Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 4 REVIEW ¤ 221
f is CU on (−∞, −0.12) and (1.24, ∞) and CD on (−0.12, 1.24);andf has inflection points at about (−0.12, 1.98) and
(1.24, −12.1).
39. From the graph, we estimate the points of inflection to be about (±0.82, 0.22).
41. f(x) =
f(x) =e −1/x2 ⇒ f 0 (x) =2x −3 e −1/x2 ⇒
f 00 (x) =2[x −3 (2x −3 )e −1/x2 + e −1/x2 (−3x −4 )] = 2x −6 e −1/x2 2 − 3x 2 .
This is 0 when 2 − 3x 2 2
=0 ⇔ x = ± ,sotheinflection points
3
are ±
2
3 ,e−3/2 .
cos 2 x
√
x2 + x +1 , −π ≤ x ≤ π ⇒ f 0 (x) =− cos x [(2x +1)cosx +4(x2 + x +1)sinx]
2(x 2 + x +1) 3/2 ⇒
f 00 (x) =− (8x4 +16x 3 +16x 2 +8x +9)cos 2 x − 8(x 2 + x + 1)(2x +1)sinx cos x − 8(x 2 + x +1) 2 sin 2 x
4(x 2 + x +1) 5/2
f(x) =0 ⇔ x = ± π 2 ; f 0 (x) =0 ⇔ x ≈−2.96, −1.57, −0.18, 1.57, 3.01;
f 00 (x) =0 ⇔ x ≈−2.16, −0.75, 0.46,and2.21.
The x-coordinates of the maximum points are the values at which f 0 changes from positive to negative, that is, −2.96, −0.18,
and 3.01. Thex-coordinates of the minimum points are the values at which f 0 changes from negative to positive, that is,
−1.57 and 1.57. Thex-coordinates of the inflection points are the values at which f 00 changes sign, that is, −2.16, −0.75,
0.46,and2.21.
43. The family of functions f(x) =ln(sinx + C) all have the same period and all
have maximum values at x = π +2πn. Since the domain of ln is (0, ∞), f has
2
a graph only if sin x + C>0 somewhere. Since −1 ≤ sin x ≤ 1, this happens
if C>−1,thatis,f has no graph if C ≤−1. Similarly, if C>1,then
sin x + C>0 and f is continuous on (−∞, ∞).AsC increases, the graph of
f is shifted vertically upward and flattens out. If −1