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*) = 2*x*.
*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*) = *x*^{2} has 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. | |