Q:

Proving that $f(x)= \frac {(x+1)}{(x-1)}$ is onto

How would I go about proving that the function is onto for:
$f(x)= \frac {(x+1)}{(x-1)}$ if x does not equal to 1. And $f(x)=1$ if $x = 1$

A:

$$y=\frac{x+1}{x-1}$$
thus
$$xy-y=x+1$$
or
$$xy-x=y+1$$
in other words
$$x=\frac{y+1}{y-1}\in\mathbb{R}-\{1\}$$
we can write
$$f^{-1}(x)=\begin{cases}\frac{x+1}{x-1}\,,\, x\ne 1\\
1\,\quad,\, \,x=1
\end{cases}$$

