> [!theorem] > > Let $f$ be a [[Derivative|differentiable]] [[Function|function]] on an [[Open Set|open]] interval $(a, b)$. If $f$ has a [[Extrema|maximum]] or minimum value at some point $c \in (a, b)$, then $f^\prime(c) = 0$. > > *Proof*. Since $c \in (a, b)$, two [[Sequence|sequences]] $(x_n)$ and $(y_n)$ may be constructed with $x_n < c < y_n \forall n \in \nat$, and since $f(c) \ge f(y_n)$, > $ > \frac{df}{dx}(c) = \limv{n}\frac{f(y_n) - f(c)}{y_n - c} \le 0 > $ > by [[Order Limit Theorem|order limit theorem]]. Similarly, > $ > \frac{df}{dx}(c) = \limv{n}\frac{f(x_n) - f(c)}{x_n - c} \ge 0 > $ > and therefore $\frac{df}{dx}(c) = 0$.