Math 170: Assignment 7 Solutions
Part I
Section 2.6
15. f 0 (x) = 2ex
16. f 0 (x) = ex + exe−1
17. f 0 (x) = (ln 2)2x + 2x
18. f 0 (x) = (2 ln 3)3x−1
19. f 0 (x) = −2/x
20. f 0 (x) =
3
x−1
ln 2
21. f 00 (x) = 2ex
22. f 00 (x) = ex + e(e − 1)xe−2
23. f 00 (x) = (ln 2)2 2x + 2
24. f 00 (x) = 2(ln 3)2 3x−1
25. f 00 (x) = 2/x2
26. f 00 (x) = − ln32 x−2
27. F (x) = 2ex + πx + C
28. F (x) = ex +
xe+1
e+1
x3
3
29. F (x) =
2x
ln 2
30. F (x) =
2
3x
3 ln 3
+
+ ex + C
+ 2x + C
+C
32. Since y 0 (x) = 1/x, y 0 (1) = 1. Thus, the line tangent to the curve y = ln x at the point
(1, 0) is y = x − 1.
Section 2.7
8. Every antiderivative of F must have the form F (x) = x2 /2 + sin x + C. If F (0) = 0,
it follows that C = 0. Thus, F (x) = x2 /2 + sin x.
17. f 0 (x) = 2 cos x
18. f 0 (x) = −3 sin x
21. f 0 (x) = 3 cos x + 2 sin x
22. f 0 (x) = −2 sin x − 3 cos x
23. f 00 (x) = −2 sin x
24. f 00 (x) = −3 cos x
27. f 00 (x) = −3 sin x + 2 cos x
28. f 00 (x) = −2 cos x + 3 sin x
29. F (x) = −2 cos x + C
30. F (x) = 3 sin x + C
33. F (x) = −3 cos x − 2 sin x + C
34. F (x) = 2 sin x + 3 cos x + C
Section 3.1
22. f 0 (x) = 2x cos x − x2 sin x
24. f 0 (x) = (3x2 + x3 )ex
26. f 0 (x) = cos2 x − sin2 x
√
28. f 0 (x) = 23 x sin x + x3/2 cos x
30. f 0 (x) = − (x
32. f 0 (x) =
2 +e2x ) sin x+(2x+2e2x ) cos x
(x2 +e2x )2
(cos x)(cos(2x))+2(sin x)(sin(2x))
cos2 (2x)
(Mistakenly assigned: requires chain rule.)
34. f 0 (x) = sec(x + 3) tan(x + 3) (Mistakenly assigned: requires chain rule.)
36. f 0 (x) = 4 sec2 (x + 5) (Mistakenly assigned: requires chain rule.)
Part II
Section 2.7
h−cos 0
2. limh→0 cos hh−1 can be rewritten as limh→0 cos h−0
= limh→0 g(h)−g(0)
. The latter
h−0
0
0
expression is, by definition, g (0). In this case, g (0) = − sin 0 = 0.
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40. A graph of f (x) over the interval [0, 2π] looks like this:
2
1.6
1.2
0.8
0.4
0
0.8
1.6
2.4
3.2
4
4.8
5.6
-0.4
-0.8
-1.2
-1.6
-2
Since f 0 (x) = cos x − sin x, local extrema occur where cos x − sin x = 0. In other
words, where cos x = sin x. On the √
given interval, this occurs at x = π/4 √
and 5π/4.
By inspection, we know that (π/4, 2) is a local maximum and (5π/4, − 2) is a
local minimum. We could also determine that the first point is a local maximum by
observing that f 00 (π/4) < 0, and the second point is a local minimum because
f 00 (5π/4) > 0. (These points are also absolute extrema on the given interval.)
Inflection points occur where the second derivative is zero: f 00 (x) = −(sin x + cos x)
= 0. That is, where sin x = − cos x. The inflection points are thus located at
(3π/4, 0) and (7π/4, 0).
44. I should not have assigned this problem because finding f 0 (x) requires the chain rule,
which we had not yet covered before this assignment was due. My apologies.
58. (a) Let h(x) = f (x) − g(x), so the vertical distance between the curves f and g at x
is |h(x)|. Since the stationary points of h on the interval [0, 2π] occur when
h0 (x) = − cos x + sin x = 0, it follows that the maximum value of |h(x)| must
occur when x = 0, x = π/4, x = 5π/4, or x = 2π. (Instructor’s note: There may
be other critical points |h(x)| on the interval due to the absolute value issue, but
if you think about you, you’ll see that those have to exist when h(x) = 0, and
those clearly aren’t
√ all four values gives us |h(0)| = 0,
√ of interest to us.) Checking
|h(π/4)| = |1 − x| ≈ 0.414, |h(5π/4)| = |1 + 2| ≈ 2.414, and |h(2π)| = 0.√Thus,
the maximum distance between the functions on the interval [0, 2π] is 1 + 2.
