Mean value theorem for a $C^1$ function Let $f:\mathbb{R^+} \to \mathbb{R}$ be $C^1$ and let's define for all $x>0$
$$c(x)=\inf \left\{ (f')^{-1}\left(\left\{\frac{f(x)-f(0)}{x}\right\}\right) \bigcap \,[0,x]\right\}$$
i.e $c(x)$ is the smallest $y$ in $[0,x]$ such that $f'(y)=\frac{f(x)-f(0)}{x}$. The mean value theorem guarantees that there is at least one such $y$.
My question is what can we say about $c(x)$? Intuitively, it feels like $c(x)$ should be at least continuous.
 A: It may be non-continuous, and here is a counterexample. 
Idea behind the example: The mean value can start decreasing and then increase again towards infinity. Then, when it goes beyond the first maximum function $c$ 'jumps' forwards, because the new mean values hadn't been reached before.
Now the counterexample (not very elegant, indeed):
Consider $f(x)=x^3-2x^2$. Then 
$$f'(t)=3t^2-4t$$
and
$$\frac{f(x)-f(0)}x=x^2-2x$$
Then, for $x>0$, $c(x)$ is the least positive root (for $t$) of
$$3t^2-4t-x^2+2x=0$$
that is
$$c(x)=\frac{4\pm\sqrt{12x^2-24x+16}}6=\frac{2\pm\sqrt{3x^2-6x+4}}3$$
where the sign $\pm$ means that it can be $+$ or $-$, depending on $x$.
We see that these roots have different sign if and only if $-x^2+2x<0$, that is, for $x>2$ (remember that $x>0$).
If the roots have the same sign, both must be positive since their sum is $4$. Therefore
$$c(x)=\begin{cases}\dfrac{2-\sqrt{3x^2-6x+4}}3\text{  if }0<x<2\\[2mm]\dfrac{2+\sqrt{3x^2-6x+4}}3\text{  if }2\le x\end{cases}$$
which is not continuous at $x=2$.
