contestada

Consider the following function. f ( x ) = 1 − x 2 / 3 Find f ( − 1 ) and f ( 1 ) . f ( − 1 ) = f ( 1 ) = Find all values c in ( − 1 , 1 ) such that f ' ( c ) = 0 . (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) c = Based off of this information, what conclusions can be made about Rolle's Theorem? This contradicts Rolle's Theorem, since f ( − 1 ) = f ( 1 ) , there should exist a number c in ( − 1 , 1 ) such that f ' ( c ) = 0 . This does not contradict Rolle's Theorem, since f ' ( 0 ) = 0 , and 0 is in the interval ( − 1 , 1 ) . This does not contradict Rolle's Theorem, since f ' ( 0 ) does not exist, and so f is not differentiable on ( − 1 , 1 ) . This contradicts Rolle's Theorem, since f is differentiable, f ( − 1 ) = f ( 1 ) , and f ' ( c ) = 0 exists, but c is not in ( − 1 , 1 ) . Nothing can be concluded.

Respuesta :

Answer:

(E)Nothing can be concluded.

Step-by-step explanation:

Given the function [tex]f(x)=1-x^{\frac{2}{3}}[/tex]

[tex]f(-1)=1-(-1)^{\frac{2}{3}}=1.5-0.86603i\\f(1)=1-1^{\frac{2}{3}}=1-1=0[/tex]

[tex]f'(x)=-\dfrac{2}{3}x^{-\frac{1}{3}}\\f'(x)=-\dfrac{2}{3\sqrt[3]{x} }[/tex]

If the derivative is set equal to zero, the function is undefined.

Nothing can be concluded since [tex]f(1)\neq f(-1)[/tex] and no such c in (-1,1) exists such that [tex]f'(c)=0[/tex]

THEOREM

Rolle's theorem states that any real-valued differentiable function that attains equal values at two distinct points must have at least one stationary point somewhere between them—that is, a point where the first derivative is zero.

Otras preguntas