( True/False) $f(x)=0$ has no positive solution if $f(0)=0$ and $f'(0)>0$

Question

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?

Answer 1

Yes, you are right!. This statement is true since $f$ is increasing

Result: If $f$ is convex and $f$ is differentiable at $a$, then the graph of $f$ lies above the tangent line through $(a,f(a))$ except at $(a,f(a))$ itself.

Here $f(0)=0$ and $f'(0)>0$ and $f$ is convex, since $f''(x)>0$.

Thus, tangent line through $(0,f(0))$ means line through $(0,f(0))$ with slope $f'(0)$

Hence the graph lies above the positive $x$ axis and so $f$ has no positive solution

Answer 2

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.