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Find the domain of the function. Will appreciate an explanation. Thanks in advance. g(x)=3x/x^2-1
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\[g(x)= \frac{ 3x }{ x^2-1 }\]
oh and answer in interval notation. I got a really weird answer.....
the bottom of fraction cant be 0 so find out what makes this possible and exclude it bottom of fraction = x^2−1 so we need to make sure that x^2-1 is not equal to 0 so set it equal to zero and solve x^2-1 = 0 (x-1)(x+1) = 0 when is this true? when x = 1, or x = -1 so x CANT be -1 or 1 that's your domain all real numbers such that x is NOT -1 or 1
hope this helps
got it thanks!
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will u come help me when u finish please
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