Let f(x) = x^{2}. 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) = 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, just that it is somewhere.

Example 2

For the function f shown below, determine if we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0. If not, explain why not.

No. We aren't allowed to use Rolle's Theorem here, because the function f is not continuous on [a, b].

Example 3

Let . Determine if Rolle's Theorem guarantees the existence of some c in (-1, 1) with f ' (c) = 0. If not, explain why not.

Since f is not continuous on [-1,1] (f is discontinuous at x = 0), we aren't allowed to use Rolle's Theorem here.

Example 4

Let f(x) = x^{2} – x. Does Rolle's Theorem guarantees the existence of some c in (0, 1) with f ' (c) = 0? If not, explain why not.

Yes.

f is a polynomial, so f is continuous on [0, 1].

f is differentiable (its derivative is 2x – 1).

f(0) = 0 and f(1) = 0, so f has the same value at the start point and end point of the interval.

Thus Rolle's Theorem says there is some c in (0, 1) with f ' (c) = 0. The exact value of c is 0.5, but we found that by studying the derivative. Rolle's Theorem doesn't tell us the actual value of c that gives us f ' (c) = 0.