2.15 One-to-One Functions

We learned that a function is a rule that assigns one and only one value to each element of its domain. However a function may assign the same value to two or more elements in its domain. For example $f(x)=x^2$ assigns 4 to both $x=2$ and $x=-2$. Or $f(x)=|x|$ assigns the same value to both $x=a$ and $x=-a$. Or $f(x)=c$ where $c$ is a constant, takes on the same value $c$ at all $x$. But some functions assign distinct values to distinct elements of their domains. For example, $f(x)=2x+3$ takes on a different value at each value of $x$. Such functions are called one-to-one or 1-1 functions.

•  Equivalently we can say that a function is one-to-one whenever $x_1\ne x_2$ in $A$, then $f(x_1)\ne f(x_2)$.
  $$x_1\ne x_2\Rightarrow f(x_1)\ne f(x_2).(\text{b})$$
•  The above definition states that a one-to-one function $y=f(x)$ takes on each value in its range only once. If the graph of the function is cut by a horizontal line $y=c$ at more than one point, the value
  $y=c$ will correspond to more than one value of $x$, and the function will not be one-to-one. We may use the following test to specify whether or not a function is one-to-one.

Solution

Method 1: If $m{x}_1+b=m{x}_2+b$ then $x_1=x_2$. Thus $f$ is a one-to-one function.
Method 2: Each horizontal line $y=c$ intersects the graph of $f$ exactly once (not twice or more). So $f$ is a one-to-one function.

Solution

Method 1: If $x_1^2+1=x_2^2+1$ then $x_1=x_2$ or $x_1=-x_2$. Therefore, $f$ is not a one-to-one function.
Method 2: Because a horizontal line $y=c$ where $c>1$ intersects the graph of $f$ twice, it is not a one-to-one function.

 

Note that in the above example, although $f$ is not a one-to-one function on its entire natural domain $\mathbb{R}$, if we restrict the domain to $x\ge 0$ , i.e. $f:[0,\mathrm{\infty })\to \mathbb{R}$ with $f(x)=x^2+1$, then the horizontal line test is passed and the function becomes one-to-one. Remark that the original and restricted functions are not the same functions because their domains are different. However, the two functions assume the same values on $[0,\mathrm{\infty })$.

•  The function $f:(-\mathrm{\infty },0]\to \mathbb{R},$ $f(x)=x^2+1$ is also one-to-one.

Solution

Method 1: Yes, it is because if $1/x_1=1/x_2$ then $x_1=x_2$.
Method 2: Each horizontal line $y=c$ ($c\ne 0$) intersects the graph of $f$ once and $y=0$ never intersects the graph of $f$.

 

We can figure out whether or not a function is one-to-one just by looking at their graphs. In the following table, we have investigated some common functions:

function	natural domain	1-1 on natural domain?	Graphs
$f(x)=x^n$  ($n$ is even)	$\mathbb{R}$	no
$f(x)=x^n$  ($n$ is odd)	$\mathbb{R}$	yes
$f(x)=\sqrt[n]{x}$ ($n$ is even)	$[0,\mathrm{\infty })$	yes
$f(x)=\sqrt[n]{x}$ ($n$ is odd)	$\mathbb{R}$	yes
$f(x)={\displaystyle \frac{1}{x^n}}$ ($n$ is even)	$\mathbb{R}-0=\{x|x\ne 0\}$	no
$f(x)={\displaystyle \frac{1}{x^n}}$ ($n$ is odd)	$\mathbb{R}-0=\{x|x\ne 0\}$	yes

• Note that every monotonic function is one-to-one. Recall that a monotonic function is a function that is increasing or decreasing. Suppose $f$ is an increasing function because if $x_1<x_2$ then $f(x_1)<f(x_2)$ and if $x_2<x_1$ then $f(x_2)<f(x_1)$. That is if $x_1\ne x_2$ then $f(x_1)\ne f(x_2)$. Similarly, we can show that if $f$ is a decreasing function, then it is one-to-one.