# Derivative at Maximum or Minimum

## Theorem

Let $f$ be a real function which is differentiable on the open interval $\openint a b$.

Let $f$ have a local minimum or local maximum at $\xi \in \openint a b$.

Then:

$\map {f'} \xi = 0$

## 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.

$\blacksquare$

## Historical Note

The Derivative at Maximum or Minimum was first arrived at by Pierre de Fermat, anticipating the more formal techniques used in differential calculus.

This result was given by Gottfried Wilhelm von Leibniz in his $1684$ article Nova Methodus pro Maximis et Minimis, published in Acta Eruditorum.