ds
110.
L
sec
u u tan du 3 3
1  cos 2u x b dx a Hint: sin2 u = 4 2 x dx 2
3 a x  xb dx
114.
1 a et  et b dt L 2 L
108.
116.
u 1  p du
118.
3 dx L 2x 2x2  1
120.
5 2 a 2 + 2 b dx x + 1 L x L
(5s + s5) ds
L
2p + r dr
du L 216  u 2
Initial Value Problems Solve the initial value problems in Exercises 121–124. dy x2 + 1 121. = , y(1) =  1 dx x2 dy 1 2 = ax + x b , y(1) = 1 122. dx
3 d 2r = 15 2t + ; r′(1) = 8, r (1) = 0 dt 2 2t d 3r 124. 3 = cos t; r″(0) = r′(0) = 0, r (0) =  1 dt
123.
Newton’s Method 95. Let ƒ(x) = 3x  x3. Show that the equation ƒ(x) =  4 has a solution in the interval 32, 34 and use Newton’s method to find it.
97.
107.
98. 100. 102. 104. 106.
L
a8t 3 
L 2 2t a
1
2
t + tb dt 2 
3 b dt t4
6 dr L 1 r  22 23
u du L 27 + u 2 L
(2  x)3>5 dx
Applications and Examples
125. Can the integrations in (a) and (b) both be correct? Explain. a. b.
dx = sin1 x + C L 21  x2
dx dx =  =  cos1 x + C L 21  x2 L 21  x2
126. Can the integrations in (a) and (b) both be correct? Explain. a. b.
dx dx =  =  cos1 x + C L 21  x2 L 21  x2
dx du = 2 L 21  x L 21  ( u)2
x = u dx =  du
L 21  u2 du
=
= cos1 u + C
= cos1 ( x) + C
u = x
127. The rectangle shown here has one side on the positive yaxis, one side on the positive xaxis, and its upper righthand vertex
Chapter 4 Additional and Advanced Exercises
2
on the curve y = ex . What dimensions give the rectangle its largest area, and what is that area? y 1
y = e−x
2
295
2
132. g(x) = e23  2x  x
T 133. Graph the following functions and use what you see to locate and estimate the extreme values, identify the coordinates of the inflection points, and identify the intervals on which the graphs are concave up and concave down. Then confirm your estimates by working with the functions’ derivatives. a. y = (ln x)> 2x 2
x
0
128. The rectangle shown here has one side on the positive yaxis, one side on the positive xaxis, and its upper righthand vertex on the curve y = (ln x)>x2. What dimensions give the rectangle its largest area, and what is that area? y y=
0.2
ln x x2
0.1 0
x
1
In Exercises 129 and 130, find the absolute maximum and minimum values of each function on the given interval. 129. y = x ln 2x  x,
c
b. y = ex
c. y = (1 + x)ex T 134. Graph ƒ(x) = x ln x. Does the function appear to have an absolute minimum value? Confirm your answer with calculus. T 135. Graph ƒ(x) = (sin x)sin x over 30, 3p4. Explain what you see.
136. A round underwater transmission cable consists of a core of copper wires surrounded by nonconducting insulation. If x denotes the ratio of the radius of the core to the thickness of the insulation, it is known that the speed of the transmission signal is given by the equation y = x2 ln (1>x). If the radius of the core is 1 cm, what insulation thickness h will allow the greatest transmission speed? Insulation
1 e , d 2e 2
x= r h
2
130. y = 10x(2  ln x), (0, e 4
Core
h r
In Exercises 131 and 132, find the absolute maxima and minima of the functions and say where they are assumed. 4
131. ƒ(x) = ex> 2x
+1
Chapter 4
Additional and Advanced Exercises
Functions and Derivatives 1. What can you say about a function whose maximum and minimum values on an interval are equal? Give reasons for your answer.
