Domain of g(x) = (x-2)/(x^2-9) |x|≠3

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The domain of the function g(x) = (x-2)/(x^2-9) is determined by ensuring the denominator does not equal zero. The critical values are x = 3 and x = -3, as both make the denominator zero. The confusion arises from the interpretation of squaring negative numbers, where (-3)^2 equals 9, confirming that both -3 and 3 are excluded from the domain. Thus, the domain of g(x) is all real numbers except x = 3 and x = -3.

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for g(x) = (x-2) / (x^2 -9)

the domain of the function would be simply to have the denominator not equal to 0, which in this case would be x = 3.

but.. the solution states that it is -3 and 3 cannot be used.

this is confusing seeing that -3^2 -9 = -18 which is not 0.

maybe it means |-3|?
 
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(-3)^2 - 9 = 9-9 = 0

there is a difference,

-3^2 = (-1)*3^2 = -1*9 = -9, but (-3)^2 = (-1*3)^2 = (-1)^2 * 3^2 = 1*9 = 9
 
i get it. thanks
 
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