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2 DERIVATIVES
2.1 Derivatives and Rates of Change
1. (a) This is just the slope of the line through two points: = ∆ ∆
= ^ ()^ −^ ^ (3)
(b) This is the limit of the slope of the secant line as approaches : = lim → 3
2. The curve looks more like a line as the viewing rectangle gets smaller. 3. (a) (i) Using Definition 1 with () = 4 − ^2 and (1 3),
= lim →
= lim → 1
(4 − ^2 ) − 3
= lim → 1
−(^2 − 4 + 3)
= lim → 1
= lim → 1
(ii) Using Equation 2 with () = 4 − ^2 and (1 3),
= lim → 0
= lim → 0
= lim → 0
4(1 + ) − (1 + )^2
= lim → 0
4 + 4 − 1 − 2 − ^2 − 3
= lim → 0
−^2 + 2
= lim → 0
= lim → 0
(b) An equation of the tangent line is − () = 0 ()( − ) ⇒ − (1) = 0 (1)( − 1) ⇒ − 3 = 2( − 1), or = 2 + 1.
(c) The graph of = 2 + 1 is tangent to the graph of = 4 − ^2 at the
point (1 3). Now zoom in toward the point (1 3) until the parabola and the tangent line are indistiguishable.
4. (a) (i) Using Definition 1 with () = − ^3 and (1 0),
= lim → 1
= lim → 1
− ^3
= lim → 1
(1 − ^2 )
= lim → 1
= lim → 1 [−(1 + )] = −1(2) = − 2
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100 ¤^ CHAPTER 2 DERIVATIVES
(ii) Using Equation 2 with () = − ^3 and (1 0),
= lim → 0
= lim → 0
= lim → 0
(1 + ) − (1 + )^3
= lim → 0
1 + − (1 + 3 + 3^2 + ^3 )
= lim → 0
−^3 − 3 ^2 − 2
= lim → 0
(−^2 − 3 − 2)
= lim → 0
(−^2 − 3 − 2) = − 2
(b) An equation of the tangent line is − () = ^0 ()( − ) ⇒ − (1) = 0 (1)( − 1) ⇒ − 0 = −2( − 1), or = − 2 + 2.
(c) The graph of = − 2 + 2 is tangent to the graph of = − ^3 at the
point (1 0). Now zoom in toward the point (1 0) until the cubic and the tangent line are indistinguishable.
5. Using (1) with () = 4 − 3 ^2 and (2 −4) [we could also use (2)],
= lim →
= lim → 2
4 − 3 ^2
= lim → 2
− 3 ^2 + 4 + 4
= lim → 2
= lim → 2 (− 3 − 2) = −3(2) − 2 = − 8
Tangent line: − (−4) = −8( − 2) ⇔ + 4 = − 8 + 16 ⇔ = − 8 + 12.
6. Using (2) with () = ^3 − 3 + 1 and (2 3),
= lim → 0
= lim → 0
= lim → 0
(2 + )^3 − 3(2 + ) + 1 − 3
= lim → 0
8 + 12 + 6^2 + ^3 − 6 − 3 − 2
= lim → 0
9 + 6^2 + ^3
= lim → 0
(9 + 6 + ^2 )
= lim → 0 (9 + 6 + ^2 ) = 9
Tangent line: − 3 = 9( − 2) ⇔ − 3 = 9 − 18 ⇔ = 9 − 15
7. Using (1),
= lim → 1
= lim → 1
= lim → 1
= lim → 1
Tangent line: − 1 = 12 ( − 1) ⇔ = 12 + (^12)
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102 ¤^ CHAPTER 2 DERIVATIVES
(b) The velocity of the particle is equal to the slope of the tangent line of the graph. Note that there is no slope at the corner points on the graph. On the
interval (0 1) the slope is
= 3. On the interval (2 3), the slope is
1 − 3 3 − 2 = − 2. On the interval (4 6), the slope is 3 −^1 6 − 4
12. (a) Runner A runs the entire 100 -meter race at the same velocity since the slope of the position function is constant. Runner B starts the race at a slower velocity than runner A, but finishes the race at a faster velocity.
