mean value theorem on an open interval I know that the conditions for the mean value theorem state $f$ must be continuous on $[a,b]$ and differentiable on $(a,b)$.
What happens if we change the condition to $f$ is continuous on $ (a,b)$ but not at the endpoints?

Given $a < b$, for any $\epsilon_{a} > 0$ and any $\epsilon_{b} > 0$ such that
$\epsilon_{a} + \epsilon_{b} < b - a\quad\exists\ \xi \ni$ 
$$
{{\rm f}\left(b - \epsilon_{b}\right) - {\rm f}\left(a + \epsilon_{\rm a}\right)
 \over
 b - a - \epsilon_{a} - \epsilon_{b}}
=
{\rm f}'\left(\xi\right)\,,
\qquad\mbox{where}\quad
a + \epsilon_{a} < \xi < b - \epsilon_{b}
$$
