Writing $f(x)=\left |x-2 \right |+\left |x+2 \right |$ without a modulus Express $f(x)=\left |x-2  \right |+\left |x+2  \right |$ in the non-modulus form. Hence sketch the graph and determine the range of $f$.
Can someone give me some ideas for solving this question? Thanks.
 A: Break up the two absolute values. By definition:
$|x-2| =
\left\{
 \begin{array}{ll}
  x-2  & \mbox{if } x-2 \geq 0 \\
  2-x & \mbox{if } x-2 < 0 \end{array}
\right.$
And:
$|x+2| =
\left\{
 \begin{array}{ll}
  x+2  & \mbox{if } x+2 \geq 0 \\
  -x-2 & \mbox{if } x+2 < 0
 \end{array}
\right.$
If you look at those pieces, you see there's basically three ranges we have to consider:
$f(x) =
\left\{
 \begin{array}{ll}
  ???  & \mbox{if } x \geq 2 \\
  ???  & \mbox{if } x \lt -2 \\
  ??? & \mbox{if } x \in [-2,2)
 \end{array}
\right.$
Simply fill in the $???$s based on the two components. 
A: To solve this question, I advise to you do a kind of "clothesline signals", in other words, for which values $ |x-2| $ is positive or negative, and for which values $ |x+2| $ is positive or negative, like this way : $$ x-2=0 $$ $$x=2$$, for $x<2$, $ F(x)=x-2 $ show negative values, and for $x>2$, $ F(x)=x-2 $ show positive values, then you do the same thing for $x+2$ and analyze the signal of entire equation.
