Question

Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.

Answer 1

Recall that

$|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

There are three cases to consider:

(1) When $x+2$, we have $|x+2|=-(x+2)$ and $|x-2|=-(x-2)$, so

$|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4$

(2) When $x+2\ge0$ and $x-2$, we get $|x+2|=x+2$ and $|x-2|=-(x-2)$, so

$|x-2|+|x+2|=-(x-2)+(x+2)=4$

(3) When $x-2\ge0$, we have $|x+2|=x+2$ and $|x-2|=x-2$, so

$|x-2|+|x+2|=(x-2)+(x+2)=2x$

So

$|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x$