(b) The distance between the runners is the greatest at the time when the largest vertical line segment fits between the two graphs—this appears to be somewhere between 9 and 10 seconds.
(c) The runners had the same velocity when the slopes of their respective position functions are equal—this also appears to be at about 9 5 s. Note that the answers for parts (b) and (c) must be the same for these graphs because as soon as the velocity for runner B overtakes the velocity for runner A, the distance between the runners starts to decrease.
13. Let () = 40 − 16 ^2.
(2) = lim → 2
= lim → 2
40 − 16 ^2
= lim → 2
− 16 ^2 + 40 − 16
= lim → 2
2 ^2 − 5 + 2
= lim → 2
= −8 lim → 2 (2 − 1) = −8(3) = − 24
Thus, the instantaneous velocity when = 2 is − 24 fts.
14. (a) Let () = 10 − 1 86 ^2.
(1) = lim → 0
= lim → 0
10(1 + ) − 1 86(1 + )^2
= lim → 0
10 + 10 − 1 86(1 + 2 + ^2 ) − 10 + 1 86
= lim → 0
10 + 10 − 1 86 − 3 72 − 1 86 ^2 − 10 + 1 86
= lim → 0
6 28 − 1 86 ^2
= lim → 0 (6 28 − 1 86 ) = 6 28
The velocity of the rock after one second is 6 28 ms.
(b) () = lim → 0
= lim → 0
10( + ) − 1 86( + )^2
− (10 − 1 86 ^2 )
= lim → 0
10 + 10 − 1 86(^2 + 2 + ^2 ) − 10 + 1 86 ^2
= lim → 0
10 + 10 − 1 86 ^2 − 3 72 − 1 86 ^2 − 10 + 1 86 ^2
= lim → 0
10 − 3 72 − 1 86 ^2
= lim → 0
= lim → 0
The velocity of the rock when = is (10 − 3 72 ) ms
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SECTION 2.1 DERIVATIVES AND RATES OF CHANGE ¤^ 103
(c) The rock will hit the surface when = 0 ⇔ 10 − 1 86 ^2 = 0 ⇔ (10 − 1 86 ) = 0 ⇔ = 0 or 1 86 = 10. The rock hits the surface when = 10 1 86 ≈ 5 4 s.
(d) The velocity of the rock when it hits the surface is
1 86
1 86
= 10 − 20 = − 10 ms.
15. () = lim → 0
= lim → 0
( + )^2
^2
= lim → 0
^2 − ( + )^2
^2 ( + )^2
= lim → 0
^2 − (^2 + 2 + ^2 )
^2 ( + )^2
= lim → 0
−(2 + ^2 )
^2 ( + )^2
= lim → 0
^2 ( + )^2
= lim → 0
^2 ( + )^2
^2 · ^2
^3
ms
So (1) =
= − 2 ms, (2) =
ms, and (3) =
ms.
16. (a) The average velocity between times and + is ( + ) − () ( + ) −
1 2 (^ +^ )
2 − 6( + ) + 23 − ^1
1 2
+ 12 ^2 − 6
fts
(i) [4 8]: = 4, = 8 − 4 = 4, so the average velocity is 4 + 12 (4) − 6 = 0 fts.
(ii) [6 8]: = 6, = 8 − 6 = 2, so the average velocity is 6 + 12 (2) − 6 = 1 fts.
(iii) [8 10]: = 8, = 10 − 8 = 2, so the average velocity is 8 + 12 (2) − 6 = 3 fts.
(iv) [8 12]: = 8, = 12 − 8 = 4, so the average velocity is 8 + 12 (4) − 6 = 4 fts.
(b) () = lim → 0
= lim → 0
= − 6 , so (8) = 2 fts.
(c)
17. ^0 (0) is the only negative value. The slope at = 4 is smaller than the slope at = 2 and both are smaller than the slope at = − 2. Thus, ^0 (0) 0 ^0 (4) ^0 (2) ^0 (−2). 18. (a) On [20 60]:
(b) Pick any interval that has the same -value at its endpoints. [0 57] is such an interval since (0) = 600 and (57) = 600.
