College Algebra, Chapter 3, 3.7, Section 3.7, Problem 42

Find the inverse of $\displaystyle f(x) = \frac{x - 2}{x + 2}$

To find the inverse of $f(x)$, we write $y = f(x)$


$
\begin{equation}
\begin{aligned}

y =& \frac{x - 2}{x + 2}
&& \text{Apply Cross Multiplication}
\\
\\
y(x + 2) =& x - 2
&& \text{Expand}
\\
\\
xy + 2y =& x - 2
&& \text{Subtract $x$ and $y$}
\\
\\
xy - x =& -2y - 2
&& \text{Factor out $x$ on the left side, add -2 from the right side}
\\
\\
x(y - 1) =& -2(y + 1)
&& \text{Divide both sides by } (y - 1)
\\
\\
x =& \frac{-2 (y + 1)}{y - 1}
&& \text{Interchange $x$ and $y$}
\\
\\
y =& \frac{-2 (x + 1)}{x - 1}
&&

\end{aligned}
\end{equation}
$


Thus, the inverse of $\displaystyle f(x) = \frac{x - 2}{x + 2}$ is $\displaystyle f^{-1} (x) = \frac{-2 (x + 1)}{x - 1}$ for $x \neq 1$.