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)