This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ) [0, +∞)
(c) none
(e) none (b) (−∞, 0]
(d) (−∞, 0), (0, +∞) 4(x + 1)
3x2/3
4(x − 2)
f (x) =
9x5/3 (a) [−1, +∞)
(c) (−∞, 0), (2, +∞)
(e) 0, 2 (b) (−∞, −1]
(d) (0, 2) 4(x − 1/4)
3x2/3
4(x + 1/2)
f (x) =
9x5/3 (a) [1/4, +∞)
(c) (−∞, −1/2), (0, +∞)
(e) −1/2, 0 (b)
(d) (a) (−∞, 0]
(c) (−∞, −1), (1, +∞)
(e) −1, 1 (b) [0, +∞)
(d) (−1, 1) 19. f (x) = 20. f (x) = 21. f (x) = −xe−x /2
2
f (x) = (−1 + x2 )e−x /2 22. f (x) = (2x2 + 1)ex
2
f (x) = 2x(2x2 + 3)ex (a) (−∞, +∞)
(c) (0, +∞)
(e) 0 (b)
(d) 23. f (x) = 2x
1 + x2
1 − x2
f (x) = 2
(1 + x2 )2 (a) [0, +∞)
(c) (−1, 1)
(e) −1, 1 (b) (−∞, 0]
(d) (−∞, −1), (1, +∞) 24. f (x) = x(2 ln x + 1)
f (x) = 2 ln x + 3 (a) [e−1/2 , +∞)
(c) (e−3/2 , +∞)
(e) e−3/2 (b) (0, e−1/2 ]
(d) (0, e−3/2 ) 25. f (x) = − sin x
f (x) = − cos x
(a) [π, 2π ]
(c) (π/2, 3π/2)
(e) π/2, 3π/2 2 2 (−∞, 1/4]
(−1/2, 0) none
(−∞, 0) 1 (b)
(d) [0, π ]
(0, π/2), (3π/2, 2π ) 0 2p 1 26. f (x) = 2 sin 4x
f (x) = 8 cos 4x
(a) (0, π/4], [π/2, 3π/4]
(b) [π/4, π/2], [3π/4, π ]
(c) (0, π/8), (3π/8, 5π/8), (7π/8, π ) (d) (π/8, 3π/8), (5π/8, 7π/8)
(e) π/8, 3π/8, 5π/8, 7π/8 1 0 p
0 Exercise Set 5.1 27. 142 f (x) = sec2 x
f (x) = 2 sec2 x tan x
(a) (−π/2, π/2)
(c) (0, π/2)
(e) 0 10 (b) none
(d) (−π/2, 0) ^ 6 10 28. f (x) = 2 − csc2 x 8 cos x
f (x) = 2 csc x cot x = 2 3
sin x
(a) [π/4, 3π/4]
(c) (0, π/2)
(e) π/2
2 (b) (0, π/4], [3π/4, π )
(d) (π/2, π )
0 p 2 29. f (x) = cos 2x
f (x) = −2 sin 2x
(a) [0, π/4], [3π/4, π ]
(c) (π/2, π )
(e) π/2 0.5 (b) [π/4, 3π/4]
(d) (0, π/2) 0 p 0.5 30. f (x) = −2 cos x sin x − 2 cos x = −2 cos x(1 + sin x)
f (x) = 2 sin x (sin x + 1) − 2 cos2 x = 2 sin x(sin x + 1) − 2 + 2 sin2 x = 4(1 + sin x)(sin x − 1/2)
Note: 1 + sin x ≥ 0
2
(a) [π/2, 3π/2]
(b) [0, π/2], [3π/2, 2π ]
(c) (π/6, 5π/6)
(d) (0, π/6), (5π/6, 2π )
(e) π/6, 5π/6
0 o 2 31. (a) (b) y (c) y 4 4 x
2 y 4 x x
2 2 143 32. Chapter 5 (a) (b) y 4 (c) y 4 4 x
2 x 34. x
2 2 (a) f (x) = 3(x − a)2 , f (x) = 6(x − a); inﬂection point is (a, 0) (b) 33. y f (x) = 4(x − a)3 , f (x) = 12(x − a)2 ; no inﬂection points For n ≥ 2, f (x) = n(n − 1)(x − a)n−2 ; there is a sign change of f (point of inﬂection) at (a, 0) if and
only if n is odd. For n = 1, y = x − a, so there is no point of inﬂection. 35. f (x) = 1/3 − 1/[3(1 + x)2/3 ] so f is increasing on [0, +∞) thus if
√
3
x
√ > 0, then f (x) > f (0) = 0, 1 + x/3 − 1 + x > 0,
3
1 + x < 1 + x/3. 2.5 0 10
