Functions, Image and Pre-image: Definition and Examples

Functions are one of the basic concepts in mathematics that have numerous applications in the real world. Be it mega skyscrapers or super-fast cars, their modeling requires methodical application of functions. Almost all real-world problems are formulated, interpreted, and solved using functions. Image and pre-image help in determining the domain and range of the function. The practical applications of image and pre-image are graphing functions, inverse functions, and database queries.

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This Story also Contains

1. Function
2. Image of a function
3. Pre-image of a function
4. Solved Examples Based On the Image and Pre-image of Functions

In this article, we will cover the concepts of function and its image and pre-image. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of eight questions have been asked on this concept, including one in 2016, two in 2021, three in 2022, and two in 2023.

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 $B$.

OR

$A$ and $B$ are two non-empty sets, then a relation from $A$ to $B$ is said to be a function if each element $x$ in $A$ is assigned a unique element $f(x)$ in $B$, and it is written as $f: A \rightarrow B$ and read as $f$ is mapping from $A$ to $B$.

Function

Function

Not a function

Not a function

Third one is not a function because $d$ is not related(mapped) to any element in $B$.

Fourth is not a function as element a in $A$ is mapped to more than one element in $B$.

Image of a function

The image of a function refers to the set of all output values it produces from its domain.

Given a function $f: A \rightarrow B$ and a subset $X \subseteq A$, the image of $X$ under $f$ is the set of all elements $f(x)$ where $x \in X$. The image of $X$ is denoted as $f(X)$ and is defined as: $f(X)=\{f(x) \mid x \in X\}$

If we consider the entire domain $A$, the image of the function $f$, also called the range, is: Image $(\mathrm{f})=\mathrm{f}(\mathrm{A})=\{f(\mathrm{a}) \mid \mathrm{a} \in \mathrm{A}\}$

Remarks:

1. It is about the image of a subset $CC$ of the domain of $AA$. Do not confuse it with the image of an element $x x$ from $AA$.
2. Therefore, do not merely say "the image." Be specific: the image of an element, or the image of a subset.
3. Better yet: include the notation $f(x) f(x)$ or $f(C) f(C)$ in the discussion.
4. While $f(x) f(x)$ is an element in the codomain, $f(C) f(C)$ is a subset of the codomain.
5. Perhaps, the most important thing to remember is:
If $y∈f(C)y∈f(C)$, then $y∈By∈B$, and there exists an $x∈Cx∈C$
such that $f(x)=yf(x)=y$.

Pre-image of a function

The pre-image of a function refers to the set of all input values that produce a given output value or set of output values.

Given a function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ and a subset $\mathrm{Y} \subseteq \mathrm{B}$, the pre-image of Y under f is the set of all elements $\mathrm{x} \in \mathrm{A}$ such that $\mathrm{f}(\mathrm{x}) \in \mathrm{Y}$. The pre-image of Y is denoted as $f^{-1}(Y)$ and is defined as: $f^{-1}(Y)=\{x \in A \mid f(x) \in Y\}$ If $y \in B$, the pre-image of $\{y\}_{\text {is: }}$

$f^{-1}(\{y\})=\{x \in A \mid f(x)=y\}$

If $f$ is a function from $A$ to $B$ and $( a , b)$ belongs to $f$, then $\mathrm{f}(\mathrm{a})=\mathrm{b}$, where ' $b$ ' is called the image of '$a$' under $f$ and '$a$' is called the pre-image of '$b$' under $f$.
In the ordered pair $(1,2)$. $1$ is the pre-image of $2$.

Number of functions from $A$ to $B$
Let set $A=\left\{x_1, x_2, x_3 \ldots \ldots \ldots \ldots, x_m\right\}$ i.e. $m$ elements
and $B=\left\{y_1, y_2, y_3 \ldots \ldots \ldots \ldots \ldots y_n\right\} \quad n$ elements

Total number of functions from $A$ to $B$ = $n^m$

(The proof of this formula requires the use of Permutation and Combination, so it will be covered later)

Remarks:

1. The preimage of $D D$ is a subset of the domain $A A$.
2. In particular, the preimage of $B B$ is always $A A$.
3. The key thing to remember is:

If $x \in f-1(D) x \in f-1(D)$, then $x \in A x \in A$, and $f(x) \in D f(x) \in D$.
4. It is possible that $f-1(D)=\varnothing f-1(D)=\varnothing$ for some subset DD. If this happens, ff is not onto.
5. Therefore, ff is onto if and only if $f-1(\{b\}) \neq \varnothing f-1(\{b\}) \neq \varnothing$ for every $b \in B$.