2. Is it true that a discontinuous function cannot have both an absolute maximum and an absolute minimum value on a closed interval? Give reasons for your answer. 3. Can you conclude anything about the extreme values of a continuous function on an open interval? On a halfopen interval? Give reasons for your answer. 4. Local extrema Use the sign pattern for the derivative dƒ = 6(x  1)(x  2)2(x  3)3(x  4)4 dx to identify the points where ƒ has local maximum and minimum values.
b. Suppose that the first derivative of y = ƒ(x) is y′ = 6x(x + 1)(x  2). At what points, if any, does the graph of ƒ have a local maximum, local minimum, or point of inflection? 6. If ƒ′(x) … 2 for all x, what is the most the values of ƒ can increase on 30, 64 ? Give reasons for your answer.
7. Bounding a function Suppose that ƒ is continuous on 3a, b4 and that c is an interior point of the interval. Show that if ƒ′(x) … 0 on 3a, c) and ƒ′(x) Ú 0 on (c, b4 , then ƒ(x) is never less than ƒ(c) on 3a, b4 . 8. An inequality
a. Show that 1>2 … x>(1 + x2) … 1>2 for every value of x.
b. Suppose that ƒ is a function whose derivative is ƒ′(x) = x>(1 + x2). Use the result in part (a) to show that
5. Local extrema
2
a. Suppose that the first derivative of y = ƒ(x) is y′ = 6(x + 1)(x  2)2. At what points, if any, does the graph of ƒ have a local maximum, local minimum, or point of inflection?
0 ƒ(b)  ƒ(a) 0 … 1 0 b  a 0
for any a and b.
296
Chapter 4: Applications of Derivatives
9. The derivative of ƒ(x) = x2 is zero at x = 0, but ƒ is not a constant function. Doesn’t this contradict the corollary of the Mean Value Theorem that says that functions with zero derivatives are constant? Give reasons for your answer.
Tank kept full, top open
y
h
10. Extrema and inflection points Let h = ƒg be the product of two differentiable functions of x. a. If ƒ and g are positive, with local maxima at x = a, and if ƒ′ and g′ change sign at a, does h have a local maximum at a? b. If the graphs of ƒ and g have inflection points at x = a, does the graph of h have an inflection point at a? In either case, if the answer is yes, give a proof. If the answer is no, give a counterexample. 11. Finding a function Use the following information to find the values of a, b, and c in the formula ƒ(x) = (x + a)> (bx2 + cx + 2). i) The values of a, b, and c are either 0 or 1. ii) The graph of ƒ passes through the point (1, 0). iii) The line y = 1 is an asymptote of the graph of ƒ.
Exit velocity = " 64(h − y)
y
Ground
0
x
Range
16. Kicking a field goal An American football player wants to kick a field goal with the ball being on a right hash mark. Assume that the goal posts are b feet apart and that the hash mark line is a distance a 7 0 feet from the right goal post. (See the accompanying figure.) Find the distance h from the goal post line that gives the kicker his largest angle b. Assume that the football field is flat. Goal posts
12. Horizontal tangent For what value or values of the constant k will the curve y = x3 + kx2 + 3x  4 have exactly one horizontal tangent?
b
Optimization 13. Largest inscribed triangle Points A and B lie at the ends of a diameter of a unit circle and point C lies on the circumference. Is it true that the area of triangle ABC is largest when the triangle is isosceles? How do you know?
Goal post line
a
h b u
14. Proving the second derivative test The Second Derivative Test for Local Maxima and Minima (Section 4.4) says: a. ƒ has a local maximum value at x = c if ƒ′(c) = 0 and ƒ″(c) 6 0 b. ƒ has a local minimum value at x = c if ƒ′(c) = 0 and ƒ″(c) 7 0. To prove statement (a), let P = (1>2) 0 ƒ″(c) 0 . Then use the fact that ƒ″(c) = lim
hS0
ƒ′(c + h)  ƒ′(c) ƒ′(c + h) = lim S h h h 0
to conclude that for some d 7 0, 0 6 0h0 6 d
1
Football
17. A maxmin problem with a variable answer Sometimes the solution of a maxmin problem depends on the proportions of the shapes involved. As a case in point, suppose that a right circular cylinder of radius r and height h is inscribed in a right circular cone of radius R and height H, as shown here. Find the value of r (in terms of R and H) that maximizes the total surface area of the cylinder (including top and bottom). As you will see, the solution depends on whether H … 2R or H 7 2R.
