One to One Function - Graph, Examples, Definition

What is One-to-One Function?

A relation from a set $A$ to a set $B$ is said to be a function from $A$ to $B$ if every element of set $A$ has one and only one image in set $B$. One-to-one function is one of the type of functions in which every element in domain has a unique image in codomain. Now, let us look in detail about what is one-to-one function.

One-to-one function Definition

An injective function, mostly referred to as a one-to-one function, is one in which distinct elements of $A$ have distinct images with $B$. If a function has distinct images, it can only be one-to-one if the pre-images are different. On similar logic, if the elements in $B$ set differ, it can only be one-to-one if the elements in $A$ set had different pre-images.

A function $f: X \rightarrow Y$ is called a one-one (or injective) function, if different elements of $X$ have different images in $B$. i.e. no two elements of set $X$ can have the same image.

Consider,
$f: X \longrightarrow Y$, function given by $y=f(x)=x$, and
$X=\{-2,2,4,6\}$ and $Y=\{-2,2,4,6\}$

The relevant one-to-one function graphs are formed below:

Now, consider, $\mathrm{X}_1=\{1,2,3\}$ and $\mathrm{X}_2=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}$

$
\mathrm{f}: \mathrm{X}_1 \longrightarrow \mathrm{X}_2
$

How to Identify a One-to-One Function?

A one-to-one function can be identified by verifying the condition of one-to-one function, $f\left(x_1\right)=f\left(x_2\right) \Longrightarrow x_1=x_2$.

For example, Let $f(x)=2 x+3$. We'll check if it's one-to-one using the definition:

Let us assume that $f\left(x_1\right)=f\left(x_2\right)$, meaning two outputs of the function are equal.

$
f\left(x_1\right)=f\left(x_2\right) \Longrightarrow 2 x_1+3=2 x_2+3
$

$
\Longrightarrow 2 x_1=2 x_2
$

$
\Longrightarrow x_1=x_2
$

Therefore, the function $f(x) = x+3$ is a one-to-one function.

Another way to identify one-to-one function is by horizontal line test.

One-to-One Function Horizontal Line Test

An injective or one to one function can be identified by the horizontal line test which is also known famously as geometric test.The process is explained below:

1. If a horizontal line intersects the graph of the function more than once or at multiple points, then we conclude that the function is not one-to-one.
2. If a horizontal line can intersect the graph of the function at only a single place or a unique point, then the function is identified as one-to-one.

Examples of One-to-One Functions

Here are a few famous one-to-one function examples:

Identity Function: The identity function is a simple example of a one-to-one function. The input and output values remain the same.In other words, we get the same output as the input itself, with no change at all. For any real number $x$, the identity function is defined as:

$f(x)=x$

Note: Every distinct input x corresponds to a unique output $f(x)$, making it a one-to-one function.

Absolute Value Function: The absolute value function $f(x)=∣x∣$ is also a one-to-one function. For any natural number $x$, the absolute value function returns a non-negative output, and if we continuously change the values of $x$, we get different positive outputs each time, being unique every single time. But this doesn't work in the case of integers as $-x$ and $x$ will have the same output.

Is Parabola a one to one function?

No, a parabola is not a one-to-one function. This conclusion can be proved by the horizontal line test.

Now, if we draw the horizontal lines, then it will intersect the parabola at exactly two points in the graph upon observation. Hence, for each value of $x$, there will be two outputs for a single input.

Properties of One-to-One Functions

1. Uniqueness: Each element of the domain maps to a unique element in the codomain.That means there is exactly one unique value in the output range(y) for every single value of input range(x).
2. Inverse Function: If f is a one-to-one function, then there always exists an inverse function $f^{-1}$ for it, which is also a function.
3. Horizontal Line Test: A function $f$ is one-to-one if and only if no horizontal line intersects the graph of $f$ more than once.

One-to-One Function Inverse

For every one-to-one function equation, there exist a inverse. If $f$ is a function defined as $y=f(x)$, then the inverse function of $f$ is $x$ $=f^{-1}(y)$ that is $f^{-1}$ defined from $y$ to $x$. In the inverse function, the codomain of $f$ is the domain of $f^{-1}$ and the domain of $f$ is the co-domain of $f^{-1}$.

Only one-to-one functions have its inverse as these functions have one to one mappings, that is each element from the range corresponds to one and only one domain element.

Let a function $\mathrm{f}: \mathrm{A}\rightarrow B$ is defined, then f is said to be invertible if there exists a function $\mathrm{g}: \mathrm{B}\rightarrow A$ in such a way that if we operate $\mathrm{f}\{\mathrm{g}(\mathrm{x})\}$ or $\mathrm{g}\{\mathrm{f}(\mathrm{x})\}$ we get the starting point or value.

Example:

Show that the function $f: X\rightarrow Y$, such that $f(x)=30 x+7$,
For all $x, y \in N$ is invertible.

