Question

Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = 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 = Based off of this information, what conclusions can be made about Rolle's Theorem? 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). This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64). This contradicts Rolle's Theorem, since f(−64) = f(64), there should exist a number c in (−64, 64) such that f '(c) = 0. This does not contradict Rolle's Theorem, since f '(0) does not exist, and so f is not differentiable on (−64, 64). Nothing can be concluded.

Answer 1

Upon evaluation, f(-64) and f(64) both equal 0. No value of c exists within the interval (-64, 64) such that f'(c) = 0, as the derivative does not exist for x = 0. Therefore, this scenario does not contradict Rolle's Theorem.

Step-by-step explanation:

Finding f(-64) and f(64)

To find f(-64) and f(64), we simply plug these values into the function f(x) = 16 -x2/3.

f(-64) = 16 - (-64)2/3 = 16 - (64)2/3 = 16 - 16 = 0

f(64) = 16 - (64)2/3 = 16 - 16 = 0

Therefore, f(-64) = f(64) = 0.

Finding all values c such that f'(c) = 0

To find the value of c that makes the derivative of the function zero, we first need to find the derivative f'(x). Using the power rule, we have f'(x) = -2/3 * x-1/3. Setting the derivative equal to zero, we solve for c.

f'(x) = -2/3 * x-1/3 = 0
This equation has no solution since x-1/3 can never be zero.

Hence, c = DNE (Does Not Exist)

Conclusion on Rolle's Theorem

This does not contradict Rolle's Theorem, since f'(0) does not exist and so f is not differentiable on (-64, 64). According to Rolle's Theorem, for it to apply, the function must be continuous on the closed interval and differentiable on the open interval. Since the function is not differentiable at x = 0, the conditions for Rolle's Theorem are not met.