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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.

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    $\begingroup$ Composition of 1-1 functions is also 1-1. $\endgroup$
    – Nick Alger
    Jan 24 '12 at 15:19
  • $\begingroup$ See this previous question $\endgroup$ Jan 24 '12 at 15:38
  • $\begingroup$ I think the kernal of the function can help determine the nature of a function. There's are theorem or two involving it, but i don't remember the details. $\endgroup$
    – Pixel
    Feb 5 '19 at 22:37
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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:

  1. 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).

  1. 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.)

enter image description here

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  • $\begingroup$ What if the equation in question is the square root of x? Obviously it is 1:1 but I always end up with the absolute value of x being equal to the absolute value of y. What have I done wrong? $\endgroup$ Jul 17 '14 at 9:37
  • $\begingroup$ @louiemcconnell The domain of the square root function is the set of non-negative reals. Keep this in mind when solving $|x|=|y|$ (you actually solve $x=|y|$, $x\ge 0$). $\endgroup$ Jul 17 '14 at 9:43
  • $\begingroup$ The second function given by the OP was $f(x) = \frac{x-3}{x^3}$ , not $f(x) = \frac{x-3}{3}$ $\endgroup$
    – john
    Nov 9 '18 at 22:22
  • $\begingroup$ $f(x)=x^3$ is a 1-1 function even though its derivative is not always positive. More precisely, its derivative can be zero as well at $x=0$. $\endgroup$ Aug 10 at 8:07
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    $\begingroup$ @WhoSaveMeSaveEntireWorld Thanks. I edited the answer for clarity. $\endgroup$ Aug 10 at 8:59
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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 $$ \begin{eqnarray*} f(x) =f(y)\Leftrightarrow x^{2}=y^{2} \Rightarrow x=y\quad \text{or}\quad x=-y. \end{eqnarray*} $$

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    $\begingroup$ Excellent answer. Points up! $\endgroup$ Jan 24 '12 at 17:30
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    $\begingroup$ @spryno724: Thanks! $\endgroup$ Jan 24 '12 at 20:49
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    $\begingroup$ Since your answer was so thorough, I'll +1 your comment! $\endgroup$ Jan 24 '12 at 21:45
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    $\begingroup$ The second function given by the OP was $f(x) = \frac{x-3}{x^3}$ , not $f(x) = \frac{x-3}{3}$ $\endgroup$
    – john
    Nov 9 '18 at 22:20
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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 $f(x) = f(y)$.

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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: $ $Non-differentiable$ - $One-one$
$CaseII:$ $Differentiable$ - $Many-one$

First Second

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    $\begingroup$ Note that the first function isn't differentiable at $02$ so your argument doesn't work. $\endgroup$
    – Thomas
    Jan 9 '18 at 19:28
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    $\begingroup$ A function doesn't have to be differentiable anywhere for it to be 1 to 1. Consider the function given by f(1)=2, f(2)=3. It is defined only at two points, is not differentiable or continuous, but is one to one. $\endgroup$ Feb 27 '18 at 21:18
  • $\begingroup$ @JonathanShock , i get what you're saying. thank you for pointing out the error. i'll remove the solution asap. $\endgroup$ Feb 28 '18 at 3:03
  • $\begingroup$ @Thomas , i get what you're saying. thank you for pointing out the error. i'll remove the solution asap. $\endgroup$ Feb 28 '18 at 3:03
  • $\begingroup$ Please rewrite your answer with MathJax for readability and searchability. $\endgroup$
    – M. Winter
    Jun 20 '18 at 7:23
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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 $f(x) = f(y)$.

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