(c) On [40 60]: ^ (60)^ −^ ^ (40) 60 − 40
=^700 −^200
=^500
On [40 70]:
Since 25 23 13 , the average rate of change on [40 60] is larger.
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SECTION 2.1 DERIVATIVES AND RATES OF CHANGE ¤^ 105
25. We begin by drawing a curve through (0 1) with a slope of 1 to satisfy (0) = 1
and ^0 (0) = 1. We round off our figure at = − 2 to satisfy ^0 (−2) = 0. As
→ − 5 +, → ∞, so we draw a vertical asymptote at = − 5. As → 5 −,
→ 3 , so we draw a dot at (5 3) [the dot could be open or closed].
26. We begin by drawing an odd function (symmetric with respect to the origin)
through the origin with slope − 2 to satisfy 0 (0) = − 2. Now draw a curve starting
at = 1 and increasing without bound as → 2 −^ since lim → 2 −^
() = ∞. Lastly,
reflect the last curve through the origin (rotate 180 ◦) since is an odd function.
27. Using (4) with () = 3^2 − ^3 and = 1,
0 (1) = lim → 0
= lim → 0
[3(1 + )^2 − (1 + )^3 ] − 2
= lim → 0
(3 + 6 + 3^2 ) − (1 + 3 + 3^2 + ^3 ) − 2
= lim → 0
3 − ^3
= lim → 0
(3 − ^2 )
= lim → 0 (3 − ^2 ) = 3 − 0 = 3
Tangent line: − 2 = 3( − 1) ⇔ − 2 = 3 − 3 ⇔ = 3 − 1
28. Using (5) with () = ^4 − 2 and = 1,
^0 (1) = lim → 1
= lim → 1
(^4 − 2) − (−1)
= lim → 1
^4 − 1
= lim → 1
(^2 + 1)(^2 − 1)
= lim → 1
(^2 + 1)( + 1)( − 1)
= lim → 1
[(^2 + 1)( + 1)] = 2(2) = 4
Tangent line: − (−1) = 4( − 1) ⇔ + 1 = 4 − 4 ⇔ = 4 − 5
29. (a) Using (4) with () = 5(1 + ^2 ) and the point (2 2), we have
0 (2) = lim → 0
= lim → 0
1 + (2 + )^2
= lim → 0
^2 + 4 + 5
= lim → 0
5 + 10 − 2(^2 + 4 + 5)
^2 + 4 + 5
= lim → 0
− 2 ^2 − 3
(^2 + 4 + 5)
= lim → 0
(^2 + 4 + 5)
= lim → 0
^2 + 4 + 5
So an equation of the tangent line at (2 2) is − 2 = − 35 ( − 2) or = − 35 + 165.
(b)
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106 ¤^ CHAPTER 2 DERIVATIVES
30. (a) Using (4) with () = 4^2 − ^3 , we have
^0 () = lim → 0
= lim → 0
[4( + )^2 − ( + )^3 ] − (4^2 − ^3 )
= lim → 0
4 ^2 + 8 + 4^2 − (^3 + 3^2 + 3^2 + ^3 ) − 4 ^2 + ^3
= lim → 0
8 + 4^2 − 3 ^2 − 3 ^2 − ^3
= lim → 0
(8 + 4 − 3 ^2 − 3 − ^2 )
= lim → 0
(8 + 4 − 3 ^2 − 3 − ^2 ) = 8 − 3 ^2
At the point (2 8), ^0 (2) = 16 − 12 = 4, and an equation of the tangent line is − 8 = 4( − 2), or = 4. At the point (3 9), ^0 (3) = 24 − 27 = − 3 , and an equation of the tangent line is − 9 = −3( − 3), or = − 3 + 18
(b)
31. Use (4) with () = 3^2 − 4 + 1.
0 () = lim → 0
= lim → 0
[3( + )^2 − 4( + ) + 1] − (3^2 − 4 + 1)]