ƒ′(c + h) 6 ƒ″(c) + P 6 0. h
Thus, ƒ′(c + h) is positive for  d 6 h 6 0 and negative for 0 6 h 6 d. Prove statement (b) in a similar way. 15. Hole in a water tank You want to bore a hole in the side of the tank shown here at a height that will make the stream of water coming out hit the ground as far from the tank as possible. If you drill the hole near the top, where the pressure is low, the water will exit slowly but spend a relatively long time in the air. If you drill the hole near the bottom, the water will exit at a higher velocity but have only a short time to fall. Where is the best place, if any, for the hole? (Hint: How long will it take an exiting droplet of water to fall from height y to the ground?)
r H
h
R
Chapter 4 Additional and Advanced Exercises
18. Minimizing a parameter Find the smallest value of the positive constant m that will make mx  1 + (1>x) greater than or equal to zero for all positive values of x. Limits 19. Evaluate the following limits.
a. lim
xS0
2 sin 5x b. lim sin 5x cot 3x 3x xS0
c. lim x csc2 22x d. lim (sec x  tan x) xS0
x S p>2
x  sin x sin x2 e. lim x  tan x f. lim S S x x 0 x 0 sin x
sec x  1 x3  8 h. lim 2 2 xS0 xS2 x  4 x
g. lim
20. L’Hôpital’s Rule does not help with the following limits. Find them some other way. a. lim
x Sq
2x + 5
2x + 5
2x b. lim x Sq x + 7 2x
Theory and Examples 21. Suppose that it costs a company y = a + bx dollars to produce x units per week. It can sell x units per week at a price of P = c  ex dollars per unit. Each of a, b, c, and e represents a positive constant. (a) What production level maximizes the profit? (b) What is the corresponding price? (c) What is the weekly profit at this level of production? (d) At what price should each item be sold to maximize profits if the government imposes a tax of t dollars per item sold? Comment on the difference between this price and the price before the tax.
22. Estimating reciprocals without division You can estimate the value of the reciprocal of a number a without ever dividing by a if you apply Newton’s method to the function ƒ(x) = (1>x)  a. For example, if a = 3, the function involved is ƒ(x) = (1>x)  3. a. Graph y = (1>x)  3. Where does the graph cross the xaxis? b. Show that the recursion formula in this case is xn + 1 = xn(2  3xn), so there is no need for division. q
23. To find x = 2a, we apply Newton’s method to ƒ(x) = xq  a. Here we assume that a is a positive real number and q is a positive integer. Show that x1 is a “weighted average” of x0 and a>x0q  1, and find the coefficients m0, m1 such that x1 = m0 x0 + m1 a
a b, x0 q  1
m0 7 0, m1 7 0, m0 + m1 = 1.
What conclusion would you reach if x0 and a>x0 q  1 were equal? What would be the value of x1 in that case? 24. The family of straight lines y = ax + b (a, b arbitrary constants) can be characterized by the relation y″ = 0. Find a similar relation satisfied by the family of all circles
297
25. Free fall in the fourteenth century In the middle of the fourteenth century, Albert of Saxony (1316–1390) proposed a model of free fall that assumed that the velocity of a falling body was proportional to the distance fallen. It seemed reasonable to think that a body that had fallen 20 ft might be moving twice as fast as a body that had fallen 10 ft. And besides, none of the instruments in use at the time were accurate enough to prove otherwise. Today we can see just how far off Albert of Saxony’s model was by solving the initial value problem implicit in his model. Solve the problem and compare your solution graphically with the equation s = 16t 2. You will see that it describes a motion that starts too slowly at first and then becomes too fast too soon to be realistic. T 26. Group blood testing During World War II it was necessary to administer blood tests to large numbers of recruits. There are two standard ways to administer a blood test to N people. In method 1, each person is tested separately. In method 2, the blood samples of x people are pooled and tested as one large sample. If the test is negative, this one test is enough for all x people. If the test is positive, then each of the x people is tested separately, requiring a total of x + 1 tests. Using the second method and some probability theory it can be shown that, on the average, the total number of tests y will be 1 y = N a1  qx + x b.
