document.write( "Question 147400: State the domain of m(x)= 5/x^2-9\r
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Algebra.Com's Answer #107771 by mangopeeler07(462) $\"\"$ $\"About$

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The domain of x would be whatever x is that won't make the denominator zero. So let's start like this: $\"m%28x%29=+5%2Fx%5E2-9\"$ . 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 $\"%28x-3%29%28x%2B3%29\"$ . Now take each expression separately as $\"x-3\"$ and $\"x%2B3\"$ . Set them both equal to zero. $\"x-3=0\"$ and $\"x%2B3=0\"$ . 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. \n" ); document.write( "

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