Let f be a differentiable function such that $\lim_{x\to +∞} f(x)=-∞$ and $\lim_{x\to -∞} f(x)=-∞$ , then f has a stationary point My proof idea was that the function must intuitively be increasing for some $x_1<0$ and decreasing for some $x_2>0$, so this would mean that $f'(x_1)>0$ and $f'(x_2)<0$. Using the intermediate zero theorem, I can then conclude that there exists $x_3∈[x_1,x_2]$ such that $f'(x_3)=0$. The problem is I'm stuck trying to prove the fact that those two points with negative and positive derivatives exist.
 A: Here's a way to prove the existence of your $x_1$ and $x_2$.
Because $\lim_{x\rightarrow -\infty} f(x) = -\infty$, there is an $a<0$ with the property that $f(x) < f(0)-1$ for all $x < a$.
Similarly, because $\lim_{x\rightarrow \infty} f(x) = -\infty$, there is a $b>0$ with the property that $f(x) < f(0) - 1$ for all $x > b$.
Then $f(a) \leq f(0) - 1$.  So, by the Mean Value theorem, there is an $
x_1\in (a,0)$ with $f'(x_1) = \frac{f(0) - f(a)}{0-a}$, and $f'(x_1) > 0$ since $-a > 0$ and $f(0) -f(a)\geq 1$.
Similarly, $f(b) \leq f(0) - 1$, so by the Mean Value theorem, there is an $x_2\in(0,b)$ with $f'(x_2) = \frac{f(b) - f(0)}{b-0}$, and $f'(x_2)< 0$ since $f(b) - f(0) \leq -1$ and $b>0$.
A: Your way is correct and we can refer to Extreme value theorem (EVT) to conclude that $f:[x_1,x_2]\to \mathbb R$ as a maximum which is also a stationary point since the function is differentiable.
In a simpler way we can show that $\exists a,b$ such that $f(a)=f(b)$ and then refer to [Rolle's theorem][2].

Following your idea, we have that $\exists a_1$ such that $f(a_1)<0$ then $\exists a_2<a_1$ such that $f(a_2)<f(a_1)<0$, then we can conclude by MVT that $\exists a\in (a_2,a_1)$ such that $f'(a)\>0$.
A: Let $F: [-\pi/2, \pi/2] \to \mathbb{R}$ be the function defined by:
$$F(t) :=\begin{cases} \arctan f(\tan t) &\text{, if } - \pi/2 < t < \pi/2 \\ -\pi/2 &\text{, if } t = \pm\pi/2\end{cases}$$
then $F$ Is continuous on $[-\pi/2, \pi/2]$, differentiable on $]-\pi/2, \pi/2[$ and takes the same value at the endpoints.
By Rolle's Theorem, there exists $\tau \in ]\pi/2, \pi/2[$ s.t. $F^\prime (\tau) =0$; but:
$$F^\prime (t) = \frac{1}{1 + f^2 (\tan t)}\ f^\prime (\tan t)\ \frac{1}{\cos^2 t}$$
hence $f^\prime (\tan \tau) =0$.
