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Question 98347: state the domain of the following:
m(x) = 5 / (x^2 + 9)
I know that the domain will be all x where the denominator does not equal zero, but when I set the denominator to zero, I get stuck:
x^2 + 9 = 0
x^2 = -9
x = sqr root -9 ----> you can't take the squareroot of a negative number, so I'm stuck.
Please help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! You are stuck because x^2+9 cannot be zero in the Real
Number System.
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Graph y=x^2+9 and you will see it is always above the x-axis.
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Or consider:
If x is Real, x^2 is always greater than or equal to zero;
so if you add 9 the result is always greater that or
equal to 9.
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Cheers,
Stan H.
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