f(x) = 2x+1
g(x) = 3x-5
find g^-1 of f^-1 of 2
Let $\displaystyle g(x) = y_1 = 3x - 5$
for $\displaystyle g^{-1}$ switch $\displaystyle x$ and $\displaystyle y$ and solve for $\displaystyle y$
so, $\displaystyle x = 3y_1 - 5$
$\displaystyle y_1 = \frac {x + 5}3$
thus, $\displaystyle g^{-1}(x) = \frac {x + 5}3$
similarly, $\displaystyle f^{-1}(x) = \frac {x - 1}2$
now can you find $\displaystyle g^{-1} \circ f^{-1}(x) = g^{-1} \left( f^{-1} (x) \right)$?