## Calculus 8th Edition

The inverse function is $$f^{-1}(x)=\frac{3x+1}{4-2x},\quad x\neq 2.$$
First, we have to exclude $2x+3=0\Rightarrow x=-\frac{3}{2}$ from the domain of the function because the denominator must be different than zero. That same value must be excluded from the range of the inverse function as well. Now, let $y=f(x)=\frac{4x-1}{2x+3}$ to find the inverse function we need to express $x$ in terms of $y$ and then we will have $x=f^{-1}(y).$ $$y=\frac{4x-1}{2x+3}\Rightarrow y(2x+3)=4x-1\Rightarrow 2xy+3y=4x-1\\ 4x-2xy=3y+1\Rightarrow x=\frac{3y+1}{4-2y},$$ so $$x=f^{-1}(y)=\frac{3y+1}{4-2y}.$$ Renaming $y$ back to $x$ we get $$f^{-1}(x)=\frac{3x+1}{4-2x}$$ where $x\neq 2$.