Graphing the sum of two absolute values

Question

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 - $$x<-2, -2<x<-1, -1<x<0, 0<x<1, 1<x<2, 2<x$$ i think i did it right. Here is how I got to these cases:

1. |x|={x, x>0 & -x, x<0}
2. |x-1|={x-1, x>1 & -x-1, x<1}
3. |-x-1|={-x-1, x<-1 & x+1, x>-1}
4. |x+2|={x+2, x>-2 & -x-2, x<-2}
5. |-x+2|={-x+2, x<2 & 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.

Answer 1

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 ...

Answer 2

Just plot the function or by hand!

• The minimum value is $3$
• At large $x >0$ 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 $x=\pm 1$