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)"
Best Answer
Let $x_1\in I$. Assume $f(x_1)<f(x_0)$ and wlog. $x_1>x_0$. Since $x_0$ is a local minimum, there is an $\epsilon>0$ such that $f(x)\ge f(x_0)$ for $x_0<x<x_0+\epsilon$. Select $x_2$ with $x_0<x_2<\min\{x_0+\epsilon,x_1\}$. Then $f(x_2)\ge f(x_0)>f(x_1)$ implies that there is $x_3\in[x_2,x_1)$ with $f(x_3)=f(x_0)$ and finally Rolle implies that there is $x_4\in(x_0,x_3)$ such that $f'(x_4)=0$, i.e. $x_4$ is a critical point different from $x_0$. Therefore the assumption $f(x_1)<f(x_0)$ must be wrong.