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Given that $f(x) = 3x -1$ and $g(x)= x(x-1)$ then $g(f^{-1}(x))$?

I got $f^{-1}(x): y = \frac{x+1}{3}$; and I put it in $g(x)$ as, $(\frac{x+1}{3})\cdot(\frac{x+1}{3} -1)$ I simplified it as $\frac{(x+1)(x-2)}{9}$ but the answer given is $\frac{x}{3}(x+1)$

The options are:

A) x/3(x+1)

B) x/3(x-1)

C) 3x(3x-1)

D) x/3(x/3+1)

MR. Raindrop
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  • You get the answer in the book if you fail to form a common denominator in the term $\frac{x + 1}{3} - 1$. Your answer is correct. – N. F. Taussig Jul 11 '18 at 08:54

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