If f''(x) > 0 does it mean it can possess at most 1 point of minimum?

If you have two local minima $x_1\ne x_2$ then $f'(x_1)=f'(x_2)=0$. Then by Rolle's theorem, there exists $c\in(x_1,x_2)$ with $f''(c)=0$, contradiction.

EDIT: The above holds if your domain of $f$ is an open interval (possibly $\mathbb R$). If your domain is a closed (or semi-closed) interval, say $[a,b]$, you have to prove that $a$ and $b$ are not local minima. But again, assuming $f(a)\le f(x)$ on a neighborhood (at the right of $a$) one gets $f(x)-f(a)\ge0$, so $f'(a)\ge 0$. On the other side, $f''(a)> 0$ implies $f'(x)>f'(a)\ge 0$ on a neighborhood (at the right of $a$), so $f$ is increasing, so $a$ cannot be local mimimum. A similar argument holds for $b$.

Obviously, if your domain is not connected, then you might have $f''>0$ and multiple minima.

Comment

Is it possible to explain MVT/Rolle's thm with reference to graph?