Why when $f$ has a local minimizer that is not global minimizer, then f have another critical point?

Question

Consider $f:\, \mathbb{R} \rightarrow \mathbb{R}$ be a univariate, real-valued function with continuous derivative. Show that if $f$ has a local minimizer that is not global minimizer, then f must have another critical point?

Answer 1

If $f$ has no other critical point, the derivative only vanishes at the local minimum. Since it is a local minimum, the derivative is negative to its left and positive to the right. Since the derivative is continuous and has no other zero, it always remains positive to the right and negative to the left (here we apply Bolzano's theorem). So the function always decreases to the left and increases to the right of the point, hence it is a global minimum.

Answer 2

Let $x$ be be the local minimum which is not global, there exists $u<x<v$ such that $f(u)>f(x), f(v)>f(x)$ and $f(x)$ is minimum on $f([u,v])$ since $x$ is a local minimum. Let $y$ such that $f(y)<f(x)$, suppose that $y>v$, since $f$ is continuous, $f([v,z])$ is an interval, we deduce that there exists $z\in [v,y]$ such that $f(z)=f(x)$, $f(z)-f(x)=0=f'(c)(z-x)$, where $c\in (x,z)$ implies that $f'(c)=0$. If $z<u$, $f(u)-f(z)=f'(c)(u-z)=0$ implies that $f'(c)=0$.