$Iff\left(x\right)=|x-1|+|x-2|+|x-3|,f:(2,3)\to Rthenfis$

$\begin{array}{l}f\left(x\right)=|x-1|+|x-2|+|x-3|andf:(2,3)\to R\\ If2<x<3then\\ f\left(x\right)=(x-1)+(x-2)-(x-3)\\ =x-1+x-2-x+3\\ =x\\ Let{x}_1,x_2\in (2,3)thenf\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2\\ fis1-1\\ Rangeoffis(2,3)\ne R\\ \therefore fisnotonto\end{array}$