Recall that
[tex]|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x<0\end{cases}[/tex]
There are three cases to consider:
(1) When [tex]x+2<0[/tex], we have [tex]|x+2|=-(x+2)[/tex] and [tex]|x-2|=-(x-2)[/tex], so
[tex]|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4[/tex]
(2) When [tex]x+2\ge0[/tex] and [tex]x-2<0[/tex], we get [tex]|x+2|=x+2[/tex] and [tex]|x-2|=-(x-2)[/tex], so
[tex]|x-2|+|x+2|=-(x-2)+(x+2)=4[/tex]
(3) When [tex]x-2\ge0[/tex], we have [tex]|x+2|=x+2[/tex] and [tex]|x-2|=x-2[/tex], so
[tex]|x-2|+|x+2|=(x-2)+(x+2)=2x[/tex]
So
[tex]|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x<-2\\4&\text{if }-2\le x<2\\2x&\text{if }x\ge2\end{cases}[/tex]
