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Given g(x) = x+1 x−1 , solve g(x) = 0
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Since g(x) = 0 g(x) = x+1 0 = x+ 1 Solve.
can you rewrite it ? what is x-1 ?
Sorry given g(x)= x+1/x-1, solve g(x)=0
\[g(x) = \frac{x+1}{x-1}\] \[0 = \frac{x+1}{x-1}\]
ok than first of all you need push the restriction for x when the denominator will be equal zero
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because than the fraction will be undefined
ok ?
so what will be ?
Rose ! are you there ?
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