Consider the following function. f(x) = 16-x^(2/3) Find f(-64) and f(64). f(-64) = 0 f(64) = 0 Find all values c in (-64, 64) such that f '(c) = 0. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) c = -64,64 Based off of this information, what conclusions can be made about Rolle's Theorem?
a. This contradicts Rolle's Theorem, since f is differentiable, f(-64) = f(64), and f '(c) = 0 exists, but c is not in (-64, 64).
b. This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (-64, 64).
c. This contradicts Rolle's Theorem, since f(-64) = f(64), there should exist a number c in (-64, 64) such that f '(c) = 0.
d. This does not contradict Rolle's Theorem, since f '(0) does not exist, and so f is not differentiable on (-64, 64).
e. Nothing can be concluded.

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flightbath flightbath · Answer 1 · 2018-11-13T08:52:47+01:00

given function is f(x)=16-x^2/3

to find the values of c we will convert function in terms of c

f(c)= 16-c^2/3

differentiate the function with respect to c

f'(c)= -2/3 c^-1/3 (1)

since f'(c)=0 is given

we can equate (1) to zero

-2/3 c^ -1/3 =0

c^-1/3=0 implies c=0

Option B can be concluded from the given question.

NOTE: it does not violate the Rolle's theorem hence, option A,C are discarded

option D is discarded because function is Differentiable .