MHB Sketching absolute value graph

[GusGus335]

Sketch the region in the plane consisting of all points (x,y) such that
|x-y|+|x|-|y|<=2

 

[anemone]

Sketch the region in the plane consisting of all points (x,y) such that
|x-y|+|x|-|y|<=2

His GusGus335! Welcome to MHB!

I believe there are many different methods to sketch the wanted region as stated by the given inequality, and here is one of the methods that you could adopt:

First, rewrite the given inequality as $|x-y|\le 2+|y|-|x|$. And we then exploit the fact that $|x|-|y|\le |x-y|$, we get $|x|-|y|\le 2+|y|-|x|$, solving it for $|y|$, we see that we have:

$|y|\ge |x|-1$, which then suggests we have to shade the regions where $y\ge |x|-1$ and $y\le -(|x|-1)$ respectively on the Cartesian plane. Like showed in the diagram below:

[desmos="-10,10,-10,10"]y\ge \left|x\right|-1;;y\le -\left(\left|x\right|-1\right)[/desmos]

But things don't end there. We need to weed out the unwanted region(s) that don't satisfy the given inequality. We could do so by considering four cases:

Case I ($x\ge 0$ and $y\ge 0$ that correspond to $y\ge |x|-1$):

For this part, we have $-|x|+|y|\le |x-y| \le 2+|y|-|x|$, which it then gives us $0\le 2$. That means the area shaded in this area must be correct.

Case II ($x\le 0$ and $y\ge 0$ that correspond to $y\ge |x|-1$):

For this part, we have $|x|+|y|\le |x-y| \le 2+|y|-|x|$, which it then gives us $|x|\le 1$, i.e. $-1\le x \le 1$. We therefore should only shade the region of $y\ge |x|-1$ that covers the interval $-1\le x \le 0$.

i.e. we should get the shaded region shown as the picture below:

[desmos="-10,10,-10,10"]y\ge \left|x\right|-1\left\{x\ge -1\right\}[/desmos]

I will leave the other two cases for you to work them out, and I encourage you to post back to see if you get the drift of my message.