Answer
$f^{-1}$(x) = $\frac{x + 1}{x - 1}$.
Work Step by Step
1. Switch f(x) and x in the equation.
x = $\frac{f(x) + 1}{f(x) - 1}$
2. Isolate f(x) in one side of the equation.
x*( f(x) - 1 ) = f(x) + 1
f(x)*x - x = f(x) + 1
f(x)*x - f(x) = x + 1
f(x)*( x - 1 ) = x + 1
f(x) = $\frac{x + 1}{x - 1}$
3. Rewrite f(x) as $f^{-1}$(x)
$f^{-1}$(x) = $\frac{x + 1}{x - 1}$
4. Test it
f(3) = $\frac{3 + 1}{3 - 1}$
f(3) = 2
$f^{-1}$(2) = $\frac{2 + 1}{2 - 1}$
$f^{-1}$(2) = 3