Recommended Video Related to Image and Pre-image of Functions

Solved Examples Based On the Image and Pre-image of Functions

Example 1: A real-valued function $f(x) {\text {satisfies the functional equation }} f(x-y)=f(x) f(y)-f(a-x) f(a+y)_{\text {where 'a' is a given constant and }} f(0)=1, f(2 a-x)$ is equal to
1) $f(x)$
2) $-f(x)$
3) $f(-x)$
4) $f(a)+f(a-x)$

Solution:

$\begin{aligned}
& f(x-y)=f(x) f(y)-f(a-x) f(a+y) \\
& f(0)=1, \quad f(2 a-x)=? \\
& \text { Put } x=y=0 \\
& f(0)=f(0) \times f(0)-f(a) \times f(a) \\
& 1=1-f^2(a) \\
& f^2(a)=0 \\
& f(a)=0
\end{aligned}$

Now $f(2 a-x)=f(a+a-x)=f(a-(x-a))$
Where $x \rightarrow a$

$y \rightarrow x-a$
Therefore, $f($ a) $f(x-a)-f(a-a) f(a+x-a)=0-1 \times f(x)=-f(x)$

Hence, the answer is the option 2.

Example 2: If $f(x)+2 f\left(\frac{1}{x}\right)=3 x, x \neq 0$, and $\stackrel{\text { S }}{S}=\{x \in R: f(x)=f(-x)\}$; then $\mathrm{s}:$
1) is an empty set
2) contains exactly one element
3) contains exactly two elements
4) contains more than two elements

Solution:

$f(x)+2 f\left(\frac{1}{x}\right)=3 x$

Put $x$ at the place of

$\begin{aligned}
& f\left(\frac{1}{x}\right)+2 f(x)=\frac{3}{x} \\
& 2 f\left(\frac{1}{x}\right)+f(x)=3 x
\end{aligned}$

Multiplying (i) by $2$

$\begin{aligned}
& 2 f\left(\frac{1}{x}\right)+4 f(x)=\frac{6}{x} \\
& 2 f\left(\frac{1}{x}\right)+f(x)=3 x
\end{aligned}$

$\begin{aligned}
& 3 f(x)=\frac{6}{x}-3 x \\
& f(x)=\frac{2}{x}-x \\
& \text { and } \quad f(-x)=\frac{2}{-x}+x \\
& \therefore \frac{2}{x}-x=-\frac{2}{x}+x \\
& \Rightarrow \frac{4}{x}-2 x=0 \\
& \Rightarrow \frac{4-2 x^2}{x}=0 \\
& \Rightarrow 4=2 x^2 \\
& \Rightarrow x^2=2 \\
& x= \pm \sqrt{2}, x \neq 0
\end{aligned}$
Hence, the answer is the option 3.

Example 3: Which of the following relations is not a function?

1) $\{(1,2) ;(1,3) ;(2,3) ;(1,4)\}$
2) $\{(1,2) ;(2,2) ;(3,2) ;(5,4)\}$
3) $\{(1,2) ;(3,4) ;(5,6)\}$
4) $\{(1,2) ;(3,4)\}$

Solution:
In option 1, the element $1$ has $2$ images $2$ and $3$. Hence it is not a function.
Hence, the answer is the option 1.

Example 4: if $n(A)=4$ and $n(B)=3$, then the total number of functions from $A$ to $B$ is

Solution:
As we learned,
The number of functions:

$\begin{aligned}
& f: A \rightarrow B \\
& n(A)=m \\
& n(B)=n
\end{aligned}$
Total number of functions $=n^m$

$\begin{aligned}
\text { Number of functions } & =(n(B))^{n(A)} \\
& =3^4=81
\end{aligned}$

Hence, the answer is $81$.

Example 5: If $n(A)=n(B)$ and total functions from $A$ to $B$ are $3125$, then what is the value of $n(A)+n(B)$ ?

Solution:
As we learned
The number of functions:

$\begin{aligned}
& f: A \rightarrow B \\
& n(A)=m \\
& n(B)=n
\end{aligned}$

Total number of functions $=n^m$
Since $3125=5^5$
Thus, $n(A)=n(B)=5$
Thus, $n(A)+n(B)=10$

Hence, the answer is $10$.