Graphing the sum of two absolute values I have an inequality $$||x|-1|+||x|+2|=3$$. Id like to graph it. I've done with just one absolute value within another absolute value, but I can't seem to grasp how to do this one. Also, I'm having some trouble setting cases for the algebraic route: I've done 6 cases in total - $$x&lt;-2, -2&lt;x&lt;-1, -1&lt;x&lt;0, 0&lt;x&lt;1, 1&lt;x&lt;2, 2&lt;x$$ i think i did it right. Here is how I got to these cases:


|x|={x, x>0 &amp; -x, x&lt;0}
|x-1|={x-1, x>1 &amp; -x-1, x&lt;1}
|-x-1|={-x-1, x&lt;-1 &amp; x+1, x>-1}
|x+2|={x+2, x>-2 &amp; -x-2, x&lt;-2}
|-x+2|={-x+2, x&lt;2 &amp; x-2, x>2}



Did I set my cases correctly? Also, I know how the graph is supposed to look like, I just don't know how to get to it.
Note that:  $||x| + 2| = |x| + 2$ so you can get rid of the second absolute value immediately.   

Case 1:  

$|x| \lt 1$

Case 2:  

$|x| \ge 1$ 

... and you just continue analyzing in each of these cases ...
Just plot the function or by hand!


The minimum value is $3$ 
At large $x &gt;0$ the slope is $+2$, since the individual terms each have a slope of $+1$
A large negative $x$, the slope is $-2$, for similar reasons
The transition points must be when $x=\pm 1$