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

Theorem

Let

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

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

holds.

Proof

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

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

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

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

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

$\Box$

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 ${}\mathbb {R} \rightarrow \mathbb {R} ,x\mapsto x^{3}$, which is strictly increasing and its derivative is zero at the zero point. We will provide a sufficient criterion in fact, see also fact.