# 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.

• consider $x\le -2$,$x\ge 2$ and $-2\lt x \lt2$ – Mohamad Misto Jun 4 '15 at 14:24
• Pick some values - large, medium and small - plug them into the formula and see what happens - say $-10, -5, -2, -1, 0, 1, 2, 5, 10$ After a few you should understand how it all comes together. – Mark Bennet Jun 4 '15 at 14:25
• What happens with $|x-2|$ when $x>2$ and when $x<2$? What happens with $|x+2|$ when $x>-2$ and when $x<-2$? – User Jun 4 '15 at 14:27
• openstudy.com/updates/4fb76440e4b05565342d0132 – Jeffrey L. Jun 4 '15 at 14:31

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.

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.