= lim → 0
3 ^2 + 6 + 3^2 − 4 − 4 + 1 − 3 ^2 + 4 − 1
= lim → 0
6 + 3^2 − 4
= lim → 0
= lim → 0
32. Use (4) with () = 2^3 + .
0 () = lim → 0
= lim → 0
[2( + )^3 + ( + )] − (2^3 + )
= lim → 0
2 ^3 + 6^2 + 6^2 + 2^3 + + − 2 ^3 −
= lim → 0
6 ^2 + 6^2 + 2^3 +
= lim → 0
(6^2 + 6 + 2^2 + 1)
= lim → 0
(6^2 + 6 + 2^2 + 1) = 6^2 + 1
33. Use (4) with () = (2 + 1)( + 3).
0 () = lim → 0
= lim → 0
= lim → 0
= lim → 0
(2^2 + 6 + 2 + 6 + + 3) − (2^2 + 2 + 6 + + + 3)
= lim → 0
= lim → 0
( + 3)^2
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108 ¤^ CHAPTER 2 DERIVATIVES
38. By (4), lim → 0
2 3+^ − 8
= 0 (3), where () = 2^ and = 3.
39. By Equation 5 , (^) lim→ 2
^6 − 64
= 0 (2), where () = ^6 and = 2.
40. By Equation 5 , lim → 1 4
= 0 (4), where () =
and =
41. By (4), lim → 0
cos( + ) + 1 = 0 (), where () = cos and = .
Or : By (4), lim → 0
cos( + ) + 1 = 0 (0), where () = cos( + ) and = 0.
42. By Equation 5 , lim → 6
sin − (^12) −
= ^0
, where () = sin and =
43. (4) = ^0 (4) = lim → 0
= lim → 0
80(4 + ) − 6(4 + )^2
80(4) − 6(4)^2
= lim → 0
(320 + 80 − 96 − 48 − 6 ^2 ) − (320 − 96)
= lim → 0
32 − 6 ^2
= lim → 0
= lim → 0 (32 − 6 ) = 32 m/s
The speed when = 4 is | 32 | = 32 ms.
44. (4) = ^0 (4) = lim → 0
= lim → 0
= lim → 0
= lim → 0
= lim → 0
= lim → 0
m/s.
The speed when = 4 is
(^) = 95 ms.
45. The sketch shows the graph for a room temperature of 72 ◦^ and a refrigerator
temperature of 38 ◦. The initial rate of change is greater in magnitude than the rate of change after an hour.
46. The slope of the tangent (that is, the rate of change of temperature with respect
to time) at = 1 h seems to be about
≈ − 0 7 ◦Fmin.
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SECTION 2.1 DERIVATIVES AND RATES OF CHANGE ¤^ 109
47. (a) (i) [1 0 2 0]:
g/dL h
(ii) [1 5 2 0]:
g/dL h
(iii) [2 0 2 5]: (25)^ −^ (2) 2 5 − 2
=^0 ^012 −^0 ^018
= −^0 ^006
= − 0 012 g/dL h
(iv) [2 0 3 0]: (3)^ −^ (2) 3 − 2
=^0 ^007 −^0 ^018
= − 0 011 g/dL h
(b) We estimate the instantaneous rate of change at = 2 by averaging the average rates of change for [1 5 2 0] and [2 0 2 5]: − 0 012 + (− 0 012) 2
g/dL h
. After 2 hours, the BAC is decreasing at a rate of 0 012 (gdL)h. 48. (a) (i) [2006 2008]:
= 2120 locationsyear
(ii) [2008 2010]:
= 89 locationsyear.
The rate of growth decreased over the period from 2006 to 2010.
(b) [2010 2012]:
= 604 locationsyear.
Using that value and the value from part (a)(ii), we have
= 346 5 locationsyear.
(c) The tangent segment has endpoints (2008 16 ,250) and (2012 17 ,500). An estimate of the instantaneous rate of growth in 2010 is 17 , 500 − 16 , 250 2012 − 2008
= 312 5 locations/year.
49. (a) [1990 2005]:
= 1169 6 thousands of barrels per day per year. This means that oil
consumption rose by an average of 1169 6 thousands of barrels per day each year from 1990 to 2005.
(b) [1995 2000]:
[2000 2005]:
An estimate of the instantaneous rate of change in 2000 is 12 (1337 + 14586) = 1397 8 thousands of barrels per day per year.
