Find the inverse of $f(x) = 1 + \frac{1}{x}, x \gt 0$ I'm tasked to find the inverse of the function $$f(x) = 1 +  \frac{1}{x}, x \gt 0$$
The book offers a solution, simply to set $$1 + \frac{1}{x} = s$$
and solve $$x = \frac{1}{s-1}$$
and I think I understand why that works. But why won't this work? 
$f_i$ is the function f(x)'s inverse
$$f_i(f(x)) = x  $$
$$f_i ( 1 + \frac{1}{x} ) = x$$
$f_i$ produces the result $x$ with input $1 + \frac{1}{x}$ if: $$f_i(x) = (x-1)\times x^2$$
So why is $(x-1)\times x^2$ not the inverse? 

You need to replace $x$ with $1+\frac1x$ in both parts of the formula.  So
$$f_i\left(1+\frac1x\right)=\left(1+\frac1x-1\right)\left(1+\frac1x\right)^2
=\frac1x\left(1+\frac1x\right)^2$$
