How to determine if a function is one-to-one? I am looking for the "best" way to determine whether a function is one-to-one, either algebraically or with calculus. I know a common, yet arguably unreliable method for determining this answer would be to graph the function. However, this can prove to be a risky method for finding such an answer at it heavily depends on the precision of your graphing calculator, your zoom, etc...
What is the best method for finding that a function is one-to-one? 
In your description, could you please elaborate by showing that it can prove the following:


*

*$\frac{x-3}{x+2}$ is one-to-one.

*$\frac{x-3}{x^3}$ is not one-to-one.
 A: To show that $f$ is 1-1, you could show that $$f(x)=f(y)\Longrightarrow x=y.$$
So, for example, for $f(x)={x-3\over x+2}$:
Suppose ${x-3\over x+2}= {y-3\over y+2}$. Then:
\begin{align*}
&{x-3\over x+2}= {y-3\over y+2}  \\
\Longrightarrow& (y+2)(x-3)= (y-3)(x+2)\\
\iff& yx+2x-3y-6= yx-3x+2y-6\\
\iff&2x-3y =-3x+2y\\
\iff&2x+3x =2y+3y\\
\iff&5x =5y\\
\iff&x=y
\end{align*}
So $f(x)={x-3\over x+2}$ is 1-1.
I'll leave showing that $f(x)={{x-3}\over 3}$ is 1-1 for you.
Alternatively, to show that $f$ is 1-1, you could show that $$x\ne y\Longrightarrow f(x)\ne f(y).$$
Or, for a differentiable $f$ whose derivative is either always positive or always negative, you can conclude $f$ is 1-1 (you could also conclude that $f$ is 1-1 for certain functions whose derivatives do have zeros; you'd have to insure that the derivative never switches sign and that $f$ is constant on no interval).

You would discover that a function $g$ is not 1-1, if, when using the first method above, you find that the equation is satisfied for some $x\ne y$. For example, take $g(x)=1-x^2$. Then
$$
\eqalign{
&g(x)=g(y)\cr
\iff&{1-x^2}= {1-y^2}  \cr
\iff&-x^2= -y^2\cr
\iff&x^2=y^2\cr}
$$
The above equation has $x=1$, $y=-1$ as a solution.  So, there is $x\ne y$ with $g(x)=g(y)$; thus $g(x)=1-x^2$ is not 1-1.
Of course, to show $g$ is not 1-1, you need only find two distinct values of the input value  $x$ that give $g$ the same output value.

Although you rightfully point out that the graphical method is unreliable; it is still instructive to consider the methods used and why they work:
Graphically, you can use either of the following:

*

*Use the "Horizontal Line Test":

$f$ is 1-1 if and only if every horizontal line intersects the graph
of $f$ in at most one point.  Note that this is just the graphical
interpretation of "if $x\ne y$ then $f(x)\ne f(y)$"; since the
intersection points of a horizontal line with the graph of $f$ give
$x$ values for which $f(x)$ has the same value (namely the $y$-intercept of the line).


*Use the fact that a continuous $f$ is 1-1 if and only if $f$ is either
strictly increasing or strictly decreasing. This, of course, is the case if $f$ is differentiable and the derivative is always positive or always negative (with perhaps being zero at "isolated" points).
(Note this method  applies to only the green function below.)


A: For your modified second function $f(x) = \frac{x-3}{x^3}$, you could note that
$$f(x) - f(y) = \frac{(x-y)((3-y)x^2 +(3y-y^2) x + 3 y^2)}{x^3 y^3}$$
As a quadratic polynomial in $x$, the factor $
(3-y)x^2 +(3y-y^2) x + 3 y^2$ has discriminant $y^2 (9+y)(y-3)$.  So when either $y > 3$ or $y < -9$ this produces two distinct real $x$ such that $f(x) = f(y)$.
A: A one-to-one function is an injective function. A function $f:A\rightarrow B$ is an injection if $x=y$ whenever $f(x)=f(y)$.
Both functions $f(x)=\dfrac{x-3}{x+2}$ and $f(x)=\dfrac{x-3}{3}$ are injective.
Let's prove it for the first one
$$
\begin{eqnarray*}
f(x) &=&f(y)\Leftrightarrow \frac{x-3}{x+2}=\frac{y-3}{y+2} \\
&\Rightarrow &\left( y+2\right) \left( x-3\right) =\left( y-3\right)
\left( x+2\right)  \qquad(\text{for  }x\neq-2,y\neq -2)\\
&\Rightarrow &xy-3y+2x-6=xy+2y-3x-6 \\
&\Rightarrow &-3y+2x=2y-3x\Leftrightarrow 2x+3x=2y+3y \\
&\Rightarrow &5x=5y\Rightarrow x=y.
\end{eqnarray*}$$
So we concluded that $f(x) =f(y)\Rightarrow x=y$, as stated in the definition.
As for the second, we have
$$
\begin{eqnarray*}
f(x) =f(y)\Leftrightarrow \frac{x-3}{3}=\frac{y-3}{3} \Rightarrow &x-3=y-3\Rightarrow x=y.
\end{eqnarray*}
$$
An example of a non injective function is $f(x)=x^{2}$ because
$$
\begin{eqnarray*}
f(x) =f(y)\Leftrightarrow x^{2}=y^{2} \Rightarrow x=y\quad \text{or}\quad x=-y.
\end{eqnarray*}
$$
A: Before putting forward my answer, I would like to say that I am a student myself, so I don't really know if this is a legitimate method of finding the required or not. It would be a good thing, if someone points out any mistake, whatsoever.
$f(x)$ is the given function. 
$f'(x)$ is it's first derivative. 
By equating $f'(x)$ to 0, one can find whether the curve of $f(x)$ is differentiable at any real x or not.
$CaseI: $      $Non-differentiable$  -    $One-one$
$CaseII:$      $Differentiable$      - $Many-one$


A: As far as I remember a function $f$ is 1-1 it is bijective thus
$f$ is surjective
$f$ is injective
By definition let $f$ a function from set $X$ to $Y$. $f$ is surjective if for every $y$ in $Y$ there exists an element $x$ in $X$ such that $f(x)=y$.
$f$ is injective if the following holds $x=y$ if and only if $f(x) = f(y)$.
