Real function/Open interval/Local extrema/Differentiable/Derivative zero/Fact/Proof

Proof

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 ${}f'(c)=0$.

To fact

To fact