# Differentiable function/R/Local extrema/Derivative/Introduction/Section

## Theorem

Let

${\displaystyle f\colon {]a,b[}\longrightarrow \mathbb {R} }$

be a function which attains in ${\displaystyle {}c\in {]a,b[}}$ a local extremum, and is differentiable there. Then ${\displaystyle {}f'(c)=0}$ holds.

### Proof

We may assume that ${\displaystyle {}f}$ attains a local maximum in ${\displaystyle {}c}$. This means that there exists an ${\displaystyle {}\epsilon >0}$, such that ${\displaystyle {}f(x)\leq f(c)}$ holds for all ${\displaystyle {}x\in [c-\epsilon ,c+\epsilon ]}$. Let ${\displaystyle {}{\left(s_{n}\right)}_{n\in \mathbb {N} }}$ be a sequence with ${\displaystyle {}c-\epsilon \leq s_{n}, tending to ${\displaystyle {}c}$ ("from below“). Then ${\displaystyle {}s_{n}-c<0}$, and so ${\displaystyle {}f(s_{n})-f(c)\leq 0}$, and therefore the difference quotient

${\displaystyle {}{\frac {f(s_{n})-f(c)}{s_{n}-c}}\geq 0\,.}$

Due to fact, this relation carries over to the limit, which is the derivative. Hence, ${\displaystyle {}f'(c)\geq 0}$. For another sequence ${\displaystyle {}{\left(t_{n}\right)}_{n\in \mathbb {N} }}$ with ${\displaystyle {}c+\epsilon \geq t_{n}>c}$, we get

${\displaystyle {}{\frac {f(t_{n})-f(c)}{t_{n}-c}}\leq 0\,.}$

Therefore, also ${\displaystyle {}f'(c)\leq 0}$ and thus ${\displaystyle {}f'(c)=0}$.

${\displaystyle \Box }$

We remark that the vanishing of the derivative is only a necessary, but not a sufficient, criterion for the existence of an extremum. The easiest example for this phenomenon is the function ${\displaystyle {}\mathbb {R} \rightarrow \mathbb {R} ,x\mapsto x^{3}}$, which is strictly increasing and its derivative is zero at the zero point. We will provide a sufficient criterion in fact, see also fact.