Find the lower and upper bounds

Question

I'm stuck with this question:

$-2 < x < 6$ and $-4<y<-2$ What are the bounds of $x^2-y^2$?

I thought that they are $(-2)^2-(-4)^2 = -12$ and $6^2-(-2)^2 = 32$, but apparently they are $-16$ and $32$. How are they calculated?

Answer 1

Since $$-2\lt x\lt 6\Rightarrow 0\le x^2\lt 6^2$$ and $$-4\lt y\lt -2\Rightarrow (-2)^2\lt y^2\lt (-4)^2\Rightarrow -(-4)^2\lt -y^2\lt -(-2)^2,$$ one has $$-16=0-(-4)^2\lt x^2-y^2\lt 6^2-(-2)^2=32.$$