## Calculus, 10th Edition (Anton)

$f^{-1}$(x) = $\frac{x + 1}{x - 1}$.
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