In mean value theorem over $[x_0,x]$, will $x^* \to+\infty$ always as $x \to +\infty$? 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 $[x_0,+\infty)$, do we always have (or we can let, in cases like $f(x)$ is constant) $x^* \to+\infty$ as $x \to +\infty$? If not, on what condition this can happen?

This is false. Consider $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  0&{x \geqslant 0} \\ 
  {{e^{\frac{1}{x}}}}&{x < 0} 
\end{array}} \right.$
It is easy to verify $f(x)$ is differentiable over $\Bbb R$ and ${f'}\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
0&{x \ge 0}\\
{ - {x^{ - 2}}{e^{{x^{ - 1}}}}}&{x < 0}
\end{array}} \right.$
We can choose any $x_0<0$, and notice $\frac{{f\left( x \right) - f({x_0})}}{{x - {x_0}}} < 0$ for any $x>x_0$, implying $x^*<0$ for any $x_0<x<+∞$, and we actually have $x^*→0$ as $x→+∞$.
