Rolle's Theorem/Proof 2

Proof
First take the case where:
 * $\forall x \in \openint a b: \map f x = 0$

Then:
 * $\forall x \in \openint a b: \map {f'} x = 0$

Otherwise:
 * $\exists c \in \openint a b: \map f c \ne 0$

Let $\map f c > 0$.

Then there exists an absolute maximum at a point $\xi \in \openint a b$.

Hence:

As $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.

That is:
 * $\map {f'} \xi = 0$

Similarly, let $\map f c < 0$.

Then there exists an absolute minimum at a point $\xi \in \openint a b$.

Hence:

Again, as $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.

That is:
 * $\map {f'} \xi = 0$

Hence the result.