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? 
 A: You're confusing the input to $f_i$ with the input to $f$. To make this more clear, let's say we have $f(x)=1+\frac 1 x$ and then make $f_i(s)$ its inverse so that $f(x)=s$ and $f_i(s)=x$. As you said, we have that:
$$f_i\left(1+\frac 1 x\right)=x$$
Then, you say that we can subtract $1$ from the input and then multiply $x^2$ to get $x$. Since the input is $s$, this gives us:
$$f_i(s)=(s-1)\cdot x^2$$
This is where you made a mistake: You mixed up $x$ and $s$ by saying they're the same when they're not. In reality, $s=f(x)=1+\frac 1 x$ and $x=f_i(s)$. Therefore, while the above is correct, it really means:
$$f_i(s)=(s-1)\cdot f(i)^2$$
which means we still need to solve for $f_i(s)$ in terms of $s$. If we divide both sides by $f(i)^2$ and take the reciprocal, we get:
$$f_i(s)=\frac{1}{s-1}$$
A: $$f_i(1+x^{-1})=x$$
Replace $x$ with $x^{-1}$ to get
$$f_i(1+x)=x^{-1}$$
Replace $x$ with $x-1$ to get
$$f_i(x)={1\over x-1}$$
A: You need to replace $x$ with $1+\frac1x$ in both parts of the formula.  So
$$f_i\left(1+\frac1x\right)=\left(1+\frac1x-1\right)\left(1+\frac1x\right)^2
=\frac1x\left(1+\frac1x\right)^2$$
