Determine whether Rolle's Theorem can be applied to f on the closed interval [a, b]. (Select all that apply.)

Question

f(x) = x^2/3 − 3,    [−8, 8]

Yes, Rolle's Theorem can be applied

No, because f is not continuous on the closed interval [a, b].

No, because f is not differentiable in the open interval (a, b).

No, because f(a) ≠ f(b).

Answer 1

Rolle's Theorem states that within an interval (open) there exists a value "c" such that f'(c) = 0 such that evaluation of the differentiable function f(x) obtains the same value as defined by the endpoints in question.

So, differentiable functions are continuous by definition. Is f(x) differentiable? Well,

f(x) = x^(2/3} - 3, both monomial components are differentiable, so f(x) is also differentiable except at x = 0. f'(x) is not defined at x = 0 due to the singularity (cusp) as seen in the graph of f(x)

f(-8) = 1

f(8) = 1

So we may have found a maximum at some point, however upon taking the derivative of f(x) = 2/3x^{-1/3} we see that point would have been at x = 0 (where f'(x) is undefined) which is included in the open interval defined. This is in violation of the definition of Rolle's Theorem, so the answer must be there is NO "c" which satisfies Rolle's Theorem