Prove that if $I \subset \mathbb{R}$ is an open interval and $f: I \to \mathbb{R}$ differentiable, and $f$ has only one critical point $x_0$ and this critical point is a local minimum, then $x_0$ is also the absolute min of $f$, using Rolle's and IVT.
To me this seems "obvious", because if $f$ has only one critical point, then there can be no other point where $f' = 0$, meaning over its image $f'(x_0)$ is the only point where $f' = 0$. If this is the case then since $x_0$ is a local min, it is smaller than $f(x)$ for all points in a local interval, but since there are no other points $x$ in the entire image where $f'(x) = 0$, it is decreasing in the entire interval before it, and increasing in the entire interval after it. How do I go about formalizing this logic using IVT and Rolle's? Thanks in advance.
(edit) Description of problem also found on wikipedia: "For example, if a bounded differentiable function f defined on a closed interval in the real line has a single critical point, which is a local minimum, then it is also a global minimum (use the intermediate value theorem and Rolle's theorem to prove this by reductio ad absurdum)"