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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.

TheSlavSQuat
  • 85

2 Answers2

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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 ...

peter.petrov
  • 12,568
-1

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 $x=\pm 1$

enter image description here

David G. Stork
  • 29,774
  • id like to do it by hand, but i am not doing something good as i cant seem to get the graph like in your picture. I know that i can use software, but there is no point, as its just a complete product and it doesn't show me how it got the result. – TheSlavSQuat Apr 22 '20 at 19:01
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    I didn't down vote, but $||x|-1|$ could be $0$ (when $x=\pm1$) – J. W. Tanner Apr 26 '20 at 03:19