Solução_Calculo_Stewart_6e

F.
154 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
15. f(x) =(x − 3) −2 ⇒ f 0 (x) =−2(x − 3) −3 . f(4) − f(1) = f 0 (c)(4 − 1) ⇒ 1 1 − 1
2 (−2) = −2
2 (c − 3) · 3 ⇒
3
3
4 = −6 ⇒ (c − 3) 3 = −8 ⇒ c − 3=−2 ⇒ c =1, which is not in the open interval (1, 4). This does not
(c − 3) 3
contradict the Mean Value Theorem since f is not continuous at x =3.
17. Let f(x) =1+2x + x 3 +4x 5 .Thenf(−1) = −6 < 0 and f(0) = 1 > 0. Since f is a polynomial, it is continuous, so the
Intermediate Value Theorem says that there is a number c between −1 and 0 such that f(c) =0. Thus, the given equation has
a real root. Suppose the equation has distinct real roots a and b with a