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

Question

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 $x = 1$

Answer 1

$$y=\frac{x+1}{x-1}$$ 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 $$f^{-1}(x)=\begin{cases}\frac{x+1}{x-1}\,,\, x\ne 1\\ 1\,\quad,\, \,x=1 \end{cases}$$

Answer 2

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 $\mathbb{R}$.

Answer 3

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 $$f^{-1}=f.$$

Answer 4

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 $\Bbb R$.

Answer 5

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.

Answer 6

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.

Answer 7

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 $2$.