Solução_Calculo_Stewart_6e

F.
180 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
decreasing on 2nπ + π 2 , (2n +1)π for each integer n.
TX.10
H.
F. Local maximum values f 2nπ + π 2
=0, no local minimum.
G. f 00 (x) =− csc 2 x 0 ⇔ x 2 < 1 2
⇔ |x| < 1 √
2
,sof is increasing on
and decreasing on −∞, − √ 1
2
and
√
1
1
2
, ∞ . F. Local maximum value f √
2
=1/ √ 2e, local minimum
value f − √ 1
2
= −1/ √ 2e G. f 00 (x) =−2xe −x2 (1 − 2x 2 ) − 4xe −x2 =2xe −x2 (2x 2 − 3) > 0 ⇔
3
x> or − 3
2e −2x [multiply by e 2x ] ⇔
H.
e 5x > 2 3
⇔ 5x >ln 2 3
⇔ x> 1 5 ln 2 3 ≈−0.081. Similarly, f 0 (x) < 0 ⇔
x< 1 5 ln 2 3 . f is decreasing on −∞, 1 5 ln 2 3
and increasing on
1
5 ln 2 3 , ∞ .
F. Local minimum value f 1
ln 2
5 3 = 2
3/5
+
2 −2/5
≈ 1.96; no local maximum.
3
3
G. f 00 (x) =9e 3x +4e −2x ,so f 00 (x) > 0 for all x,andf is CU on (−∞, ∞). NoIP
53. m = f(v) =
m 0
1 − v2 /c .Them-intercept is f(0) = m 0.Therearenov-intercepts. lim f(v) =∞,sov = c is a VA.
2 v→c− f 0 (v) =− 1 2 m0(1 − v2 /c 2 ) −3/2 (−2v/c 2 )=
increasing on (0,c). There are no local extreme values.
m 0 v
c 2 (1 − v 2 /c 2 ) 3/2 =
f 00 (v)= (c2 − v 2 ) 3/2 (m 0 c) − m 0 cv · 3
2 (c2 − v 2 ) 1/2 (−2v)
[(c 2 − v 2 ) 3/2 ] 2
= m 0c(c 2 − v 2 ) 1/2 [(c 2 − v 2 )+3v 2 ]
(c 2 − v 2 ) 3 = m 0c(c 2 +2v 2 )
(c 2 − v 2 ) 5/2 > 0,
so f is CU on (0,c). Therearenoinflection points.
m 0 v
c 2 (c 2 − v 2 ) 3/2
c 3 =
m 0 cv
> 0,sof is
(c 2 − v 2 )
3/2