Answer
$f(^{-1}(x)=\dfrac{4x+2}{3(x+1)},\text{ }x\ne-1$
Work Step by Step
Let $y=f(x)$. Then the given one-to-one function, $
f(x)=\dfrac{-3x+2}{3x-4},\text{ }x\ne\dfrac{4}{3}
,$ becomes
\begin{align*}\require{cancel}
y&=\dfrac{-3x+2}{3x-4},\text{ }x\ne\dfrac{4}{3}
.\end{align*}
To find the inverse, interchange the $x$ and $y$ variables and then solve for $y$. That is,
\begin{align*}
x&=\dfrac{-3y+2}{3y-4}
&(\text{interchange $x$ and $y$})
\\\\
(3y-4)(x)&=\left(\dfrac{-3y+2}{\cancel{3y-4}}\right)(\cancel{3y-4})
&(\text{solve for $y$})
\\\\
3xy-4x&=-3y+2
\\
3xy+3y&=4x+2
\\
3y(x+1)&=4x+2
\\\\
\dfrac{\cancel3y(\cancel{x+1})}{\cancel3(\cancel{x+1})}&=\dfrac{4x+2}{3(x+1)}
\\\\
y&=\dfrac{4x+2}{3(x+1)},\text{ }x\ne-1
.\end{align*}
Hence, the inverse of $
f(x)=\dfrac{-3x+2}{3x-4},\text{ }x\ne\dfrac{4}{3}
$ is $
f(^{-1}(x)=\dfrac{4x+2}{3(x+1)},\text{ }x\ne-1
$.
