Rolle theorem on infinite interval We have :
$f(x)$ is continuous  on $[1,\infty]$ and differentiable on $(1,\infty)$
$\lim\limits_{x \to \infty}f(x) = f(1)$
we have to prove that : there is $b\in(1,\infty)$ such that $f'(b) = 0$
I'm sure we have to use Rolle's theorem so, I tried using Mean Value theorem and using the limit definition  at $\infty$
Any ideas how can I use them ?
Update :
after seeking the answers that I've got :
I'm having trouble finding $\boldsymbol x_{\boldsymbol 1}\neq\boldsymbol x_{\boldsymbol 2}$ such that $\boldsymbol{f(x}_{\boldsymbol 1}\boldsymbol {)=f(x}_{\boldsymbol 2}\boldsymbol )$
*I need a formal solution

Let $g(x)=f\bigl(\frac 1x\bigr)$ for $x\in (0,1]$ and $g(0)=f(1)$. Then $g$ is continuous in $[0,1]$ and derivable in $(0,1)$. By Rolle's Theorem there exists $c\in (0,1)$ such that $g'(c)=0$, hence
$$0=g'(c)=-f'\Bigl(\frac 1c\Bigr)\frac 1{c^2}$$
thus if $b=\frac 1c$, then $f'(b)=0$ and $b\in (1,+\infty)$.
