Solution
= Solution
\b[True.] If $|f'(x)|\le K$ everywhere, the <mean value theorem> applied between any two distinct points gives
$$
|f(x)-f(y)|=|f'(\xi)||x-y|\le K|x-y|.
$$
Thus $f$ is <Lipschitz continuous> and hence <uniformly continuous>. No continuity of $f'$ is needed: differentiability of $f$ gives continuity on each finite closed interval and is exactly the hypothesis needed for the <mean value theorem> there.