I would like some help on this problem I've been struggling to solve for a while.
Let $f$ be a continuous function on $[0,+\infty)$ and differentiable on $(0,+\infty)$ such that $\lim \limits_{x \to +\infty}f(x) = f(0)$.
Prove that there exists $c>0$ such that $f'(c) = 0$.
So this actually reminds me of Rolle's theorem but I can't use it since it has to be used on a closed interval. My intuition is that $f$ should be bounded and thus it changes its monotonicity which makes its derivative equal to 0 at a certain point.
The problem is no matter how I tried I wasn't able to write a full rigorous proof so if you could help me on that I would greatly appreciate it!
Best Answer
You actually can use Rolle’s theorem.
Consider the function
$$g(x) = \begin{cases} f(\tan(x)) & x \in [0, \pi / 2)\\ f(0) & x = \pi / 2 \end{cases}$$
Then $g$ is continuous on $[0, \pi/2]$ and differentiable on $(0, \pi/2)$. Moreover, $g(0) = g(\pi/2)$. By Rolle’s theorem, there is some $k \in (0, \pi/2)$ with $g’(k) = 0$. By the chain rule, $g’(k) = \sec^2(k) f’(\tan(k)) = 0$. Thus, $f’(\tan(k)) = 0$. Let $c = \tan(k)$.