Bounded derivative implies uniform continuity on an open interval

Suppose $f : (a,b) \to \mathbb{R}$ such that $f’$ exists and is bounded on $(a,b)$.

Then is $f$ uniformly continuous?

I have a hint to use the mean value theorem, but I’m not sure I can apply it to an open interval like this? Does the bounded derivative imply that $f$ is continuous on $[a,b]$ somehow?

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Use the MVT to check that $f$ is a Lipschitz function.

Let $x,y \in (a,b)$. By the MVT, there is $z \in (a,b)$ between $x$ and $y$ with $f(x)-f(y) = f'(z)(x-y)$. Since $|f’| \leq M$ for some $M \geq 0$, we have: $$|f(x)-f(y)| = |f'(z)(x-y)| \leq M|x-y|.$$

Then check that every Lipschitz function is uniformly continuous.

Let $\epsilon > 0$ and choose $\delta = \epsilon/M > 0$. So, if $|x-y| < \delta$, we get: $$|f(x)-f(y)| \leq M|x-y| < M \frac{\epsilon}{M} = \epsilon,$$and $\delta$ does not depend on the points $x$ and $y$.