Interior Extremum Theorem

This is also known as Fermat's Theorem.

Let $f : (a, b) \to R$ . Suppose $f$ has a maximum or minimum at some $c \in (a, b)$ , and suppose $f$ is differentiable at $c$ . Then $f^{'} (c) = 0$ .

Proof

Without loss of generality, suppose $f (c)$ is a maximum. Then $\forall x \in (a, b)$ then $f (x) \leq f (c)$ . Now $a < c < b$ , so then we can make a sequence on the right and left of $c$ . Namely, since $f$ is differentiable at $c$ , then:

\begin{aligned} f^{'} (c) & = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \\ = lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c} \\ \geq 0 \end{aligned}

Because $f (x) - f (x) \geq 0$ (using $f (c)$ is a maximum) and via the Limits and Order (Order Limit Theorem) (extending it to limits of functions). Similarly:

\begin{aligned} f^{'} (c) & = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \\ = lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c} \\ \leq 0 \end{aligned}

As a result then $0 \leq f^{'} (c) \leq 0$ thus $f^{'} (c) = 0$ .

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