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
$$f^{-1}(x)=\begin{cases}\frac{x+1}{x-1}\,,\, x\ne 1\\
1\,\quad,\, \,x=1
\end{cases}$$
A: Write $$f(x) = 1 + \frac{2}{x-1}$$
The fraction takes on all values except $0$, since it's simply $\frac 1x$ scaled and shifted horizontally. Hence $f$ takes on all values except $1$. The piecewise definition guarantees that $1$ is in the image of $f$. Therefore, $f$ is surjective onto $\mathbb{R}$.  
A: We have
$f(f(1))=1$
and
$\forall x\neq1$
$$f(f(x))=f(1+\frac{2}{x-1})$$
$$=1+\frac{2}{\frac{2}{x-1}}$$
$$=1+x-1=x.$$
thus $f$ is one to one from $\mathbb R \to \mathbb R,$ with $$f^{-1}=f.$$
A: Let $y$ be an arbitrary number and find a $x$ such that $f(x)=y$ (i.e. solve an equation). If you can do that for all $y$, then you have shown that $f$ is onto $\Bbb R$.
A: Let $y$ be a real.
we look for $x$ such that
$$f(x)=y.$$
if $y=1$ then $x=1$.
assume $y\neq1$.
$$\frac{x+1}{x-1}=y  \implies$$
$$1+\frac{2}{x-1}=y  \implies$$
$$\frac{1}{1-y}=\frac{1-x}{2}  \implies$$
$$x=1+\frac{2}{y-1} =f(y) \implies $$
$f$ is onto.
A: If you have some theorems about continuity as your disposal, you can proceed as follows, without doing any algebra to speak of:
The function $f$ is continuous on $(1,\infty)$.  We have $\lim_{x\to1^+}f(x)=\infty$ and $\lim_{x\to\infty}f(x)=1$, hence $f$ takes on every value greater than $1$, by the Intermediate Value Theorem.
Similarly, $f$ is continuous on $(-\infty,1)$, and we have $\lim_{x\to1^-}f(x)=-\infty$ and $\lim_{x\to-\infty}f(x)=1$, hence $f$ takes on every value less than $1$, by IVT.
Finally, since $f(1)=1$ (by definition), the function $f$ takes on every real value, hence $f:\mathbb{R}\to\mathbb{R}$ is onto.
A: Here is more of a meta-answer.
When trying to prove surjectivity, it's often worth trying to prove that at least it hits some particular value. For example, does it hit $2$? 
Well, that amounts to finding $x$ such that $\frac{x+1}{x-1} = 2$, which hopefully you can see how to do.
Once you've done that, you can see if your method works for arbitrary $2$.
