( True/False) $f(x)=0$ has no positive solution if $f(0)=0$ and $f'(0)>0$ 
True/False:
Let $f$ be a twice differentiable function on $\mathbb{R}$ with $f''(x)>0$ for all $x\in \mathbb{R}$. If $f(0)=0$ and $f'(0)>0$, then $f(x)=0$ has no positive solution

Attempt  [trying to show that this statement is TRUE]
$f''(x)>0$ implies that $f'(x)$ is an increasing function.
Let us assume on the contrary that $f(x)$ has a positive solution say $a>0, f(a)=0$
Consider the interval $[0,a]$,

If a real-valued function f is continuous on a proper closed interval
$[a, b]$, differentiable on the open interval $(a, b)$, and $f (a) = f (b),$
then there exists at least one c in the open interval $ (a, b)$ such that
$ {\displaystyle f'(c)=0}$.

so we get a point $c$ such that $f'(c)=0$ and that contradicts the fact that $f'$ is increasing. We are also using the fact that $f'(0)>0$
So this statement is TRUE?
Am I correct?

The result is true.
As $f^{\prime \prime}(x) > 0$ for all $x \in \mathbb R$, $f^\prime$ is an increasing map. As $f^\prime(0) > 0$, you have $f^\prime(x) > 0$ for all $x > 0$. Hence $f$ is a strictly increasing map.
As $f(0) =0$, you have $f(x) > 0$ for all $x>0$ proving the desired result.
