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.

Question

Mathematics

Bridger Cline

14 April, 07:37

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.

+3

Answers (1)

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Dante Morgan · Answer 1

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.