I know the mean value theorem hypotheses are:
If $f: [a,b] \mapsto R$ is continuous on $[a,b]$ and differentiable on $(a,b)$.
But can I still invoke MVT under the hypothesis that:
$f: (a,b) \mapsto R$ is differentiable
Does my domain necessarily need to be closed and continuous at those points? I am proving that $f: (a,b) \mapsto R$ is differentiable and $|f'(x)| \leq M$ for all $x \in (a,b)$ then $f$ is uniformly continuous on $(a,b)$. I am pretty certain I need to use MVT for the proof but does it need any special treatment since my hypothesis is a bit different?
Edit: this is my proof
Suppose $f:(a,b) \mapsto \mathbb{R}$ is differentiable and $|f'(x)| \leq M$ for all $x \in (a,b)$. By the Mean-Value Theorem $\frac{f(b) - f(a)}{b-a} \leq M$. Fix $\epsilon >0$ and let $\delta = \frac{\epsilon}{M}$. Now if we have $|b-a|<\delta$ then $f(b)-f(a) \leq M|b-a| < M\delta = \epsilon$.
Yes you can still apply the MVT, just not for the points $a$ and $b$. If $f$ is differentiable on $(a,b)$, then it is continuous on $[c,d]$ and differentiable on $(c,d)$ for all $c,d \in (a,b)$. Thus you can you the MVT for any points $c<d$ in the interval $(a,b)$.