Solution
= Solution
The function $f$ is <continuous> at $y_0$ when for every $\varepsilon>0$ there is a $\delta>0$ such that
$$
|x-y_0|<\delta\quad\Longrightarrow\quad|f(x)-f(y_0)|<\varepsilon.
$$
Suppose $f$ has a <local minimum> at $c$ and is differentiable there. For sufficiently small $h>0$,
$$
\frac{f(c+h)-f(c)}h\geq0,
\qquad
\frac{f(c-h)-f(c)}{-h}\leq0.
$$
Both one-sided limits equal $f'(c)$, so it is both nonnegative and nonpositive. Therefore
$$
\boxed{f'(c)=0}.
$$