Question 115061
Since you cannot take the square root of a negative number, set the inner expression greater than zero


{{{x^2-9>0}}}


Now if you graph {{{x^2-9}}}



{{{ graph( 500, 500, -10, 10, -10, 10, x^2-9) }}}


we can see that the graph is greater than zero when {{{x<-3}}} or {{{x>3}}}



So the domain of {{{f(x)=sqrt(x^2-9)}}} is *[Tex \LARGE \left(-\infty,-3\right)\cup\left(3,\infty\right)]