Verifying Inverse Functions

Verify that the inverse of \(f(x) = \dfrac{1}{x-1}\) is \(f^{-1}(x) = \dfrac{1}{x}+1\). For what values of \(x\) is \(f^{-1} (f( x)) = x?\) For what values of \(x\) is \(f (f^{-1}(x)) = x?\)

Solution The domain of \(f\) is \(\{x|x\neq 1\}\) and the domain of \(f^{\,-1}\) is \(\{ x|x\neq 0\}\). Now \[ \begin{array}{*{20}c} \;\;\;\;\;\;\;\;\;\;{f^{ - 1} (f(x)) = f^{ - 1} \left( {\frac{1}{{x - 1}}} \right) = \frac{1}{{\left( {\frac{1}{{x - 1}}} \right)}} + 1 = x - 1 + 1 = x,} & {{\rm{provided }}\;\;x \ne \;1} \\ {f(f^{ - 1} (x)) = f\left( {\frac{1}{x} + 1} \right) = \frac{1}{{\left( {\frac{1}{x} + 1} \right) - 1}} = \frac{1}{{\frac{1}{x}}} = x,} & {{\rm{provided}}\;x\; \ne 0} \\ \end{array} \]