With q = 0.99 and N = 1000, find the integer value of x that minimizes y. Also find the integer value of x that maximizes y. (This second result is not important to the reallife situation.) The group testing method was used in World War II with a savings of 80% over the individual testing method, but not with the given value of q. 27. Assume that the brakes of an automobile produce a constant deceleration of k ft>sec2. (a) Determine what k must be to bring an automobile traveling 60 mi>hr (88 ft>sec) to rest in a distance of 100 ft from the point where the brakes are applied. (b) With the same k, how far would a car traveling 30 mi>hr go before being brought to a stop? 28. Let ƒ(x), g(x) be two continuously differentiable functions satisfying the relationships ƒ′(x) = g(x) and ƒ″(x) =  ƒ(x). Let h(x) = ƒ2(x) + g2(x). If h(0) = 5, find h(10). 29. Can there be a curve satisfying the following conditions? d 2y>dx2 is everywhere equal to zero and, when x = 0, y = 0 and dy>dx = 1. Give a reason for your answer. 30. Find the equation for the curve in the xyplane that passes through the point (1, 1) if its slope at x is always 3x2 + 2. 31. A particle moves along the xaxis. Its acceleration is a = t 2. At t = 0, the particle is at the origin. In the course of its motion, it reaches the point x = b, where b 7 0, but no point beyond b. Determine its velocity at t = 0. 32. A particle moves with acceleration a = 2t  1 1> 2t 2. Assuming that the velocity y = 4>3 and the position s =  4>15 when t = 0, find
(x  h)2 + (y  h)2 = r 2,
a. the velocity y in terms of t.
where h and r are arbitrary constants. (Hint: Eliminate h and r from the set of three equations including the given one and two obtained by successive differentiation.)
b. the position s in terms of t. 33. Given ƒ(x) = ax2 + 2bx + c with a 7 0. By considering the minimum, prove that ƒ(x) Ú 0 for all real x if and only if b2  ac … 0.
298
Chapter 4: Applications of Derivatives
34. Schwarz’s inequality
B
a. In Exercise 33, let d2
ƒ(x) = (a1 x + b1)2 + (a2 x + b2)2 + g + (an x + bn)2, and deduce Schwarz’s inequality: (a1 b1 + a2 b2 + g + an bn)2 … 1 a1 2 + a2 2 + g + an 2 21 b1 2 + b2 2 + g + bn 2 2. b. Show that equality holds in Schwarz’s inequality only if there exists a real number x that makes ai x equal bi for every value of i from 1 to n.
35. The best branching angles for blood vessels and pipes When a smaller pipe branches off from a larger one in a flow system, we may want it to run off at an angle that is best from some energysaving point of view. We might require, for instance, that energy loss due to friction be minimized along the section AOB shown in the accompanying figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe upstream from B, and O is the point where the branching occurs. A law due to Poiseuille states that the loss of energy due to friction in nonturbulent flow is proportional to the length of the path and inversely proportional to the fourth power of the radius. Thus, the loss along AO is (kd1)>R4 and along OB is (kd2)>r 4, where k is a constant, d1 is the length of AO, d2 is the length of OB, R is the radius of the larger pipe, and r is the radius of the smaller pipe. The angle u is to be chosen to minimize the sum of these two losses: L = k
Chapter 4
b = d 2 sin u u
A d1
O a
C d 2 cos u
In our model, we assume that AC = a and BC = b are fixed. Thus we have the relations d1 + d2 cos u = a d2 sin u = b, so that d2 = b csc u, d1 = a  d2 cos u = a  b cot u. We can express the total loss L as a function of u: L = ka
b csc u a  b cot u + b. R4 r4
a. Show that the critical value of u for which dL>du equals zero is uc = cos1
r4 . R4
b. If the ratio of the pipe radii is r>R = 5>6, estimate to the nearest degree the optimal branching angle given in part (a).
