Section 2.8 The Derivative as a Function
138 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
3. (a)0=II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then negative again. The actual function values in graph II follow the same pattern.
(b)0=IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)0=I, since the slopes of the tangents to graph (c) are negative for 0 and positive for 0, as are the function values of graph I.
(d)0=III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in graph III follow the same pattern.
Hints for Exercises 4–11: First plot-intercepts on the graph of0for any horizontal tangents on the graph of. Look for any corners on the graph of—there will be a discontinuity on the graph of0. On any interval wherehas a tangent with positive (or negative) slope, the graph of0will be positive (or negative). If the graph of the function is linear, the graph of0will be a horizontal line.
4. 5.
6. 7.
° 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
138 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
3. (a)0=II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then negative again. The actual function values in graph II follow the same pattern.
(b)0=IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)0=I, since the slopes of the tangents to graph (c) are negative for 0 and positive for 0, as are the function values of graph I.
(d)0=III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in graph III follow the same pattern.
Hints for Exercises 4–11: First plot-intercepts on the graph of0for any horizontal tangents on the graph of. Look for any corners on the graph of—there will be a discontinuity on the graph of0. On any interval wherehas a tangent with positive (or negative) slope, the graph of0will be positive (or negative). If the graph of the function is linear, the graph of0will be a horizontal line.
4. 5.
6. 7.
° 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
142 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 24. 0() = lim
→0
( + ) − ()
= lim
→0
4 + 8( + ) − 5( + )2
− (4 + 8 − 52)
= lim
→0
4 + 8 + 8 − 5(2+ 2 + 2) − 4 − 8 + 52
= lim
→0
8 − 52− 10 − 52+ 52
= lim
→0
8 − 10 − 52
= lim
→0
(8 − 10 − 5)
= lim
→0(8 − 10 − 5)
= 8 − 10
Domain of = domain of 0= R.
25. 0() = lim
→0
( + ) − ()
= lim
→0
( + )3− 3( + ) + 5
− (3− 3 + 5)
= lim
→0
3+ 32 + 32+ 3− 3 − 3 + 5
−
3− 3 + 5
= lim
→0
32 + 32+ 3− 3
= lim
→0
32+ 3 + 2− 3
= lim
→0
32+ 3 + 2− 3
= 32− 3
Domain of = domain of 0= R.
26. 0() = lim
→0
( + ) − ()
= lim
→0
+ +√
+
− ( +√
)
= lim
→0
+
√ + −√
·
√ + +√
√
+ +√
= lim
→0
1 + ( + ) −
√
+ +√
= lim
→0
1 + 1
√ + +√
= 1 + 1
√ +√
= 1 + 1 2√
Domain of = [0 ∞), domain of 0= (0 ∞).
27. 0() = lim
→0
( + ) − ()
= lim
→0
9 − ( + ) −√ 9 −
9 − ( + ) +√ 9 −
9 − ( + ) +√ 9 −
= lim
→0
[9 − ( + )] − (9 − )
9 − ( + ) +√
9 − = lim
→0
−
9 − ( + ) +√ 9 −
= lim
→0
−1
9 − ( + ) +√
9 − = −1 2√
9 − Domain of = (−∞ 9], domain of 0= (−∞ 9).
° 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
SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤ 143
28. 0() = lim
→0
( + ) − ()
= lim
→0
( + )2− 1
2( + ) − 3−2− 1 2 − 3
= lim
→0
[( + )2− 1](2 − 3) − [2( + ) − 3](2− 1) [2( + ) − 3](2 − 3)
= lim
→0
(2+ 2 + 2− 1)(2 − 3) − (2 + 2 − 3)(2− 1)
[2( + ) − 3](2 − 3)
= lim
→0
(23+ 42 + 22− 2 − 32− 6 − 32+ 3) − (23+ 22 − 32− 2 − 2 + 3)
(2 + 2 − 3)(2 − 3)
= lim
→0
42 + 22− 6 − 32− 22 + 2
(2 + 2 − 3)(2 − 3) = lim
→0
(22+ 2 − 6 − 3 + 2)
(2 + 2 − 3)(2 − 3)
= lim
→0
22+ 2 − 6 − 3 + 2
(2 + 2 − 3)(2 − 3) =22− 6 + 2 (2 − 3)2 Domain of = domain of 0= (−∞32) ∪ (32 ∞).
29. 0() = lim
→0
( + ) − ()
= lim
→0
1 − 2( + )
3 + ( + ) −1 − 2
3 +
= lim
→0
[1 − 2( + )](3 + ) − [3 + ( + )](1 − 2) [3 + ( + )](3 + )
= lim
→0
3 + − 6 − 22− 6 − 2 − (3 − 6 + − 22+ − 2)
[3 + ( + )](3 + ) = lim
→0
−6 −
(3 + + )(3 + )
= lim
→0
−7
(3 + + )(3 + ) = lim
→0
−7
(3 + + )(3 + )= −7 (3 + )2 Domain of = domain of 0= (−∞ −3) ∪ (−3 ∞).
