Answer
$f^{-1}(f(x))$ = $f^{-1}(f(x))$ = $x$
Work Step by Step
$f(x)$ = $y$ = $1+\frac{1}{x}$
$y-1$ = $\frac{1}{x}$
$x$ = $\frac{1}{y-1}$
so
$f^{-1}(x)$ = $\frac{1}{x-1}$
$f^{-1}(f(x))$ = $f^{-1}(f(x))$
= $f^{-1}(1+\frac{1}{x})$
= $\frac{1}{(1+\frac{1}{x})-1}$
= $x$
$f(f^{-1}(x))$ = $f(\frac{1}{x-1})$
= $1+\frac{1}{(\frac{1}{x-1})}$
= $1+x-1$
= $x$
