I would like some input in my proof on this question:
Let $f:\mathbb{R} \to \mathbb{R}$ smooth, with local minimum at $x=0$, and suppose that this minimum it is not global. Show that there is another critical point besides $x=0$.
My answer:
Suppose $x=0$ is not a global minimum, hence, $\exists x_{1} \in \mathbb{R}:f(x_{1}) \lt f(0).$ Without loss of generality, suppose $x_{1} \lt 0$. By the Mean Value Theorem, $\exists c \in (x_{1},0):f'(c)=\frac{f(0)-f(x_{1})}{x_{1}} \gt 0$.
But $x=0$ is local minimum, therefore, $\exists \delta \gt 0:y \in (-\delta,\delta) \implies f(y) \geq f(0)$.
Consider the closed interval $[-\delta,0]$. By the Mean Value Theorem, $\exists k \in (-\delta,0): f'(k)=\frac{f(0)-f(-\delta)}{\delta} \lt 0$.
Since $f$ is smooth, it is continuous in $\mathbb{R}$. Therefore, we have $c,k \in (x_{1},0)$ and $f'(c) \gt 0$ and $f'(k) \lt 0$, hence, $\exists z \in (c,k): f'(z) =0$ and $z$ is a critical point of $f$.
It is almost perfect. The only small problem is that you should apply the mean value theorem to $f'$ and therefore, at the final sentence, when you write that $f$ is continuous, you should have written that $f'$ is continuous.