Solution:

Let $y \in N \rightarrow y=f(x)=30 x+2$ for $x \in N$

$
x=(y-7) / 30
$

If we define $h: Y \rightarrow X$ by $h(y)=\frac{(y-2)}{30}$

Again $h \circ f(x)=h[f(x)]=h\{30 x+2\}=\frac{30(y-2)}{30+2=x}$

And $f \circ h(y)=f[h(y)]=f(\frac{(y-2)}{30})=\frac{30(y-2)}{30+2=y}$

Hence $f$ is an invertible function and $h$ is the inverse of $f$.

Properties of the Inverse of One to One Function

Here are the properties of the inverse of one to one function:

• - The function $f$ has an inverse function if and only if $f$ is a one to one function.
• If $f$ and $g$ are inverses of each other then the domain of $f$ is equal to the range of $g$ and the range of $g$ is equal to the domain of $f$.
• If $f$ and $g$ are inverses of each other then their graphs will make reflections of each other on the line $y=x$.
• If the point ( $c, d$ ) is on the graph of $f$ then point $(d, c)$ is on the graph of $f^{-1}$
• If the functions g and f are inverses of each other then, both these functions can be considered as one to one functions.
• If $f$ and $g$ are inverses of each other if and only if ( $f \circ g$ ) $(x)=x, x$ in the domain of $g$ and $(g \circ f)(x)=x, x$ in the domain of $f$. Here it is the composition function that has '$f$' composed with '$g$'.

Important Points on One-to-One Function

• One to one functions are functions in which there are equal numbers of items in the domain and in the range, or one can only be paired with another item.
• It helps us to understand various types of equations and find out inverses.
• One can easily determine if a function is one to one geometrically and algebraically as well.

One-to-One vs Many-One Function

One-to-One Function Real Life Examples

One-to-one function examples in real life includes

• Phone number assigned to a individual
• Exam number of a student
• Bank account number of a bank holder and so on.

Recommended Video Based on One-to-One Function

One-to-One Function Example Problems with Solutions

Example1: The function
$f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]_{\text {defined as }} f(x)=\frac{x}{1+x^2}$, is
1) injective but not surjective
2) surjective but not injective
3) neither injective nor surjective
4) invertible

Solution:

$
\begin{aligned}
& f(x)=\frac{x}{1+x^2} \\
& f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right] \\
& f^{\prime}(x)=\frac{\left(1+x^2\right) \times 1-x \times 2 x}{\left(1+x^2\right)^2}=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2} \\
& \therefore \text { So that } \frac{-\left(x^2-1\right)}{\left(x^2+1\right)^2}
\end{aligned}
$

So that it is not strictly increasing or decreasing function.
So that it is not one-one.
So, the given function is surjective but not injective.
Hence, the answer is the option 2.

Example 2: Which of the following functions are one - one functions?
1) $f(x)=x^2$
2) $f(x)=x^4$
3) $f(x)=\cos x$
4) $f(x)=\sqrt{x}$

Solution:
A line parallel to the $x$-axis cuts the curve of one-one function at at most one point.

$
f(x)=\sqrt{x}
$
Clearly, this function is one - one function.
Hence, the answer is the option (4).

Example 3: Which of the following functions are one-one functions?
1) $\sin (\cos x)$
2) $\cos (\sin x)$
3) $\sin (\tan x)$
4) None of these

Solution:
As we learned
In the case of composite functions,
If both $f(x)$ and $g(x)$ are one-one functions, then fog $(x)$ and $g \circ f(x)$ are both one-one functions.
Hence, the answer is the option 4 .

Example 4: Let $A=x \in \mathrm{R}: \mathrm{x}$ is not a positive integer. Define a function $f: A \rightarrow R_{\text {as }}$ $f(x)=\frac{2 x}{x-1}$ then f is:
1) injective but not surjective
2) neither injective nor surjective
3) not injective
4) surjective but not injective

Solution:
One - One or Injective functions -
A function in which every element of the range of function corresponds to exactly one element.
- wherein

A line parallel to the $x$-axis cuts the curve at most one point.

$
f(x)=\frac{2 x}{x-1}
$
This can be written as

$
\begin{aligned}
f(x) & =2\left(1+\frac{1}{x-1}\right) \\
f^{\prime}(x) & =-\frac{2}{(x-1)^2}
\end{aligned}
$

$\Rightarrow \mathrm{f}$ is one-one i.e injective but not surjective.
Hence, the answer is option 1 .

Example 5: Which of the functions $\mathrm{f}(\mathrm{x})$ will be one-one functions if $f^{\prime}(x)$ is given
1) $f^{\prime}(x)=\sin x, x \equiv R$
2) $f^{\prime}(x)=e^x, x \equiv R$
3) $f^{\prime}(x)=\ln x, x \equiv R^{+}$
4) $f^{\prime}(x)=\cos x, x \equiv R$

Solution:

When $f^{\prime}=e^x>0$
Then $f(x)$ is an increasing function, so it is a one-one function.
Hence, the answer is option 2.

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