If for $x\in(\frac{1}{2},\infty)$ we have $f'(x)=(e^x-1)(x-2)(x-3)$. Show that there exist exactly two roots of $f''(x)=0$ in the given domain

Question

Let $f:\left(\frac{1}{2},\infty \right)\to \mathbb{R}$ be a function such that $f'(x)=(e^x-1)(x-2)(x-3)$. Show that there exist exactly two roots of $f''(x)=0$ in the given domain.

My Attempt

I evaluated $f''(x)=e^x(x-2)(x-3)+(e^x-1)(x-2)+(e^x-1)(x-3)$ and then plugged in the values $1,2$ and $3$ and obtained

$f''(1)=3-e>0$

$f''(2)=(e^2-1)(-1)<0$

$f''(3)=e^3-1>0$

So one can observe that there is at least one root on both of the intervals $(1,2)$ and $(2,3)$.

But how does one claim that there are exactly two roots in the given domain