Rolle's Theorem/Proof 2

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Theorem

Let $f$ be a real function which is:

continuous on the closed interval $\closedint a b$

and:

differentiable on the open interval $\openint a b$.

Let $\map f a = \map f b$.


Then:

$\exists \xi \in \openint a b: \map {f'} \xi = 0$


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:

\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) \(\le\) \(\ds 0\) for $\xi < \xi + h < b$
\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) \(\ge\) \(\ds 0\) for $a < \xi + h < \xi$

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:

\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) \(\ge\) \(\ds 0\) for $\xi < \xi + h < b$
\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) \(\le\) \(\ds 0\) for $a < \xi + h < \xi$

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.

$\blacksquare$


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