30. 0() = lim
→0
( + ) − ()
= lim
→0
( + )32− 32
= lim
→0
[( + )32− 32][( + )32+ 32]
[( + )32+ 32]
= lim
→0
( + )3− 3
[( + )32+ 32] = lim
→0
3+ 32 + 32+ 3− 3
[( + )32+ 32] = lim
→0
32+ 3 + 2
[( + )32+ 32]
= lim
→0
32+ 3 + 2
( + )32+ 32 = 32
232 = 3212
Domain of = domain of 0= [0 ∞). Strictly speaking, the domain of 0is (0 ∞) because the limit that defines 0(0)does not exist (as a two-sided limit). But the right-hand derivative (in the sense of Exercise 64) does exist at 0, so in that sense one could regard the domain of 0to be [0 ∞).
° 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
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
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
148 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 53. 0() = lim
→0
( + ) − ()
= lim
→0
[3( + )2+ 2( + ) + 1] − (32+ 2 + 1)
= lim
→0
(32+ 6 + 32+ 2 + 2 + 1) − (32+ 2 + 1)
= lim
→0
6 + 32+ 2
= lim
→0
(6 + 3 + 2)
= lim
→0(6 + 3 + 2) = 6 + 2
00() = lim
→0
0( + ) − 0()
= lim
→0
[6( + ) + 2] − (6 + 2)
= lim
→0
(6 + 6 + 2) − (6 + 2)
= lim
→0
6
= lim
→06 = 6
We see from the graph that our answers are reasonable because the graph of
0is that of a linear function and the graph of 00is that of a constant function.
54. 0() = lim
→0
( + ) − ()
= lim
→0
[( + )3− 3( + )] − (3− 3)
= lim
→0
(3+ 32 + 32+ 3− 3 − 3) − (3− 3)
= lim
→0
32 + 32+ 3− 3
= lim
→0
(32+ 3 + 2− 3)
= lim
→0(32+ 3 + 2− 3) = 32− 3
00() = lim
→0
0( + ) − 0()
= lim
→0
[3( + )2− 3] − (32− 3)
= lim
→0
(32+ 6 + 32− 3) − (32− 3)
= lim
→0
6 + 32
= lim
→0
(6 + 3)
= lim
→0(6 + 3) = 6
We see from the graph that our answers are reasonable because the graph of
0is that of an even function ( is an odd function) and the graph of 00is that of an odd function. Furthermore, 0= 0when has a horizontal tangent and 00= 0when 0has a horizontal tangent.
55. 0() = lim
→0
( + ) − ()
= lim
→0
2( + )2− ( + )3
− (22− 3)
= lim
→0
(4 + 2 − 32− 3 − 2)
= lim
→0(4 + 2 − 32− 3 − 2) = 4 − 32
00() = lim
→0
0( + ) − 0()
= lim
→0
4( + ) − 3( + )2
− (4 − 32)
= lim
→0
(4 − 6 − 3)
= lim
→0(4 − 6 − 3) = 4 − 6
000() = lim
→0
00( + ) − 00()
= lim
→0
[4 − 6( + )] − (4 − 6)
= lim
→0
−6
= lim
→0(−6) = −6
° 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
SECTION 2.8 THE DERIVATIVE AS A FUNCTION ¤ 149
(4)() = lim
→0
000( + ) − 000()
= lim
→0
−6 − (−6)
= lim
→0
0
= lim
→0(0) = 0
The graphs are consistent with the geometric interpretations of the derivatives because 0has zeros where has a local minimum and a local maximum, 00has a zero where 0has a local maximum, and 000is a constant function equal to the slope of 00.
56. (a) Since we estimate the velocity to be a maximum at = 10, the acceleration is 0 at = 10.
(b) Drawing a tangent line at = 10 on the graph of , appears to decrease by 10 ms2over a period of 20 s. So at = 10 s, the jerk is approximately −1020 = −05 (ms2)s or ms3.
57. (a) Note that we have factored − as the difference of two cubes in the third step.
0() = lim
→
() − ()
− = lim
→
13− 13
− = lim
→
13− 13
(13− 13)(23+ 1313+ 23)
= lim
→
1
23+ 1313+ 23 = 1
323 or13−23 (b) 0(0) = lim
→0
(0 + ) − (0)
= lim
→0
√3
− 0
= lim
→0
1
23. This function increases without bound, so the limit does not exist, and therefore 0(0)does not exist.
(c) lim
→0|0()| = lim
→0
1
323 = ∞ and is continuous at = 0 (root function), so has a vertical tangent at = 0.
58. (a) 0(0) = lim
→0
() − (0)
− 0 = lim
→0
23− 0
= lim
→0
1
13, which does not exist.
(b) 0() = lim
→
() − ()
− = lim
→
23− 23
− = lim
→
(13− 13)(13+ 13) (13− 13)(23+ 1313+ 23)
= lim
→
13+ 13
23+ 1313+ 23 =213 323 = 2
313 or 23−13 (c) () = 23is continuous at = 0 and
lim→0|0()| = lim
→0
2
3 ||13 = ∞. This shows that
has a vertical tangent line at = 0.
(d)
° 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
