For all numbers x such that x is not equal 1, if g(x) is defined by

Question is \(\frac{1}{g(2)}*\frac{1}{g(x)}=\)

\(g(x)=\frac{x^2+2}{x-1}\) so \(\frac{1}{g(x)}=\frac{x-1}{x^2+2}\);

\(g(2)=\frac{2^2+2}{2-1}=6\) so \(\frac{1}{g(2)}=\frac{1}{6}\);

\(\frac{1}{g(2)}*\frac{1}{g(x)}=\frac{1}{6}*\frac{x-1}{x^2+2}=\frac{x-1}{6(x^2+2)}\).

Answer: D.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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I've got a question.

Isn't answer B exactly the same as D?

Consider this example with some numbers.

D says (As Bunual mentioned): 4/2 *2/1= 8/2 --> which would be the answer to the question

But,

B says: (4/2)/(1/2)=4 --> It's the same!!!!!

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.