d1 d2 + k 4. 4 R r
Technology Application Projects
Mathematica/Maple Modules: Motion Along a Straight Line: Position u Velocity u Acceleration You will observe the shape of a graph through dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated. Newton’s Method: Estimate P to How Many Places? Plot a function, observe a root, pick a starting point near the root, and use Newton’s Iteration Procedure to approximate the root to a desired accuracy. The numbers p, e, and 22 are approximated.
5
Integrals
OVERVIEW A great achievement of classical geometry was obtaining formulas for the
areas and volumes of triangles, spheres, and cones. In this chapter we develop a method to calculate the areas and volumes of very general shapes. This method, called integration, is a way to calculate much more than areas and volumes. The definite integral is the key tool in calculus for defining and calculating many important quantities, such as areas, volumes, lengths of curved paths, probabilities, averages, energy consumption, the weights of various objects, and the forces against a dam’s floodgates, just to mention a few. Many of these applications are studied in subsequent chapters. As with the derivative, the definite integral also arises as a limit, this time of increasingly fine approximations to the quantity of interest. The idea behind the integral is that we can effectively compute such quantities by breaking them into small pieces, and then summing the contributions from each piece. We then consider what happens when more and more, smaller and smaller pieces are taken in the summation process. As the number of terms contributing to the sum approaches infinity and we take the limit of these sums in a way described in Section 5.3, the result is a definite integral. By considering the rate of change of the area under a graph, we prove that definite integrals are connected to antiderivatives, a connection that gives one of the most important relationships in calculus.
5.1 Area and Estimating with Finite Sums The basis for formulating definite integrals is the construction of appropriate approximations by finite sums. In this section we consider three examples of this construction process: finding the area under a graph, the distance traveled by a moving object, and the average value of a function. Although we need to define precisely what we mean by the area of a general region in the plane, or the average value of a function over a closed interval, we do have intuitive ideas of what these notions mean. So in this section we begin our approach to integration by approximating these quantities with finite sums. We also consider what happens when we take more and more terms in the summation process. In subsequent sections we look at taking the limit of these sums as the number of terms goes to infinity, which then leads to precise definitions of the quantities being approximated here.
y
1 y = 1 − x2
0.5 R
0
Area
0.5
1
x
FIGURE 5.1 The area of the region R cannot be found by a simple formula.
Suppose we want to find the area of the shaded region R that lies above the xaxis, below the graph of y = 1  x2, and between the vertical lines x = 0 and x = 1 (Figure 5.1). Unfortunately, there is no simple geometric formula for calculating the areas of general shapes having curved boundaries like the region R. How, then, can we find the area of R? While we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the
299
300
Chapter 5: Integrals
y
1
y y=1−x
(0, 1)
2
1
y = 1 − x2
(0, 1) Q1 , 15R 4 16
1 3
1 3
Q2 , 4R
Q2 , 4R
3
0.5
R
7
Q4 , 16R
0.5 R
0
0.5
1
x
0
0.25
0.5
0.75
1
x
(b)
(a)
FIGURE 5.2 (a) We get an upper estimate of the area of R by using two rectangles containing R. (b) Four rectangles give a better upper estimate. Both estimates overshoot the true value for the area by the amount shaded in light red.
