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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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 ➝ 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→B and a subset X⊆A, the image of X under f is the set of all elements f(x) where x∈X. The image of X is denoted as f(X) and is defined as: f(X)={f(x)∣x∈X}

If we consider the entire domain A, the image of the function f, also called the range, is: Image(f)=f(A)={f(a)∣a∈A}

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 $\mathrm{f}:\mathrm{A}\to \mathrm{B}$ 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 $A=\{x_1,x_2,x_3\dots \dots \dots \dots ,x_m\}$ i.e. $m$ elements
and $B=\{y_1,y_2,y_3\dots \dots \dots \dots \dots y_n\}n$ 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)

Vertical Line Test

Functionality check using the graph:

If any line drawn parallel to the y-axis cuts the curve at most one point, then it is a function.

If any such line cuts the graph at more than one point, then it is not a function.

In Figure 1, any line parallel to the y-axis cuts the curve at one point only. Each value of x would have one and only one image (value of y), so Figure 1 is a function.

Whereas in Figure 2, a line parallel to the y-axis cuts the curve in three points. Here for x = x₁, we have three images i.e. y₁, y_2, and y₃. Therefore, figure 2 is not a function.

Summary

We concluded that function is a very important term in mathematics. Without this half of the mathematics is meaningless. The concepts of image and pre-image, one can gain a deeper understanding of the behavior of functions and their mappings between different sets.

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

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

Solution:

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

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

$y\to x-a$

f(a) f(x - a) - f(a - a) f(a + x - a) = 0 - 1 f(x) = - f(x)

Hence, the answer is the option 2.

Example 2: If $f(x)+2f\left(\frac{1}{x}\right)=3x,x\ne 0$, 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:

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

$\underset{―}{1}$
Put $x$ at the place of

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

Multiplying (i) by 2

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

$\begin{array}{rl}& 3f(x)=\frac{6}{x}-3x\\ & f(x)=\frac{2}{x}-x\\ & \text{and}f(-x)=\frac{2}{-x}+x\\ & \therefore \frac{2}{x}-x=-\frac{2}{x}+x\\ & \Rightarrow \frac{4}{x}-2x=0\\ & \Rightarrow \frac{4-2x^2}{x}=0\\ & \Rightarrow 4=2x^2\\ & \Rightarrow x^2=2\\ & x=\pm \sqrt{2},x\ne 0\end{array}$
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:

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

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

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:

$\begin{array}{rl}& f:A\to B\\ & n(A)=m\\ & n(B)=n\end{array}$

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.