Derivative at Maximum or Minimum

Theorem
Let $$f$$ be a real function which is differentiable on the open interval $$\left({a \, . \, . \, b}\right)$$.

Let $$f$$ have a local minimum or local maximum at $$\xi \in \left({a \, . \, . \, b}\right)$$.

Then $$f^{\prime} \left({\xi}\right) = 0$$.

Proof
By definition $$\frac {f \left({x}\right) - f \left({\xi}\right)} {x - \xi} \to f^{\prime} \left({\xi}\right)$$ as $$x \to \xi$$.


 * Suppose $$f^{\prime} \left({\xi}\right) > 0$$.

Then from Behaviour of Function Near Limit‎ it follows that:

$$\exists I = \left({\xi - h \, . \, . \, \xi + h}\right): \frac {f \left({x}\right) - f \left({\xi}\right)} {x - \xi} > 0$$

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

Now let $$x_1$$ be any number in the open interval $$\left({\xi - h \, . \, . \, \xi}\right)$$.

Then $$x_1 - \xi < 0$$ and hence from $$\frac {f \left({x_1}\right) - f \left({\xi}\right)} {x_1 - \xi} > 0$$ it follows that $$f \left({x_1}\right) < f \left({\xi}\right)$$.

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

Now let $$x_2$$ be any number in the open interval $$\left({\xi \, . \, . \, \xi + h}\right)$$.

Then $$x_2 - \xi > 0$$ and hence from $$\frac {f \left({x_2}\right) - f \left({\xi}\right)} {x_2 - \xi} > 0$$ it follows that $$f \left({x_2}\right) > f \left({\xi}\right)$$.

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


 * A similar argument can be applied to $$-f$$ to handle the case where $$f^{\prime} \left({\xi}\right) < 0$$.

The only other possibility is that $$f^{\prime} \left({\xi}\right) = 0$$, hence the result.