region R. Each rectangle has width 1>2 and they have heights 1 and 3>4, moving from left to right. The height of each rectangle is the maximum value of the function ƒ in each subinterval. Because the function ƒ is decreasing, the height is its value at the left endpoint of the subinterval of 3 0, 14 forming the base of the rectangle. The total area of the two rectangles approximates the area A of the region R, A ≈ 1#
1 3#1 7 + = = 0.875. 2 4 2 8 This estimate is larger than the true area A since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of each rectangle as the maximum (uppermost) value of ƒ(x) for a point x in the base interval of the rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1>4, which taken together contain the region R. These four rectangles give the approximation A ≈ 1#
7 # 1 25 1 15 # 1 3 # 1 + + + = = 0.78125, 4 16 4 4 4 16 4 32 which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained inside the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1>4 as before, but the rectangles are shorter and y 1
y y = 1 − x2
15 Q14 , 16 R
1
63 Q18 , 64 R 55 Q38 , 64 R
y = 1 − x2
1 3 Q2 , 4R 39 Q58 , 64 R 7 Q34 , 16 R
0.5
0.5 15 Q78 , 64 R
0
0.25
0.5 (a)
0.75
1
x
0
0.25 0.5 0.75 1 0.125 0.375 0.625 0.875
x
(b)
FIGURE 5.3 (a) Rectangles contained in R give an estimate for the area that undershoots the true value by the amount shaded in light blue. (b) The midpoint rule uses rectangles whose height is the value of y = ƒ(x) at the midpoints of their bases. The estimate appears closer to the true value of the area because the light red overshoot areas roughly balance the light blue undershoot areas.
5.1 Area and Estimating with Finite Sums
301
lie entirely beneath the graph of ƒ. The function ƒ(x) = 1  x2 is decreasing on 3 0, 14 , so the height of each of these rectangles is given by the value of ƒ at the right endpoint of the subinterval forming its base. The fourth rectangle has zero height and therefore contributes no area. Summing these rectangles with heights equal to the minimum value of ƒ(x) for a point x in each base subinterval gives a lower sum approximation to the area, A ≈
15 # 1 3 # 1 7 #1 1 17 + + + 0# = = 0.53125. 4 32 16 4 4 4 16 4
This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums: 0.53125 6 A 6 0.78125. y
By considering both lower and upper sum approximations, we get not only estimates for the area, but also a bound on the size of the possible error in these estimates, since the true value of the area lies somewhere between them. Here the error cannot be greater than the difference 0.78125  0.53125 = 0.25. Yet another estimate can be obtained by using rectangles whose heights are the values of ƒ at the midpoints of their bases (Figure 5.3b). This method of estimation is called the midpoint rule for approximating the area. The midpoint rule gives an estimate that is between a lower sum and an upper sum, but it is not quite so clear whether it overestimates or underestimates the true area. With four rectangles of width 1>4 as before, the midpoint rule estimates the area of R to be
1 y = 1 − x2
1
0
x
(a)
A ≈
63 # 1 55 # 1 39 # 1 15 # 1 172 # 1 + + + = = 0.671875. 64 4 64 4 64 4 64 4 64 4
In each of our computed sums, the interval 3 a, b4 over which the function ƒ is defined was subdivided into n subintervals of equal width (also called length) ∆x = (b  a)>n, and ƒ was evaluated at a point in each subinterval: c1 in the first subinterval, c2 in the second subinterval, and so on. The finite sums then all take the form
y 1 y = 1 − x2
ƒ(c1) ∆x + ƒ(c2) ∆x + ƒ(c3) ∆x + g + ƒ(cn) ∆x.
1
0
x
(b)
FIGURE 5.4 (a) A lower sum using 16 rectangles of equal width ∆x = 1>16. (b) An upper sum using 16 rectangles.
By taking more and more rectangles, with each rectangle thinner than before, it appears that these finite sums give better and better approximations to the true area of the region R. Figure 5.4a shows a lower sum approximation for the area of R using 16 rectangles of equal width. The sum of their areas is 0.634765625, which appears close to the true area, but is still smaller since the rectangles lie inside R. Figure 5.4b shows an upper sum approximation using 16 rectangles of equal width. The sum of their areas is 0.697265625, which is somewhat larger than the true area because the rectangles taken together contain R. The midpoint rule for 16 rectangles gives a total area approximation of 0.6669921875, but it is not immediately clear whether this estimate is larger or smaller than the true area.
