Here is a more "PS" oriented approach. If you have practiced factoring, this should take you 15 seconds with no writing or arithmetic involved.
\((\frac{1}{g(x)})\) is the reciprocal of \(g(x)\). You cannot factor out \((x-1)\) from \(x^2+2\). So flip \(\frac{(x^2+2)}{(x-1)}\) to get \(\frac{(x-1)}{(x^2+2)}\)
Furthermore \(g(x)\) is a function that only contains one variable \(x\). \(g(2)\) means you plug 2 into every x, and will therefore equal a numerical value and \(\frac{1}{g(2)}\) = numerical value.
Therefore the denominator must be \(x^2+2\). Answers B, C, E are out.
Between A and D, it comes down to whether \((\frac{1}{g(2)})\) is \(\frac{1}{6}\) or \(\frac{6}{1}\).
We know \(x-1\) < \(x^2+2\) so numerator < denominator means that \((\frac{1}{g(2)})\) <1. That eliminates A. D must be the correct answer.
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