Question 147400
The domain of x would be whatever x is that won't make the denominator zero.  So let's start like this:  {{{m(x)= 5/x^2-9}}}.  Now since the domain of x is whatever x is that won't make the denominator zero, let's focus just on the denominator.  First factor it.  It is the difference between two perfect squares, so it is {{{(x-3)(x+3)}}}.  Now take each expression separately as {{{x-3}}} and {{{x+3}}}.  Set them both equal to zero. {{{x-3=0}}} and {{{x+3=0}}}.  Solve both of them and get 3 and -3.  These values for x are what x is to make the denominator zero. Therefore, these values are what x cannot be in this function.  So the domain of x would be any value for x except those two. Domain: all x except 3 and -3.