$f\left(x\right)=(x-2)|x-3|$  is monotonically increasing in

$$\begin{gathered} f\left(x\right)=(x-2)|x-3| \\ f\left(x\right)=\left\{\begin{array}{l}-(x-2)(x-3)=-(x^2-5x+6) if x<3\\ (x-2)(x-3)=(x^2-5x+6) if x\ge 3\end{array}\right. \\ f\text{'}\left(x\right)=\left\{\begin{array}{l}5-2x if x<3\\ 2x-5 if x\ge 3\end{array}\right. \\ Now f\text{'}\left(x\right)=0\Rightarrow x=\frac{5}{2} \end{gathered}$$ 

Now, the graph of   $f\left(x\right)=(x-2)|x-3|$   is as follows

 Clearly, from the graph, $f\left(x\right)$  increases in  $(-\mathrm{\infty },5/2)\cup (3,\mathrm{\infty })$