√
√
(b) f 0 (5π/4) = − cos(5π/4) = 2/2. g 0 (5π/4) = − sin(5π/4) = 2/2.
Section 3.1
53. Since g 0 (x) = (cos x)f (x) + (sin x)f 0 (x), it follows that g 0 (0) = f (0) = 3.
54. Yes. g 0 (x) = (cos x)f (x) + (sin x)f 0 (x), so g 0 (−π/2) = −f 0 (−π/2) > 0, since
f 0 (π/2) < 0.
55. Yes. g 0 (x) = (cos x)f (x) + (sin x)f 0 (x), so g 0 (π/2) = f 0 (π/2) > 3. Since g 0 (π/2) > 0, g
is increasing at x = π/2.
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56. g 00 (x) = −(sin x)f (x) + 2(cos x)f 0 (x) + (sin x)f 00 (x). So g 00 (0) = 2f 0 (0) = 2.
57. No. g 00 (x) = −(sin x)f (x) + 2(cos x)f 0 (x) + (sin x)f 00 (x). So
g 00 (π/2) = −f (π/2) + f 00 (π/2). Since f 0 (x) > 0 on the interval [0, 2],
f (π/2) > f (0) = 3. Since the graph of f 0 is decreasing at x = π/2, f 00 (π/2) < 0.
Therefore, g 00 (π/2) < 0. This implies that g is concave down at x = π/2. [Whew!]
58. h0 (x) = ex (f (x) + f 0 (x)), so h0 (0) = f (0) + f 0 (0) = 3 + 1 = 4.
59. Yes. h0 (x) = ex (f (x) + f 0 (x)), so h0 (1) = (f (1) + f 0 (1))e1 . Since f 0 (x) > 0 on the
interval [0, 1], f (1) > f (0) = 3. Therefore, h0 (1) > 0, which implies that h is
increasing at x = 1.
60. Yes. h0 (x) = ex (f (x) + f 0 (x)), so h0 (2) = (f (2) + f 0 (2))e2 . Since f 0 (x) > 0 on the
interval [0, 2], f (2) > f (0) = 3. Therefore,h0 (2) > 0, which implies that h is increasing
at x = 2.
61. h00 (x) = ex (f (x) + 2f 0 (x) + f 00 (x)). Since f (0) = 3, f 0 (0) = 1, and f 00 (0) ≈ 2.5,
h00 (0) ≈ 7.5.
62. Yes. Since the graph of f 0 is increasing at x = 1, f 00 (1) > 0. Thus,
h00 (1) = e1 (f (1) + 2f 0 (1) + f 00 (2)) > 0. This implies g 00 (1) > 0, so g is concave up at
x = 1. (This makes use of facts already demonstrated in previous problems. It’s OK
to do that! Introduce what’s new, but there’s no need to redo what you’ve already
done.)
63. Yes. h00 (x) = ex (f (x) + 2f 0 (x) + f 00 (x)). Now, f 00 (2) ≈ −1.5, f 0 (2) = 3, and, since
f 0 (x) > 1 on the interval [0, 2], f (2) > f (0) + 2 · 1 = 5, g 00 (2) > 0. This implies that g
is concave up at x = 2.
64. k 0 (x) = 2f (x)f 0 (x), so k 0 (0) = 2f (0)f 0 (0) = 2 · 3 · 1 = 6.
65. k 0 (x) = 2f (x)f 0 (x), so k 0 (1) = 2f (1)f 0 (1). Now, f 0 (x) > 0 on the interval [0, 1], so
f (1) > f (0) = 3 and f 0 (1) = 3, so k 0 (1) = 6f (1) > 0. This implies that k is increasing
at x = 1.
66. No. k 0 (x) = 2f (x)f 0 (x), so k 0 (−1) = 2f (−1)f 0 (−1). Now, f (0) = 3 and
−2 < f 0 (x) < 1 if −1 < x < 0, so the speed limit law implies that 2 < f (−1) < 5. It
follows that k 0 (−1) < 0, so k is decreasing at x = −1.
67. k 00 (x) = 2(f 0 (x))2 + 2f (x) · f 00 (x). Therefore, k 00 (0) = 2(f 0 (0))2 + 2f (0) · f 00 (0) =
2 · (1)2 + 6f 00 (0) ≈ 2 + 6 · 2.5 = 17.
68. No. k 00 (x) = 2(f 0 (x))2 + 2f (x) · f 00 (x), so k 00 (1) = 2(f 0 (1))2 + 2f (1)f 00 (1). f (1), f 0 (1),
and f 00 (1) are all positive, so k 00 (1) > 0. This implies that k is concave up at x = 1.
72. F 0 (x) = Axex + (A + B)ex = f (x) = xex . Matching coefficients tells us that A = 1
and A + B = 0, so B = −1. (In other words, Axex + (A + B)ex = 1 · xex + 0 · ex .)
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