50. (a) (i) [4 11]:
RNA copiesmL day
(ii) [8 11]:
=^9 ^4 −^18
= −^8 ^6
≈ − 2 87 RNA copiesmL day
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SECTION 2.1 DERIVATIVES AND RATES OF CHANGE ¤^ 111
54. (a) 0 (5) is the rate of growth of the bacteria population when = 5 hours. Its units are bacteria per hour.
(b) With unlimited space and nutrients, 0 should increase as increases; so ^0 (5) ^0 (10). If the supply of nutrients is limited, the growth rate slows down at some point in time, and the opposite may be true.
55. (a) ^0 (58) is the rate at which the daily heating cost changes with respect to temperature when the outside temperature is
58 ◦F. The units are dollars ◦F.
(b) If the outside temperature increases, the building should require less heating, so we would expect ^0 (58) to be negative.
56. (a) 0 (8) is the rate of change of the quantity of coffee sold with respect to the price per pound when the price is $8 per pound.
The units for ^0 (8) are pounds(dollarspound).
(b) 0 (8) is negative since the quantity of coffee sold will decrease as the price charged for it increases. People are generally less willing to buy a product when its price increases.
57. (a) 0 ( ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mgL)◦C.
(b) For = 16◦C, it appears that the tangent line to the curve goes through the points (0 14) and (32 6). So
^0 (16) ≈
= − 0 25 (mgL)◦C. This means that as the temperature increases past 16 ◦C, the oxygen
solubility is decreasing at a rate of 0 25 (mgL)◦C.
58. (a) 0 ( ) is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature. Its units are (cms)◦C.
(b) For = 15◦C, it appears the tangent line to the curve goes through the points (10 25) and (20 32). So
^0 (15) ≈
= 07 (cms)◦C. This tells us that at = 15◦C, the maximum sustainable speed of Coho salmon is
changing at a rate of 0.7 (cms)◦C. In a similar fashion for = 25◦C, we can use the points (20 35) and (25 25) to
obtain ^0 (25) ≈ 25 −^35 25 − 20 = − 2 (cms)◦C. As it gets warmer than 20 ◦C, the maximum sustainable speed decreases
rapidly.
59. Since () = sin(1) when 6 = 0 and (0) = 0, we have
0 (0) = lim → 0
= lim → 0 sin(1) − 0 = lim → 0 sin(1). This limit does not exist since sin(1) takes the
values − 1 and 1 on any interval containing 0. (Compare with Example 1.5.4.)
60. Since () = ^2 sin(1) when 6 = 0 and (0) = 0, we have
0 (0) = lim → 0
= lim → 0 ^2 sin(1) − 0 = lim → 0 sin(1). Since − 1 ≤ sin
≤ 1 , we have
− || ≤ || sin
≤ || ⇒ − || ≤ sin
≤ ||. Because (^) lim→ 0 (− ||) = 0 and (^) lim→ 0 || = 0, we know that
lim → 0
sin^1
= 0 by the Squeeze Theorem. Thus, 0 (0) = 0.
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112 ¤^ CHAPTER 2 DERIVATIVES
61. (a) The slope at the origin appears to be 1.
(b) The slope at the origin still appears to be 1.
(c) Yes, the slope at the origin now appears to be 0.
2.2 The Derivative as a Function
1. It appears that is an odd function, so 0 will be an even function—that is, 0 (−) = 0 (). (a) 0 (−3) ≈ − 0 2 (b) 0 (−2) ≈ 0 (c) 0 (−1) ≈ 1 (d) 0 (0) ≈ 2 (e) 0 (1) ≈ 1 (f) 0 (2) ≈ 0 (g) ^0 (3) ≈ − 0 2 2. Your answers may vary depending on your estimates. (a) Note: By estimating the slopes of tangent lines on the graph of , it appears that 0 (0) ≈ 6. (b) 0 (1) ≈ 0 (c) 0 (2) ≈ − 1 5 (d) 0 (3) ≈ − 1 3 (e) 0 (4) ≈ − 0 8 (f) 0 (5) ≈ − 0 3 (g) 0 (6) ≈ 0 (h) 0 (7) ≈ 0 2 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.
