a) Z definicji wartości bezwzględnej:

$\left|x-3\right|$  

$\{\begin{array}{l}x-3\text{,}x\in \left[3,\infty \right)\\ -x+3\text{,}x\in \left(-\infty ,3\right)\end{array}$  

 

A więc:

$f\left(x\right)=\{\begin{array}{l}x\left(x-3\right)-x\text{,}x\in \left[3,\infty \right)\\ x\left(-x+3\right)-x\text{,}x\in \left(-\infty ,3\right)\end{array}$ $f\left(x\right)=\{\begin{array}{l}x\left(x-3\right)-x\text{,}x\in \left[3,\infty \right)\\ x\left(-x+3\right)-x\text{,}x\in \left(-\infty ,3\right)\end{array}$  

$f\left(x\right)=\{\begin{array}{l}x\left(x-4\right)\text{,}x\in \left[3,\infty \right)\\ x\left(-x+2\right)\text{,}x\in \left(-\infty ,3\right)\end{array}$  

$f\left(x\right)=\{\begin{array}{l}x\left(x-4\right)\text{,}x\in \left[3,\infty \right)\\ -x\left(x-2\right)\text{,}x\in \left(-\infty ,3\right)\end{array}$  

Wykres:

 

$a)Z_w=R$  

 

$b)f\left(x\right)\nearrow \text{dla}x\in \left(-\infty ,1\right]\text{i}\left[3,\infty \right)$  

$f\left(x\right)\searrow \text{dla}x\in \left[1,3\right]$  

 

$c)f\left(1\right)=1$  

$f\left(2\right)=0$  

$f\left(3\right)=-3$