f(x) = 1 − x^(2/3)

**Answer 1**

Recall that f must be continuous on [a, b], differentiable on (a, b), and f(a) = f(b) in order to apply Rolle’s Theorem.

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Here, a = -1, b = 1, and f(x) = 1 – x^(2/3).

We know that f(-1) = f(1) = 0, and f is continuous on [-1, 1] (since the exponent is not negative and cube roots are defined for all x).

However, you found that f ‘(c) = 0 has no solutions (in (-1, 1)).

Since Rolle’s Theorem does not apply, the only hypothesis of the theorem that can not be satisfied is “f is not differentiable on (−1, 1)”.

This is indeed the case, because f ‘(x) = (2/3)x^(-1/3) is not defined at x = 0. So, (D) is the answer.

I hope this helps!

**Answer 2**

The answer above is correct for WebAssign.