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114 ¤^ CHAPTER 2 DERIVATIVES
12. The slopes of the tangent lines on the graph of = () are always positive, so the -values of = 0 () are always positive. These values start out relatively small and keep increasing, reaching a maximum at about = 6. Then the -values of = 0 () decrease and get close to zero. The graph of 0 tells us that the yeast culture grows most rapidly after 6 hours and then the growth rate declines. 13. (a) ^0 () is the instantaneous rate of change of percentage of full capacity with respect to elapsed time in hours.
(b) The graph of ^0 () tells us that the rate of change of percentage of full capacity is decreasing and approaching 0.
14. (a) 0 () is the instantaneous rate of change of fuel economy with respect to speed.
(b) Graphs will vary depending on estimates of 0 , but will change from positive to negative at about = 50.
(c) To save on gas, drive at the speed where is a maximum and 0 is 0 , which is about 50 mi h.
15. It appears that there are horizontal tangents on the graph of for = 1963 and = 1971. Thus, there are zeros for those values of on the graph of
^0. The derivative is negative for the years 1963 to 1971.
The graph of the derivative looks like the graph of the cosine function.
17. (a) By zooming in, we estimate that 0 (0) = 0, 0
2
and 0 (2) = 4. (b) By symmetry, 0 (−) = −^0 (). So 0
and 0 (−2) = − 4. (c) It appears that 0 () is twice the value of , so we guess that 0 () = 2.
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SECTION 2.2 THE DERIVATIVE AS A FUNCTION ¤^ 115
(d) 0 () = lim → 0
= lim → 0
( + )^2 − ^2
= lim → 0
^2 + 2 + ^2
− ^2
= lim → 0
2 + ^2
= lim → 0
= lim → 0 (2 + ) = 2
18. (a) By zooming in, we estimate that 0 (0) = 0, 0
2
0 (1) ≈ 3 , 0 (2) ≈ 12 , and 0 (3) ≈ 27.
(b) By symmetry, ^0 (−) = 0 (). So 0
0 (−1) ≈ 3 , ^0 (−2) ≈ 12 , and 0 (−3) ≈ 27.
(c) (d) Since ^0 (0) = 0, it appears that 0 may have the form 0 () = ^2. Using ^0 (1) = 3, we have = 3, so 0 () = 3^2.
(e) 0 () = lim → 0
= lim → 0
( + )^3 − ^3
= lim → 0
(^3 + 3^2 + 3^2 + ^3 ) − ^3
= lim → 0
3 ^2 + 3^2 + ^3
= lim → 0
(3^2 + 3 + ^2 )
= lim → 0
(3^2 + 3 + ^2 ) = 3^2
19. ^0 () = lim → 0
= lim → 0
[3( + ) − 8] − (3 − 8)
= lim → 0
= lim → 0
= lim → 0 3 = 3
Domain of = domain of ^0 = R.
20. ^0 () = lim → 0
= lim → 0
[( + ) + ] − ( + )
= lim → 0
= lim → 0
= lim → 0
Domain of = domain of 0 = R.
21. ^0 () = lim → 0
= lim → 0
2 5( + )^2 + 6( + )
2 5 ^2 + 6
= lim → 0
2 5(^2 + 2 + ^2 ) + 6 + 6 − 2 5 ^2 − 6
= lim → 0
2 5 ^2 + 5 + 2 5 ^2 + 6 − 2 5 ^2
= lim → 0
5 + 2 5 ^2 + 6
= lim → 0
= lim → 0
Domain of = domain of ^0 = R.
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SECTION 2.2 THE DERIVATIVE AS A FUNCTION ¤^ 117
26. ^0 () = lim → 0
= lim → 0
( + )^2 − 1
^2 − 1
= 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
(2^3 + 4^2 + 2^2 − 2 − 3 ^2 − 6 − 3 ^2 + 3) − (2^3 + 2^2 − 3 ^2 − 2 − 2 + 3)
= lim → 0
4 ^2 + 2^2 − 6 − 3 ^2 − 2 ^2 + 2
= lim → 0
(2^2 + 2 − 6 − 3 + 2)
= lim → 0
2 ^2 + 2 − 6 − 3 + 2
2 ^2 − 6 + 2
(2 − 3)^2
Domain of = domain of 0 = (−∞ 32 ) ∪ ( 32 ∞).
