Solution
= Solution
{c}
Take any $a<u<v<b$. By <differentiability implies continuity>, $f$ is <continuous> on $[u,v]$, and it is <differentiable> on $(u,v)$. The <mean value theorem> gives $c\in(u,v)$ for which
$$
f(v)-f(u)=f'(c)(v-u)\geq0.
$$
Therefore $\boxed{f(u)\leq f(v)\text{ whenever }u<v}$: $f$ is a <nondecreasing function>, which is the non-strict meaning of increasing here.