0 36. f (x) = sec2 x − 1 so f is increasing on [0, π/2) thus if 0 < x < π/2,
then f (x) > f (0) = 0, tan x − x > 0, x < tan x. 10 0
0 37. x ≥ sin x on [0, +∞): let f (x) = x − sin x. Then f (0) = 0 and
f (x) = 1 − cos x ≥ 0, so f (x) is increasing on [0, +∞). 6 4 0 4 1 38. (a) Let h(x) = ex − 1 − x for x ≥ 0. Then h(0) = 0 and h (x) = ex − 1 ≥ 0 for x ≥ 0, so h(x) is
increasing. (b) Let h(x) = ex − 1 − x − 1 x2 . Then h(0) = 0 and h (x) = ex − 1 − x. By part (a), ex − 1 − x ≥ 0
2
for x ≥ 0, so h(x) is increasing. Exercise Set 5.1 (c) 6 0 6 2
0 39. 144 0 2
0 Points of inﬂection at x = −2, +2. Concave up on (−5, −2) and
(2, 5); concave down on (−2, 2). Increasing on [−3.5829, 0.2513] and
[3.3316, 5] , and decreasing on [−5, −3.5829] and [0.2513, 3.3316]. 250 5 5 250 40. √
√
Points of inﬂection at x = ±1/ 3. Concave √ on (−5, −1/ 3) and
up
√
√
(1/ 3, 5), and concave down on (−1/ 3, 1/ 3). Increasing on
[−5, 0] and decreasing on [0, 5]. 1 5 5 2 41. Break the interval [−5, 5] into ten subintervals and check f (x) at each endpoint. We ﬁnd f (−1) > 0
and f (0) < 0. Reﬁne [−1, 0] into ten subintervals; f (−0.2) > 0, f (−0.1) < 0; repeat, f (−0.18) > 0,
f (−0.17) < 0, so x = −0.175 is correct to two decimal places. Note also that f (1) = 0 so there are
two inﬂection points. 42. Break the interval [−5, 5] into ten subintervals and check f (x) at each endpoint. We discover
f (−1) > 0, f (0) < 0 and f (1) > 0. Reﬁne [−1, 0] into ten subintervals and we see that f (−0.6) > 0,
f (−0.5) < 0. Subdivide [−0.6, −0.5] into 10 subintervals and we see that f (−0.58) > 0 and
f (−0.57) < 0. Thus x = −0.575 is within 0.005 of the true root and is thus correct to two decimal places. For the other root we could proceed in a similar manner, but it easier to note that f (x)
is an even function and thus the other root is x = 0.575 to two decimal places.
43. 44. 45. 90x3 − 81x2 − 585x + 397
. The denominator has complex roots, so is always positive; hence
(3x2 − 5x + 8)3
the xcoordinates of the points of inﬂection of f (x) are the roots of the numerator (if it changes sign).
A plot of the numerator over [−5, 5] shows roots lying in [−3, −2], [0, 1], and [2, 3]. Breaking each of
these intervals into ten subintervals locates the roots in [−2.5, −2.4], [0.6, 0.7] and [2.7, 2.8]. Thus to
one decimal pla...
View
Full
Document
This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

Click to edit the document details