Let f(x) = x2. Prove that there is some c in (-2,2) with f'(c) = 0.
We need to check that f satisfies all the hypotheses of Rolle's Theorem.
f is a polynomial, so f is continuous on [-2,2].
f is differentiable on (-2,2), since we found that f'(x) = 2x.
f(-2) = 4 and f(2) = 4, so f(-2) = f(2).
Since f satisfies all the hypotheses of Rolle's Theorem, Rolle's Theorem says there must be some c in (-2,2) for which f'(c) = 0.
In this case, f(x) = x2has a "turn-around point" at x = 0, so f'(0) = 0. We can see this from looking at the graph or from finding f'(0), but not from Rolle's Theorem. Rolle's Theorem doesn't tell us where f' is zero, but that it is somewhere.