Is this claim true about the global minimum of a function?

Question

I have this information of the function $f(x)$ for $x>0$. $f(x)$ is positive. Its derivative $f'(x)$ has only one root $x=a$ in which $f''(a)>0$. $f(x)$ and $f'(x)$ are both continuous and differentiable. Then, can I claim that the point $x=a$ is the global minimum of the function?

Answer 1

$f(x)$ being positive when $x > 0$ doesn't seem to be relevant or necessary.

The KEY aspect is $f'$ is continuous and $f'$ has only one root.

If $f''(a)>0$ then $f'(a)=0$ and $f'(a)$ is increasing at $x=a$ so $f'(x) > 0$ for all $x > a$ and $f'(x) < 0$ for all $x < a$. (as $f'$ is continuous it can't "jump" from neg to pos without "going through" being zero, but $a$ is the only root). So for $(-\infty, a)$ $f(x)$ is decreasing and for $(a, \infty)$, $f(a)$ is increasing. so the $f(a)$ is an absolute global minimum.

But without knowing $f'$ has only one root we'd have no way of claiming $f(a)$ was a global minimum, although we would know it is a local minimum