Solução_Calculo_Stewart_6e

F.
158 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
19. f(x) =x 5 − 5x +3 ⇒ f 0 (x) =5x 4 − 5=5(x 2 +1)(x +1)(x − 1).
First Derivative Test: f 0 (x) < 0 ⇒ −1 1 or x 0 ⇒ f(1) = −1 is a local minimum value.
Preference: For this function, the two tests are equally easy.
21. f(x) =x + √ 1 − x ⇒ f 0 (x) =1+ 1 2 (1 − x)−1/2 (−1) = 1 −
1
2 √ .Notethatf is defined for 1 − x ≥ 0;thatis,
1 − x
for x ≤ 1. f 0 (x) =0 ⇒ 2 √ 1 − x =1 ⇒ √ 1 − x = 1 2
⇒ 1 − x = 1 4
⇒ x = 3 4 . f 0 does not exist at x =1,
but we can’t have a local maximum or minimum at an endpoint.
First Derivative Test: f 0 (x) > 0 ⇒ x< 3 and f 0 (x) < 0 ⇒ 3