# Derivative at Maximum or Minimum/Proof 2

## 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 local maximum:

$(1): \quad \map f {\xi + \epsilon} - \map f \xi < 0$

for sufficiently small $\epsilon \in \R_{>0}$.

Similarly, definition of local minimum:

$(2): \quad \map f {\xi + \epsilon} - \map f \xi > 0$

for sufficiently small $\epsilon \in \R_{>0}$.

Let it be assumed that $\map f {\xi + \epsilon}$ can be expanded, using Taylor's Theorem, in positive integer powers of $\epsilon$

Then:

$(3): \quad \map f {\xi + \epsilon} - \map f \xi = \epsilon \map {f'} \xi + \dfrac {\epsilon^2 \map {f' '} \xi} 2 + \map \OO {\epsilon^3}$

where $\map \OO {\epsilon^3}$ denotes big-$\OO$ notation.

That is, the quantity $\dfrac {\map \OO {\epsilon^3} } {\epsilon^3}$ is bounded.

From $(1)$ and $(2)$, at a local maximum or local minimum, the sign of $\map f {\xi + \epsilon} - \map f \xi$ is independent the sign of $\epsilon$ itself.

So from $(3)$ it follows that $\map {f'} \xi = 0$.

$\blacksquare$