James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

356 Chapter 4 Applications of Differentiation
;
22. f sxd − 2x 2 1 5
x 2 1 1
23–24 Find the antiderivative F of f that satisfies the given
condition. Check your answer by comparing the graphs of
f and F.
23. f sxd − 5x 4 2 2x 5 , Fs0d − 4
24. f sxd − 4 2 3s1 1 x 2 d 21 , Fs1d − 0
51–52 The graph of a function f is shown. Which graph is
an antiderivative of f and why?
51. y
a
f
b
c
x
52.
y
f
b
c
a
x
25–48 Find f.
25. f 0sxd − 20x 3 2 12x 2 1 6x
26. f 0sxd − x 6 2 4x 4 1 x 1 1
27. f 0sxd − 2x 1 3e x 28. f 0sxd − 1yx 2
29. f -std − 12 1 sin t 30. f -std − st 2 2 cos t
31. f 9sxd − 1 1 3sx , f s4d − 25
32. f 9sxd − 5x 4 2 3x 2 1 4, f s21d − 2
33. f 9std − 4ys1 1 t 2 d, f s1d − 0
34. f 9std − t 1 1yt 3 , t . 0, f s1d − 6
35. f 9sxd − 5x 2y3 , f s8d − 21
36. f 9sxd − sx 1 1dysx , f s1d − 5
37. f 9std − sec t ssec t 1 tan td, 2y2 , t , y2,
f sy4d − 21
38. f 9std − 3 t 2 3yt, f s1d − 2, f s21d − 1
39. f 0sxd − 22 1 12x 2 12x 2 , f s0d − 4, f9s0d − 12
40. f 0sxd − 8x 3 1 5, f s1d − 0, f 9s1d − 8
41. f 0sd − sin 1 cos , f s0d − 3, f 9s0d − 4
42. f 0std − t 2 1 1yt 2 , t . 0, f s2d − 3, f 9s1d − 2
43. f 0sxd − 4 1 6x 1 24x 2 , f s0d − 3, f s1d − 10
44. f 0sxd − x 3 1 sinh x, f s0d − 1, f s2d − 2.6
45. f 0sxd − e x 2 2 sin x, f s0d − 3, f sy2d − 0
46. f 0std − s 3 t 2 cos t, f s0d − 2, f s1d − 2
47. f 0sxd − x 22 , x . 0, f s1d − 0, f s2d − 0
48. f -sxd − cos x, fs0d − 1, f9s0d − 2, f 0s0d − 3
49. Given that the graph of f passes through the point
(2, 5) and that the slope of its tangent line at sx, f sxdd
is 3 2 4x, find f s1d.
50. Find a function f such that f 9sxd − x 3 and the line
x 1 y − 0 is tangent to the graph of f .
;
53. The graph of a function is shown in the figure. Make a
rough sketch of an antiderivative F, given that Fs0d − 1.
y
y=ƒ
0 x
1
54. The graph of the velocity function of a particle is shown
in the figure. Sketch the graph of a position function.
√
0 t
55. The graph of f 9 is shown in the figure. Sketch the graph
of f if f is continuous and f s0d − 21.
y
2
1
0 1 2
_1
y=fª(x)
56. (a) Use a graphing device to graph f sxd − 2x 2 3sx .
(b) Starting with the graph in part (a), sketch a rough graph
of the antiderivative F that satisfies Fs0d − 1.
(c) Use the rules of this section to find an expression
for Fsxd.
(d) Graph F using the expression in part (c). Compare with
your sketch in part (b).
; 57–58 Draw a graph of f and use it to make a rough sketch
of the antiderivative that passes through the origin.
57. f sxd − sin x
2
, 22 < x < 2
1 1 x
58. f sxd − sx 4 2 2x 2 1 2 2 2, 23 < x < 3
x
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