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What is one of the real number values that must be excluded from the domain of the function g(x) = (3x-6)/(x^2-7x+12)
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I factored it out to be 3(x-2)/(x-4)(x-3), but where do I go from there?
then x cant be 4 and x cant be 3
Oh, that's it?
Those are the 2 real numbers that have to be excluded from the domain of g(x).
Because if you plug in either 3 or 4, you get zero on the bottom and you dont want that
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Ok I see now. The wording of the problem got to me but I should've realized it was so simple (: It can't be 4 or 3 because it will make the fraction invalid. Thanks again!
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