27. ^0 () = lim → 0
= lim → 0
= lim → 0
[1 − 2( + )](3 + ) − [3 + ( + )](1 − 2 )
[3 + ( + )](3 + )
= lim → 0
3 + − 6 − 2 ^2 − 6 − 2 − (3 − 6 + − 2 ^2 + − 2 )
[3 + ( + )](3 + )
= lim → 0
= lim → 0
= lim → 0
= −^7
(3 + )^2
Domain of = domain of ^0 = (−∞ −3) ∪ (− 3 ∞).
28. ^0 () = lim → 0
= lim → 0
( + )^3 ^2 − ^3 ^2
= lim → 0
[( + )^3 ^2 − ^3 ^2 ][( + )^3 ^2 + ^3 ^2 ]
[( + )^3 ^2 + ^3 ^2 ]
= lim → 0
( + )^3 − ^3
[( + )^3 ^2 + ^3 ^2 ]
= lim → 0
^3 + 3^2 + 3^2 + ^3 − ^3
[( + )^3 ^2 + ^3 ^2 ]
= lim → 0
3 ^2 + 3 + ^2
[( + )^3 ^2 + ^3 ^2 ]
= lim → 0
3 ^2 + 3 + ^2
( + )^3 ^2 + ^3 ^2
3 ^2
2 ^3 ^2
= 32 ^1 ^2
Domain of = domain of ^0 = [0 ∞). Strictly speaking, the domain of ^0 is (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 62) does exist at 0 , so in that sense one could regard the domain of 0 to be [0 ∞).
29. 0 () = lim → 0
= lim → 0
( + )^4 − ^4
= lim → 0
^4 + 4^3 + 6^2 ^2 + 4^3 + ^4
− ^4
= lim → 0
4 ^3 + 6^2 ^2 + 4^3 + ^4
= lim → 0
4 ^3 + 6^2 + 4^2 + ^3
= 4^3
Domain of = domain of 0 = R.
° c 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.
INSTRUCTOR USE ONLY
118 ¤^ CHAPTER 2 DERIVATIVES
30. (a)
(b) Note that the third graph in part (a) has small negative values for its slope, 0 ; but as → 6 −, 0 → −∞.
See the graph in part (d).
(c) 0 () = lim → 0
= lim → 0
= lim → 0
[6 − ( + )] − (6 − )
(^) = lim → 0
= lim → 0
Domain of = (−∞ 6], domain of 0 = (−∞ 6).
(d)
31. (a) 0 () = lim → 0
= lim → 0
[( + )^4 + 2( + )] − (^4 + 2)
= lim → 0
^4 + 4^3 + 6^2 ^2 + 4^3 + ^4 + 2 + 2 − ^4 − 2
= lim → 0
4 ^3 + 6^2 ^2 + 4^3 + ^4 + 2
= lim → 0
(4^3 + 6^2 + 4^2 + ^3 + 2)
= lim → 0
(4^3 + 6^2 + 4^2 + ^3 + 2) = 4^3 + 2
(b) Notice that ^0 () = 0 when has a horizontal tangent, 0 () is
positive when the tangents have positive slope, and ^0 () is
negative when the tangents have negative slope.
32. (a) 0 () = lim → 0
= lim → 0
[( + ) + 1( + )] − ( + 1)
= lim → 0
( + )^2 + 1
^2 + 1
= lim → 0
[( + )^2 + 1] − ( + )(^2 + 1)
= lim → 0
(^3 + 2^2 + ^2 + ) − (^3 + + ^2 + )
= lim → 0
^2 + ^2 −
= lim → 0
(^2 + − 1)
= lim → 0
^2 + − 1
^2 − 1
^2
, or 1 −
^2
° c 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.
INSTRUCTOR USE ONLY
