Conditions for the mean value theorem the mean value theorem which most of us know starts with the conditions that $f$ is continuous on the closed interval $[a,b]$ and differentiable on the opened interval $(a,b)$, then there exists a $c \in (a,b)$, where 
$\frac{f(b)-f(a)}{b-a} = f'(c)$. 
I'm guessing we're then able to define $\forall a', b' \in [a,b]$ where $c \in (a', b')$ and the mean value theorem is correspondingly valid.
However, are we then able to start with $\forall c \in (a,b)$ and then claim that there exist $a',b' \in [a,b]$, where the mean value theorem is still valid?

The answer is No. Consider $y=f(x)=x^3$ and $c=0$. $f'(c)=0$ but no secant line has a zero slope as ${{f(r)-f(s)}\over{r-s}}=r^2+s^2+rs>0$.
