Answer
$f(^{-1}(x)=\dfrac{-2x+4}{x-1},\text{ }x\ne1$
Work Step by Step
Let $y=f(x)$. Then the given one-to-one function, $
f(x)=\dfrac{x+4}{x+2},\text{ }x\ne-2
,$ becomes
\begin{align*}\require{cancel}
y&=\dfrac{x+4}{x+2},\text{ }x\ne-2
.\end{align*}
To find the inverse, interchange the $x$ and $y$ variables and then solve for $y$. That is,
\begin{align*}
x&=\dfrac{y+4}{y+2}
&(\text{interchange $x$ and $y$})
\\\\
(y+2)(x)&=\left(\dfrac{y+4}{\cancel{y+2}}\right)(\cancel{y+2})
&(\text{solve for$y$})
\\\\
xy+2x&=y+4
\\
xy-y&=-2x+4
\\
y(x-1)&=-2x+4
\\\\
\dfrac{y(\cancel{x-1})}{\cancel{x-1}}&=\dfrac{-2x+4}{x-1}
\\\\
y&=\dfrac{-2x+4}{x-1},\text{ }x\ne1
.\end{align*}
Hence, the inverse of $
f(x)=\dfrac{x+4}{x+2},\text{ }x\ne-2
$ is $
f(^{-1}(x)=\dfrac{-2x+4}{x-1},\text{ }x\ne1
$.
