Solução_Calculo_Stewart_6e

F.
TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 163
(d) f 00 (x) = (x +1)2 e −1/(x+1) 1/(x +1) 2 − e −1/(x+1) [2(x +1)]
[(x +1) 2 ] 2
= e−1/(x+1) [1 − (2x +2)]
(x +1) 4 = − e−1/(x+1) (2x +1)
(x +1) 4 ⇒
(e)
f 00 (x) > 0 ⇔ 2x +1< 0 ⇔ x 0 ⇒ (x − 3) 5 > 0 ⇒ x − 3 > 0 ⇒ x>3. Thus,f is increasing on the interval (3, ∞).
55. (a) From the graph, we get an estimate of f(1) ≈ 1.41 as a local maximum
value, and no local minimum value.
f(x) = x +1 √
x2 +1
⇒ f 0 (x) =
1 − x
(x 2 +1) 3/2 .
f 0 (x) =0 ⇔ x =1. f(1) = 2 √
2
= √ 2 is the exact value.
(b) From the graph in part (a), f increases most rapidly somewhere between x = − 1 and x = − 1 .Tofind the exact value,
2 4
we need to find the maximum value of f 0 , which we can do by finding the critical numbers of f 0 .
f 00 (x) = 2x2 − 3x − 1
=0 ⇔ x = 3 ± √ 17
. x = 3+√ 17
corresponds to the minimum value of f 0 .
(x 2 +1) 5/2 4
4
Themaximumvalueoff 0 3 −
is at
√
17 7
4
, − √ 17
≈ (−0.28, 0.69).
6 6
57. f(x) =cosx + 1 2 cos 2x ⇒ f 0 (x) =− sin x − sin 2x ⇒ f 00 (x) =− cos x − 2cos2x
(a)
From the graph of f,itseemsthatf is CD on (0, 1),CUon(1, 2.5),CDon
(2.5, 3.7),CUon(3.7, 5.3), and CD on (5.3, 2π). The points of inflection
appear to be at (1, 0.4), (2.5, −0.6), (3.7, −0.6),and(5.3, 0.4).
(b)
From the graph of f 00 (and zooming in near the zeros), it seems that f is CD
on (0, 0.94),CUon(0.94, 2.57),CDon(2.57, 3.71),CUon(3.71, 5.35),
and CD on (5.35, 2π). Refined estimates of the inflection points are
(0.94, 0.44), (2.57, −0.63), (3.71, −0.63),and(5.35, 0.44).
59. In Maple, we define f andthenusethecommand
plot(diff(diff(f,x),x),x=-2..2);. InMathematica,wedefine f
andthenusePlot[Dt[Dt[f,x],x],{x,-2,2}]. Weseethatf 00 > 0 for
x0.0 [≈ 0.03] andf 00 < 0 for −0.6