Some "converse" of mean value theorem The mean value theorem states that if $f:\Bbb R \to \Bbb R$ is continuous over $[x_0,x]$, and differentiable over $(x_0,x)$, then there exists $x^*\in(x_0,x)$ s.t. $f'({x^*}) = \frac{{f(x) - f({x_0})}}{{x - {x_0}}}$.
Now suppose $f$ is differentiable over $[x_0,+\infty)$, for any $x^*>x_0$, can we always choose some $x>x^*$ s.t. $f'({x^*}) = \frac{{f(x) - f({x_0})}}{{x - {x_0}}}$?

Counterexample: 
For $f(x) = x^3$  we have $f'(0) = 0$, but for all $x_0 < x$ in $[-1,1]$ we have 
$$\frac{f(x) - f(x_0)}{x - x_0} = \frac{x^3 - x_0^3}{x-x_0} = x^2 + xx_0 + x_0^2 > 0.$$
In general, you need more restrictions for the converse to hold.
