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 $$\begin{array}{*{20}c} \;\;\;\;\;\;\;\;\;\;{f^{ - 1} (f(x)) = f^{ - 1} \left( {\frac{1}{{x - 1}}} \right) = \frac{1}{{\left( {\frac{1}{{x - 1}}} \right)}} + 1 = x - 1 + 1 = x,} & {{\rm{provided }}\;\;x \ne \;1} \\ {f(f^{ - 1} (x)) = f\left( {\frac{1}{x} + 1} \right) = \frac{1}{{\left( {\frac{1}{x} + 1} \right) - 1}} = \frac{1}{{\frac{1}{x}}} = x,} & {{\rm{provided}}\;x\; \ne 0} \\ \end{array}$$