Final answer:
After calculating, we found that f(-8) = f(8) = 0, and no real value of c exists in the interval (-8, 8) for which f'(c) = 0, because f'(x) is not defined at x = 0. This does not contradict Rolle's Theorem, as the function is not differentiable over the entire interval, particularly at x = 0.
Step-by-step explanation:
First, let's find the values of f(-8) and f(8) for the function f(x) = 4 - x2/3.
f(-8) = 4 - (-8)2/3 = 4 - (64)1/3 = 4 - 4 = 0
f(8) = 4 - (8)2/3 = 4 - (64)1/3 = 4 - 4 = 0
Since f(-8) = f(8), we see that the function has the same value at both ends of the interval [-8, 8]. Next, we will find the values of c in the interval (-8, 8) where the derivative f'(c) equals 0.
The derivative is f'(x) = -½ x-1/3. Setting f'(x) to 0 and solving for x gives us no real solution since x-1/3 cannot equal 0. Therefore, the list of values c where f'(c) = 0 is DNE (Does Not Exist).
Considering Rolle's Theorem, we see that it does not apply here because although f(-8) = f(8), the function is not differentiable over the entire interval [-8, 8], specifically at x = 0. Rolle's Theorem states that if a function is continuous on [a, b] and differentiable on (a, b), and f(a) = f(b), then there exists at least one c in (a, b) such that f'(c) = 0. However, because f'(0) does not exist, the function f(x) is not differentiable at x = 0, which is within the interval. Thus, the theorem's conditions are not fully met.