EXAMPLE 1 Table 5.1 shows the values of upper and lower sum approximations to the area of R, using up to 1000 rectangles. In Section 5.2 we will see how to get an exact value of the areas of regions such as R by taking a limit as the base width of each rectangle goes to zero and the number of rectangles goes to infinity. With the techniques developed there, we will be able to show that the area of R is exactly 2>3. Distance Traveled Suppose we know the velocity function y(t) of a car moving down a highway, without changing direction, and want to know how far it traveled between times t = a and t = b. The position function s(t) of the car has derivative y(t). If we can find an antiderivative F(t)
302
Chapter 5: Integrals
TABLE 5.1 Finite approximations for the area of R Number of subintervals
Lower sum
Midpoint sum
Upper sum
2 4 16 50 100 1000
0.375 0.53125 0.634765625 0.6566 0.66165 0.6661665
0.6875 0.671875 0.6669921875 0.6667 0.666675 0.66666675
0.875 0.78125 0.697265625 0.6766 0.67165 0.6671665
of y(t) then we can find the car’s position function s(t) by setting s(t) = F(t) + C. The distance traveled can then be found by calculating the change in position, s(b)  s(a) = F(b)  F(a). If the velocity function is known only by the readings at various times of a speedometer on the car, then we have no formula from which to obtain an antiderivative function for velocity. So what do we do in this situation? When we don’t know an antiderivative for the velocity function y(t), we can approximate the distance traveled with finite sums in a way similar to our estimates for area discussed before. We subdivide the interval 3 a, b4 into short time intervals on each of which the velocity is considered to be fairly constant. Then we approximate the distance traveled on each time subinterval with the usual distance formula distance = velocity * time
and add the results across 3 a, b4 . Suppose the subdivided interval looks like Δt a
t1
Δt
Δt
t2
b
t3
t (sec)
with the subintervals all of equal length ∆t. Pick a number t1 in the first interval. If ∆t is so small that the velocity barely changes over a short time interval of duration ∆t, then the distance traveled in the first time interval is about y(t1) ∆t. If t2 is a number in the second interval, the distance traveled in the second time interval is about y(t2) ∆t. The sum of the distances traveled over all the time intervals is D ≈ y(t1) ∆t + y(t2) ∆t + g + y(tn) ∆t, where n is the total number of subintervals.
EXAMPLE 2
The velocity function of a projectile fired straight into the air is ƒ(t) = 160  9.8t m>sec. Use the summation technique just described to estimate how far the projectile rises during the first 3 sec. How close do the sums come to the exact value of 435.9 m? (You will learn how to compute the exact value easily in Section 5.4.) Solution We explore the results for different numbers of intervals and different choices of evaluation points. Notice that ƒ(t) is decreasing, so choosing left endpoints gives an upper sum estimate; choosing right endpoints gives a lower sum estimate. (a) Three subintervals of length 1, withƒ evaluated at left endpoints giving an upper sum: t1
t2
t3
0
1
2
Δt
3
t
5.1 Area and Estimating with Finite Sums
303
With ƒ evaluated at t = 0, 1, and 2, we have D ≈ ƒ(t1) ∆t + ƒ(t2) ∆t + ƒ(t3) ∆t = 3 160  9.8(0) 4 (1) + 3 160  9.8(1) 4 (1) + 3 160  9.8(2) 4 (1) = 450.6. (b) Three subintervals of length 1, with ƒ evaluated at right endpoints giving a lower sum:
0
t1
t2
t3
1
2
3
t
Δt
With ƒ evaluated at t = 1, 2, and 3, we have D ≈ ƒ(t1) ∆t + ƒ(t2) ∆t + ƒ(t3) ∆t = 3 160  9.8(1) 4 (1) + 3 160  9.8(2) 4 (1) + 3 160  9.8(3) 4 (1) = 421.2. (c) With six subintervals of length 1>2, we get t1 t 2 t 3 t 4 t 5 t 6 0
1
2
t1 t 2 t 3 t 4 t 5 t 6 3
Δt
t
0
1
2
3
t
Δt
These estimates give an upper sum using left endpoints: D ≈ 443.25; and a lower sum using right endpoints: D ≈ 428.55. These sixinterval estimates are somewhat closer than the threeinterval estimates. The results improve as the subintervals get shorter. As we can see in Table 5.2, the leftendpoint upper sums approach the true value 435.9 from above, whereas the rightendpoint lower sums approach it from below. The true value lies between these upper and lower sums. The magnitude of the error in the closest entries is 0.23, a small percentage of the true value. Error magnitude = 0 true value  calculated value 0 = 0 435.9  435.67 0 = 0.23.
