Section 2.8 The Derivative as a Function
40. Suppose N is the number of people in the United States who travel by car to another state for a vacation this year when the average price of gasoline is p dollars per gallon. Do you expect dN/dp to be positive or negative? Explain.
Solution:
146 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
(c)
37. As in Exercise 35, we use one-sided difference quotients for the first and last values, and average two difference quotients for all other values.
14 21 28 35 42 49
() 41 54 64 72 78 83
0() 137 2314 1814 1414 1114 57
38. As in Exercise 35, we use one-sided difference quotients for the first and last values, and average two difference quotients for all other values. The units for 0()are grams per degree (g◦C).
155 177 200 224 244
() 372 310 198 97 −98
0() −282 −387 −453 −673 −975
39. (a) is the rate at which the percentage of the city’s electrical power produced by solar panels changes with respect to time , measured in percentage points per year.
(b) 2 years after January 1, 2000 (January 1, 2002), the percentage of electrical power produced by solar panels was increasing at a rate of 3.5 percentage points per year.
40. is the rate at which the number of people who travel by car to another province for a vacation changes with respect to the price of gasoline. If the price of gasoline goes up, we would expect fewer people to travel, so we would expect to be negative.
41. is not differentiable at = −4, because the graph has a corner there, and at = 0, because there is a discontinuity there.
42. is not differentiable at = −1, because there is a discontinuity there, and at = 2, because the graph has a corner there.
43. is not differentiable at = 1, because is not defined there, and at = 5, because the graph has a vertical tangent there.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
52. The figure shows the graphs of four functions. One is the position function of a car, one is the velocity of the car, one is its acceleration, and one is its jerk. Identify each curve, and explain your choices.
164 Chapter 2 Limits and Derivatives
53–54 Use the definition of a derivative to find f 9sxd and f 0sxd.
Then graph f , f 9, and f 0 on a common screen and check to see if your answers are reasonable.
53. fsxd − 3x212x 1 1 54. fsxd − x323x
55. If fsxd − 2x22x3, find f 9sxd, f 0sxd, f -sxd, and fs4dsxd.
Graph f , f 9, f 0, and f -on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?
56. (a) The graph of a position function of a car is shown, where s is measured in feet and t in seconds. Use it to graph the velocity and acceleration of the car. What is the acceleration at t − 10 seconds?
0 10 t
s
100
20
(b) Use the acceleration curve from part (a) to estimate the jerk at t − 10 seconds. What are the units for jerk?
57. Let fsxd −s3x.
(a) If a ± 0, use Equation 2.7.5 to find f 9sad.
(b) Show that f 9s0d does not exist.
(c) Show that y −s3x has a vertical tangent line at s0, 0d.
(Recall the shape of the graph of f . See Figure 1.2.13.) 58. (a) If tsxd − x2y3, show that t9s0d does not exist.
(b) If a ± 0, find t9sad.
(c) Show that y − x2y3 has a vertical tangent line at s0, 0d.
(d) Illustrate part (c) by graphing y − x2y3.
59. Show that the function fsxd −
|
x 26|
is not differentiable at 6. Find a formula for f 9 and sketch its graph.60. Where is the greatest integer function fsxd − v x b not differentiable? Find a formula for f 9 and sketch its graph.
61. (a) Sketch the graph of the function fsxd − x
|
x|
.(b) For what values of x is f differentiable?
(c) Find a formula for f 9.
62. (a) Sketch the graph of the function tsxd − x 1
|
x|
.(b) For what values of x is t differentiable?
(c) Find a formula for t9.
63. Recall that a function f is called even if fs2xd − f sxd for all x in its domain and odd if fs2xd − 2f sxd for all such x. Prove each of the following.
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function.
;
;
; 49. The figure shows the graphs of f , f 9, and f 0. Identify each
curve, and explain your choices.
x
y a
b c
50. The figure shows graphs of f, f 9, f 0, and f -. Identify each curve, and explain your choices.
x
y a b c d
51. The figure shows the graphs of three functions. One is the position function of a car, one is the velocity of the car, and one is its acceleration. Identify each curve, and explain your choices.
t
y a
b c
0
52. The figure shows the graphs of four functions. One is the position function of a car, one is the velocity of the car, one is its acceleration, and one is its jerk. Identify each curve, and explain your choices.
8et0208x52 08/29/13
0 t
y
a b c
d
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Solution:
SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤ 147
44. is not differentiable at = −2 and = 3, because the graph has corners there, and at = 1, because there is a discontinuity there.
