Answer
Using the inverse function property we have:
$f^{-1}(x)=\frac{5+4x}{1-3x}$
$g^{-1}(x)=\frac{x-5}{3x+4}$
Which means that $f$ and $g$ are inverse of each other.
Work Step by Step
To show it we will simply calculate inverse of the both $f$ and $g$ functions.
To calculate $f^{−1}(x)$ first we have to write it in terms of $y$ and $x$ and then replace $x$ with $y$ and vice versa. And at last solve the equation for $y$:
$f(x)=\frac{x-5}{3x+4}$
$y=\frac{x-5}{3x+4}$
$x=\frac{y-5}{3y+4}$
$x(3y+4)=y-5$
$3xy+4x=y-5$
$3xy-y=-4x-5$
$y(3x-1)=-4x-5$
$y=\frac{-4x-5}{3x-1}$ //Factor out the minus sign
$y=\frac{-(4x+5)}{-(1-3x)}$
$y=\frac{5+4x}{1-3x}$
$f^{-1}(x)=\frac{5+4x}{1-3x}$
$g(x)=\frac{5+4x}{1-3x}$
$y=\frac{5+4x}{1-3x}$
$x=\frac{5+4y}{1-3y}$
$x(1-3y)=5+4y$
$x-3xy=5+4y$
$-3xy-4y=5-x$
$y(-3x-4)=5-x$
$y=\frac{5-x}{-3x-4}$ //Factor out the minus sign
$y=\frac{-(x-5)}{-(3x+4)}$
$y=\frac{x-5}{3x+4}$
$g^{-1}(x)=\frac{x-5}{3x+4}$