Say I have smooth function $f : X \to \mathbb{R}$, where open interval $X \subset \mathbb{R}$, and for some $x_0$, I have $f'(x_0) =0$ and $f''(x_0) < 0$.
And I have $f(x) \to -\infty$ as $x \to \partial(X)$, where $\partial(X)$ : the boundary of $X$.
Can I be sure that $f$ has its global maximum at $x_0$?
This seems quite intuitive for me, but is there any counter-example against my reasoning? Any other conditions for the solution to be global?
Edited
As the answers suggest, my reasoning is not true, but is it true when $\exists! x_0 : f'(x_0)=0$?
Intuitively for me, a counterexample would be a function where $x_0$ might just be a local maximum. So we need a "wobbly" function. We take a Taylor approximation for $\cos(x)$ (say $1-\frac{x^2}{2}+\frac{x^4}{24}$) and then add a function that makes sure that we have $f(x) \to -\infty$ as $x \to \partial(X)=\{-a,a\}$ (say $-\frac{1}{a^2-x^2}$, where $a$ is chosen such that the the first function increases enough at the tips, exceeding the local maximum at $x=0$).
So we have the function $f(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{1}{16-x^2}$ on $(-4,4)$. $x_0=0$ is such a candidate but is not a global maximum as you can see.