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?
 A: 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 ${{f(r)-f(s)}\over{r-s}}=r^2+s^2+rs>0$.
A: Let $[a,b]$ be any closed interval and $f$ satisfying the mean value property for every open interval $(a',b')$ contained within $[a,b]$.  Let $a_n = a - \frac{1}{n}$ and $b_n = b - \frac{1}{n}$ for large values of $n$.  Then we have that for each $n$, there's a point $c_n$ in $(a_n, b_n)$ so that $f'(c_n) = \frac{f(b_n) - f(a_n)}{b_n - a_n}$.  It's easy to see by continuity of $f$ on $[a,b]$ that the right hand side converges to $\frac{f(b) - f(a)}{b-a}$.  I guess the question then becomes:
1) Is the sequence $\{c_n\}$ convergent and,
2) Even if it is, do we have $\lim_{n\to\infty} f'(c_n) = f'(c)$.
An imposition of continuity on $f'$ would be enough for 2).  Now for $1$ we argue as follows: $\{c_n\}_n$ is a sequence in $[a,b]$ and so it has a convergent subsequence by compactness.  Then this new sequence, say $\{c_{n_k}\}$ satisfies 
$\lim_{n_k \to \infty} f'(c_{n_k}) = \frac{f(b) - f(a)}{b-a}$, and $\lim_{n_k} c_{n_k} = c$, for some point $c \in [a,b]$.  By continuity, we have that there's a point $c \in [a,b]$ so that $f'(c) = \frac{f(b) - f(a)}{b-a}$, which isn't quite the mean value theorem.  
