How to find the inverse function of $f(x)=\frac x{x-1}$ How do I find the inverse of $f(x)=\frac x{x-1}$? I have attached my work below. I am getting stuck at $xy-x=y$. I am not sure what to do from there.

 A: $xy-x=y$
$xy-y=x$
$(x-1)y=x$
$y=\dfrac x{x-1}$
Your function is an involution.
A: \begin{align}
y&=\frac{x}{x-1}\\
y&=\frac{x-1+1}{x-1}\\
y&=1+\frac1{x-1}\\
y-1&=\frac1{x-1}\\
x-1&=\frac1{y-1}\\
x&=1+\frac1{y-1}\\
x&=\frac{y-1+1}{y-1}\\
x&=\frac{y}{y-1}
\end{align}
As it turns out, its inverse is itself!
A: $$xy - x = y$$
$$xy - y = x$$
$$y(x-1) = x$$
$$y = \frac{x}{x-1}$$
A: Normally when finding an inverse we would specify a domain, because such an inverse may not exist everywhere. In our case we should at least assume we are working on the domain $\mathbb{R} \setminus \{1\}$. With that in mind let's try to rearrange things. You correctly started with
$$ x = \frac{y}{y-1}.$$
You got to
$$xy - x= y$$
so let's go from there.
Then
$$-x = y - xy.$$
Factoring gives,
$$-x = (1-x)y.$$
So then dividing through by $(1-x)$ (this is possible since we assumed $ x\neq 1$) gives us
$$ y = - \frac{x}{1-x}.$$
So let's finally just check this. Let's call this function we just found $g(x)$ and work out $f(g(x))$.
$$
f(g(x)) = \frac{g(x)}{g(x) - 1} = \frac{\frac{x}{x-1}}{\frac{x}{x-1} -1},
$$
and I will leave you the task of checking this just gives $x$ back. So it turns out it is its own inverse.
