how to tell analytically if the function is injective How can I tell analytically if the following function is injective or not
$f(x)=\frac{3x-5}{x^{2}-1}$
I start by letting $f(x)=f(y)$ and then I would like to see that I can have $x \neq y$, this is so because according to the graph the function is not inyective:

Then I have 
$$\frac{3x-5}{x^{2}-1}=\frac{3y-5}{y^{2}-1}$$ 
$$(3x-5)(y^{2}-1)=(3y-5)(x^{2}-1)$$ 
doing the Calculus I get 
$5x^{2}-3yx^{2}+3xy^{2}-3x+3y-5y^{2}=0$
but then I don't know what to do, for instance for the function 
$f(x)=x^{2}-x$ I can do the following
Let $f(x)=x^{2}+x=f(y)=y^{2}+y$
then
$x^{2}-y^{2}=y-x$
$(x-y)(x+y)=y-x$
now assuming $x \neq y$ (I do not know If I can make this assumption) I can divide both sides and get
$x+y=-\frac{y-x}{y-x}=-1$,
so in this case the function is not injective since for every pair of numbers such that $x+y=-1$ I will have $f(x)=f(y)$ for instance $f(0)=f(-1)$
Is this the general method? Should I start letting $f(x)=f(y)$ and the by doing the Calculus I should get $x=y$ or $x$ posssibly different from $y$?
 A: It suffices to note that letting $y=f(x)=5$ leads to two different solutions for $x$: For $x\neq\pm1$, 
$$f(x)=\frac{3x-5}{x^{2}-1}=5 \Leftrightarrow 3x-5=5x^{2}-5 \Leftrightarrow 3x=5x^{2}\Leftrightarrow x=0 \ \ \textrm{or} \ \ x=\frac{3}{5}$$
Thus the function is not injective.
A: Your method is general provided you can factor polynomials or other expressions (not always easy).
In your example we have
$$
5x^{2}-3yx^{2}+3xy^{2}-3x+3y-5y^{2}=
(x-y)(-3 x y+5 x+5 y-3)
$$
and so we can have $5x^{2}-3yx^{2}+3xy^{2}-3x+3y-5y^{2}=0$ without $x-y=0$: we just need that $-3 x y+5 x+5 y-3=0$. For instance, $x=0, y=3/5$ works.
A: How can I tell analytically if $f(x)=\frac{3x-5}{x^{2}-1}$ is injective?
You have to prove that there are two real numbers $x$ and $y$ such that $x\neq y$ and $f(x)=f(y)$. For this, it's enough to prove that there is a real number $x\neq 0$ such that $f(x)=f(2x)$ (then, just take $y=2x$).
Note that
\begin{align}
f(x)=f(2x)&\quad\Leftrightarrow \quad\frac{3x-5}{x^2-1}=\frac{6x-5}{4x^2-1}\tag{if $|x|,|2x|\neq 1$}\\\\
&\quad\Leftrightarrow \quad 6 x^3-15 x^2+3 x = 0\\\\
&\quad\Leftrightarrow \quad 6 x^2-15 x+3 = 0\tag{if $x\neq 0$}\\
\end{align}
Solving the last equation we get the result.