45. As we zoom in toward (−1 0), the curve appears more and more like a straight line, so () = +
|| is differentiable at = −1. But no matter how much we zoom in toward the origin, the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So is not differentiable at = 0.
46. As we zoom in toward (0 1), the curve appears more and more like a straight line, so () = (2− 1)23is differentiable at = 0. But no matter how much we zoom in toward (1 0) or (−1 0), the curve doesn’t straighten out—we can’t eliminate the sharp point (a cusp). So is not differentiable at = ±1.
47. Call the curve with the positive -intercept and the other curve . Notice that has a maximum (horizontal tangent) at
= 0, but 6= 0, so cannot be the derivative of . Also notice that where is positive, is increasing. Thus, = and
= 0. Now 0(−1) is negative since 0is below the -axis there and 00(1)is positive since is concave upward at = 1.
Therefore, 00(1)is greater than 0(−1).
48. Call the curve with the smallest positive -intercept and the other curve . Notice that where is positive in the first quadrant, is increasing. Thus, = and = 0. Now 0(−1) is positive since 0is above the -axis there and 00(1) appears to be zero since has an inflection point at = 1. Therefore, 0(1)is greater than 00(−1).
49. = , = 0, = 00. We can see this because where has a horizontal tangent, = 0, and where has a horizontal tangent,
= 0. We can immediately see that can be neither nor 0, since at the points where has a horizontal tangent, neither nor is equal to 0.
50. Where has horizontal tangents, only is 0, so 0= . has negative tangents for 0 and is the only graph that is negative for 0, so 0= . has positive tangents on R (except at = 0), and the only graph that is positive on the same domain is , so 0= . We conclude that = , = 0, = 00, and = 000.
51. We can immediately see that is the graph of the acceleration function, since at the points where has a horizontal tangent, neither nor is equal to 0. Next, we note that = 0 at the point where has a horizontal tangent, so must be the graph of the velocity function, and hence, 0= . We conclude that is the graph of the position function.
52. must be the jerk since none of the graphs are 0 at its high and low points. is 0 where has a maximum, so 0= . is 0 where has a maximum, so 0= . We conclude that is the position function, is the velocity, is the acceleration, and is the jerk.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
63. Derivatives of Even and Odd Function Recall that a function f is called even if f (−x) = f (x) for all x in its domain and odd if f (−x) = −f (x) for all such x. Prove each of the following.
(a) The derivative of an even function is an odd function.
(b) The derivative of an odd function is an even function.
Solution:
SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤ 151 (c) () =
2 if ≥ 0
0 if 0 ⇒ 0() =
2 if 0 0 if 0
Another way of writing the formula is 0() = 1 + sgn for 6= 0.
63. (a) If is even, then
0(−) = lim
→0
(− + ) − (−)
= lim
→0
[−( − )] − (−)
= lim
→0
( − ) − ()
= − lim
→0
( − ) − ()
− [let ∆ = −]
= − lim∆
→0
( + ∆) − ()
∆ = −0() Therefore, 0is odd.
(b) If is odd, then
0(−) = lim
→0
(− + ) − (−)
= lim
→0
[−( − )] − (−)
= lim
→0
−( − ) + ()
= lim
→0
( − ) − ()
− [let ∆ = −]
= lim
∆→0
( + ∆) − ()
∆ = 0() Therefore, 0is even.
64. (a) −0(4) = lim
→0−
(4 + ) − (4)
= lim
→0−
5 − (4 + ) − 1
= lim
→0−
−
= −1 and
+0(4) = lim
→0+
(4 + ) − (4)
= lim
→0+
1
5 − (4 + )− 1
= lim
→0+
1 − (1 − )
(1 − ) = lim
→0+
1 1 − = 1
(b)
(c) () =
0 if ≤ 0
5 − if 0 4 1(5 − ) if ≥ 4
At 4 we have lim
→4− () = lim
→4−(5 − ) = 1 and lim
→4+ () = lim
→4+
1
5 − = 1, so lim
→4 () = 1 = (4)and is continuous at 4. Since (5) is not defined, is discontinuous at 5. These expressions show that is continuous on the intervals (−∞ 0), (0 4), (4 5) and (5 ∞). Since lim
→0+ () = lim
→0+(5 − ) = 5 6= 0 = lim
→0−
(), lim
→0 ()does not exist, so is discontinuous (and therefore not differentiable) at 0.
(d) From (a), is not differentiable at 4 since −0(4) 6= +0(4), and from (c), is not differentiable at 0 or 5.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
1
