$$\sqrt{x^2-9}$$
I know that the domain of square root is greater than or equal to zero. I solve for when $x^2-9<0$ and get $x^2<9$. Now I get $x<-3$ and $x<3$. I know that the domain is $(-\infty,-3] \cup [3,\infty)$ The problem is I do not understand how the $x<3$ gets flipped to $x>3$, am I doing this step properly or is there another way to do it?
Best Answer
You went a bit off track midway. Instead, let's rewrite $x^2-9<0$ as $(x+3)(x-3)<0.$ This will only be true when $x+3$ and $x-3$ are of opposite sign (why?), so since $x-3<x+3$ for all real $x,$ then we need to figure out when $x-3<0$ and $x+3>0$. Can you take it the rest of the way?