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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.

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    $\begingroup$ consider $x\le -2$,$x\ge 2$ and $-2\lt x \lt2$ $\endgroup$ – Mohamad Misto Jun 4 '15 at 14:24
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Jun 4 '15 at 14:25
  • $\begingroup$ What happens with $|x-2|$ when $x>2$ and when $x<2$? What happens with $|x+2|$ when $x>-2$ and when $x<-2$? $\endgroup$ – User Jun 4 '15 at 14:27
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    $\begingroup$ openstudy.com/updates/4fb76440e4b05565342d0132 $\endgroup$ – Jeffrey L. Jun 4 '15 at 14:31
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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.

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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.

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