Solução_Calculo_Stewart_6e

F.
220 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
35. f(x) = x2 − 1
⇒ f 0 (x) = x3 (2x) − x 2 − 1 3x 2
= 3 − x2
x 3 x 6
x 4
f 00 (x) = x4 (−2x) − 3 − x 2 4x 3
= 2x2 − 12
x 8 x 5
Estimates: From the graphs of f 0 and f 00 ,itappearsthatf is increasing on
(−1.73, 0) and (0, 1.73) and decreasing on (−∞, −1.73) and (1.73, ∞);
f has a local maximum of about f(1.73) = 0.38 and a local minimum of about
f(−1.7) = −0.38; f is CU on (−2.45, 0) and (2.45, ∞), and CD on
(−∞, −2.45) and (0, 2.45);andf has inflection points at about
(−2.45, −0.34) and (2.45, 0.34).
Exact: Now f 0 (x) = 3 − x2
x 4
⇒
is positive for 0