Derivative at Maximum or Minimum/Proof 1

Proof
By definition of derivative at a point:
 * $\dfrac {\map f x - \map f \xi} {x - \xi} \to \map {f'} \xi$ as $x \to \xi$

Suppose $\map {f'} \xi > 0$.

Then from Behaviour of Function Near Limit‎ it follows that:


 * $\exists I = \openint {\xi - h} {\xi + h}: \dfrac {\map f x - \map f \xi} {x - \xi} > 0$

provided that $x \in I$ and $x \ne \xi$.

Now let $x_1$ be any number in the open interval $\openint {\xi - h} \xi$.

Then:
 * $x_1 - \xi < 0$

and hence from:
 * $\dfrac {\map f {x_1} - \map f \xi} {x_1 - \xi} > 0$

it follows that:
 * $\map f {x_1} < \map f \xi$

Thus $f$ can not have a local minimum at $\xi$.

Now let $x_2$ be any number in the open interval $\openint \xi {\xi + h}$.

Then:
 * $x_2 - \xi > 0$

and hence from:
 * $\dfrac {\map f {x_2} - \map f \xi} {x_2 - \xi} > 0$

it follows that:
 * $\map f {x_2} > \map f \xi$

Thus $f$ can not have a local maximum at $\xi$ either.

A similar argument can be applied to $-f$ to handle the case where $\map {f'} \xi < 0$.

The only other possibility is that $\map {f'} \xi = 0$, hence the result.