Suppose $f(x)$ is continuous on $[a,b)$ and differentiable on $(a,b)$ and that $f '(x)$ tends to a finite limit $L$ as $x \to a^+$. Then $f(x)$ is right-differentiable at $x=a$ and $f '(a)=L$.
(epsilon-delta proof not needed).
This is a practice exam question.
I am having trouble translating this into a 'mathematical' statement. The MVT states that there exists $c$, $a\leq c\leq b$, such that:
$f'(c) = (f(b)-f(a))/(b-a)$
I suppose to prove that $f(x)$ is right differentiable at $x=a$, using the MVT, I need to somehow show that as $x \to a^+$, $f'(c)=f'(a)=L$ ??? Am I on the right track here? Can someone help me get started?