Solução_Calculo_Stewart_6e

F.
162 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
(d) f 00 (x) = (x2 − 1) 2 (−2) − (−2x) · 2(x 2 − 1)(2x)
[(x 2 − 1) 2 ] 2
= 2(x2 − 1)[−(x 2 − 1) + 4x 2 ]
(x 2 − 1) 4 = 2(3x2 +1)
(x 2 − 1) 3 .
The sign of f 00 (x) is determined by the denominator; that is, f 00 (x) > 0 if
|x| > 1 and f 00 (x) < 0 if |x| < 1. Thus,f is CU on (−∞, −1) and (1, ∞),
and f is CD on (−1, 1). Therearenoinflection points.
(e)
√
47. (a) lim x2 +1− x = ∞ and
x→−∞
√
lim x2 +1− x √
= lim x2 +1− x √ x 2 +1+x
√
x→∞
x→∞
x2 +1+x = lim 1
√ =0,soy =0is a HA.
x→∞ x2 +1+x
(b) f(x) = √ x 2 +1− x ⇒ f 0 x
x
(x) = √ − 1. Since√ x2 +1 x2 +1 < 1 for all x, f 0 (x) < 0,sof is decreasing on R.
(c) No minimum or maximum
(d) f 00 (x) = (x2 +1) 1/2 (1) − x · 1
2 (x2 +1) −1/2 (2x)
√
x2 +1 2
(e)
(x 2 +1) 1/2 x 2
−
(x
=
2 +1) 1/2
x 2 +1
so f is CU on R. NoIP
= (x2 +1)− x 2
(x 2 +1) 3/2 =
1
> 0,
(x 2 +1)
3/2
49. f(x) =ln(1− ln x) is defined when x>0 (so that ln x is defined) and 1 − ln x>0 [so that ln(1 − ln x) is defined].
The second condition is equivalent to 1 > ln x ⇔ x 0 ⇔ ln x