Error percentage =
0.23 ≈ 0.05,. 435.9
It would be reasonable to conclude from the table’s last entries that the projectile rose about 436 m during its first 3 sec of flight.
TABLE 5.2 Traveldistance estimates Number of subintervals
Length of each subinterval
Upper sum
Lower sum
3 6 12 24 48 96 192
1 1>2 1>4 1>8 1>16 1>32 1>64
450.6 443.25 439.58 437.74 436.82 436.36 436.13
421.2 428.55 432.23 434.06 434.98 435.44 435.67
304
Chapter 5: Integrals
Displacement Versus Distance Traveled If an object with position function s(t) moves along a coordinate line without changing direction, we can calculate the total distance it travels from t = a to t = b by summing the distance traveled over small intervals, as in Example 2. If the object reverses direction one or more times during the trip, then we need to use the object’s speed 0 y(t) 0 , which is the absolute value of its velocity function, y(t), to find the total distance traveled. Using the velocity itself, as in Example 2, gives instead an estimate to the object’s displacement, s(b)  s(a), the difference between its initial and final positions. To see why using the velocity function in the summation process gives an estimate to the displacement, partition the time interval 3 a, b4 into small enough equal subintervals ∆t so that the object’s velocity does not change very much from time tk  1 to tk. Then y(tk) gives a good approximation of the velocity throughout the interval. Accordingly, the change in the object’s position coordinate, which is its displacement during the time interval, is about
s s(5)
Height (ft)
400
256
(+)
(−)
s(2)
s(8)
144
y(tk) ∆t. The change is positive if y(tk) is positive and negative if y(tk) is negative. In either case, the distance traveled by the object during the subinterval is about
0 y(tk) 0 ∆t.
s=0
The total distance traveled over the time interval is approximately the sum
0 y(t1) 0 ∆t + 0 y(t2) 0 ∆t + g + 0 y(tn) 0 ∆t.
s(0)
FIGURE 5.5 The rock in Example 3. The height s = 256 ft is reached at t = 2 and t = 8 sec. The rock falls 144 ft from its maximum height when t = 8.
TABLE 5.3 Velocity function t
Y(t)
t
Y(t)
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
160 144 128 112 96 80 64 48 32
4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0
16 0 16 32 48 64 80 96
We revisit these ideas in Section 5.4.
EXAMPLE 3
In Example 4 in Section 3.4, we analyzed the motion of a heavy rock blown straight up by a dynamite blast. In that example, we found the velocity of the rock at any time during its motion to be y(t) = 160  32t ft>sec. The rock was 256 ft above the ground 2 sec after the explosion, continued upward to reach a maximum height of 400 ft at 5 sec after the explosion, and then fell back down to reach the height of 256 ft again at t = 8 sec after the explosion. (See Figure 5.5.) If we follow a procedure like that presented in Example 2, and use the velocity function y(t) in the summation process over the time interval 3 0, 84 , we will obtain an estimate to the rock’s 256 ft height above the ground at t = 8. The positive upward motion (which yields a positive distance change of 144 ft from the height of 256 ft to the maximum height) is canceled by the negative downward motion (giving a negative change of 144 ft from the maximum height down to 256 ft again), so the displacement or height above the ground is estimated from the velocity function. On the ot