Solução_Calculo_Stewart_6e

F.
160 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
(c) f 00 (x) =4− 12x 2 =4(1− 3x 2 ). f 00 (x) =0 ⇔ 1 − 3x 2 =0 ⇔
(d)
x 2 = 1 3
⇔ x = ±1/ √ 3. f 00 (x) > 0 on −1/ √ 3, 1/ √ 3 and f 00 (x) < 0
on −∞, −1/ √ 3 and 1/ √ 3, ∞ .Sof is concave upward on
−1/
√
3, 1/
√
3
and f is concave downward on
−∞, −1/
√
3
and
√ √ 1/ 3, ∞ . f ±1/ 3 =2+
2
− 1 = 23 . There are points of inflection
3 9 9
at ±1/ √
3, 23
9 .
37. (a) h(x) =(x +1) 5 − 5x − 2 ⇒ h 0 (x) =5(x +1) 4 − 5. h 0 (x) =0 ⇔ 5(x +1) 4 =5 ⇔ (x +1) 4 =1 ⇒
(x +1) 2 =1 ⇒ x +1=1or x +1=−1 ⇒ x =0or x = −2. h 0 (x) > 0 ⇔ x0 and
h 0 (x) < 0 ⇔ −2 −1 and
h 00 (x) < 0 ⇔ x 0 for x>−2 and A 0 (x) < 0 for −3 −3,soA is concave upward on (−3, ∞). Thereisnoinflection point.
41. (a) C(x) =x 1/3 (x +4)=x 4/3 +4x 1/3 ⇒ C 0 (x) = 4 3 x1/3 + 4 3 x−2/3 = 4 3 x−2/3 (x +1)=
4(x +1)
3 3√ x 2 . C 0 (x) > 0 if
−1