<s>[INST] Find the domain of $\sqrt{x^2-9}$ $$\sqrt{x^2-9}$$
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?
 [/INST] You need to have $x^2-9\ge 0$, not $x^2-9<0$. This is $(x-3)(x+3)\ge 0$, which is the case when 


*

*one of $x-3$ and $x+3$ is $0$,  

*both $x-3$ and $x+3$ are positive, or  

*both $x-3$ and $x+3$ are negative.


In all other cases $(x-3)(x+3)$ is negative, which is precisely what you don’t want.


*

*The first of these happens when $x=3$ or $x=-3$. 

*The second happens when $x>3$ and $x>-3$; of course if $x>3$, then automatically $x>-3$, so this happens when $x>3$.

*And the third happens when $x<3$ and $x<-3$; this time we notice that if $x<-3$, then automatically $x<3$, so this happens when $x<-3$.
Putting the pieces together, we see that $x^2-9\ge 0$ when $x\le -3$ or $x\ge 3$.
With practice, however, you should come to realize that if $a\ge 0$, then $x^2\ge a^2$ if and only if $|x|>\sqrt{a}$. Here that means that $|x|\ge 3$, which means that $x$ is at least as far away from $0$ as $3$ is, i.e., that $x\le -3$ or $x\ge 3$.
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