Interior_Extremum_Theorem

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Definition

Let $f$ be differentiable on an open interval $(a,b)$. If $f$ attains a maximum or minimum value at some point $c \in (a,b)$, then $f'(c) = 0$.

Proof

Suppose $f$ has a maximum at $c \in (a,b)$. Then for $h > 0$ small enough:

$$ \frac{f(c+h) - f(c)}{h} \leq 0 $$

since $f(c+h) \leq f(c)$. Taking $h \to 0^+$, we get $f'(c) \leq 0$.

Similarly, for $h < 0$:

$$ \frac{f(c+h) - f(c)}{h} \geq 0 $$

Taking $h \to 0^-$, we get $f'(c) \geq 0$.

Therefore $f'(c) = 0$. The proof for a minimum is similar.