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