## Function

Subject: Optional Mathematics

#### Overview

A function is a special relationship where each input has a single output. This note includes all the notes of function including its types, features and the way of representation.

##### Function

A function is a special relationship where each input has a single output. Let A and B be two sets. AB is the set of all ordered pairs (a,b) such that a ∈ A and b ∈ B . Let f:AB be a non-empty subset of AB. Then f is a relation from A to B. This f is said to be a function from A to B if f associates each element of B. So, a function is a special type of relation which associates each element of the set A with one and only element of B.

#### Image and Pre-image

Let f = {(x,y) : x ∈ A, y ∈ B} be a function from A to B. The first element x of ordered pair (x,y) is called pre-image of second element y under the function f and y is called an image of x under f. We write f(x) = y to mean y is the image of x under f and is read as f of x is y or of f of x equals y. Since y is the image of x under f, so, f(x) is an image of x under f.

#### Domain, Co-domain and Range of a function

Let f be a function a set A to a set B. Then, A is called domain of f and B is called codomain of f. The set of all images of the elements of an under f is called range of f. Range of f is denoted by f(A).

If A = {1,2,3}, B = {1,4,9}

f = {(1,1),(2,4),(3,9)}, then f is the function from A to B.

Here,

Domain of f = A = {1,2,,3}

Co-domain of f = B = {1,4,9}

Range of f = f(A) = {1,4,9}

Here, range of f and co-domain of f are equal sets. i.e. f(A) = B.

Again, consider a function g which is define as follows:

Here

Domain of g = A = {1,2,3}

Co-domain of g = {1,2,3,4}

In this example, a range of g i.e. g (A) is a proper subset of co-domain of g i.e. B.

Hence, a range of a function may be a proper subset of its co-domain or equal to co-domain.

So for a function f from A to B, we write f(A)≤B.

#### Types of Function

1. Onto function
Let f be a function from A to B. Then f is said to be an onto function, if each element of B appears as the image of at least one member of A.
Here, the element of B appears as the image of element of A. So, f is onto function.
In this example, a range of f and co-domain of f are equal.So, a function f is called an onto function if its range and co-domain are equal i.e. f(A) = B.
2. Into Function
Let f be a function A to B. Then, f is said o be an into function if there is at least one element in B which does not appear the image of any into a function.
Here, range of f = {1,2,3} = A
Co-domain of f = {1,2,3,4} = B
Here, range of f is a proper subset of its co-domain.
Hence, a function f is said to be an into function if a range of f is a proper subset of its co-domain.
3. One-to-one function
Let f be a function from A to B. Then, f is called a one-to-one function if no two different elements inA have the sane imagine in B.
4. Many-to-one function
Let f be the function from A to B. Then, f is called many to one function if at least two elements of A have the same imagine in B.
5. Equal functions
Two functions f and g defined on the same domain are said to be equal if f(a) = g(a) for every element a in the domain.
6. Independent and dependent variable
Let f be a function from A to B, then the variable x which takes on values in the domain is called the independent variable and the variable y which takes on values in the range is called the dependent variable.

#### Main features of a function

Let f be a function from A to B, then

1. to every x∈A, there exists an element y∈B such that (x,y)∈f i.e. y is an image of x under f i.e. y = f(x).
2. no element of A can have more than one image in B.
3. there may be elements of B which are not associated with any element of A.
4. distinct elements of A may have the same image in B.

#### Testing of a function

A function can be tested in various ways. We usually test it through the definition, i.e.

a. The relation is said to be a function if all the elements of the domain must have an image in co-domain. Otherwise, it will not be the function.

The function can also be tested by using a test known as vertical line test. For this test, a vertical line is drawn in the graph at any point. If the vertical line cuts the graph at only one point it is a function and if it cuts at more than one point then it is not a function.

#### Representation of a function

1. Roster form
In this form, a function is represented by the set of all ordered pairs which belong to the given function.
For example, let A = {0,2,3,5,7} and
B = {0,2,4,5,7,9,10,15}, and f be the function 'is less than' form A to B. then
f = {(0,2)(2,4)(3,5)(5,7)(7,9)}
2. Set-builder form
In this form, the function is represented as {():xA, yB, x...y}, the blank is to be replaced by the rule which associates x and y.
For example, let, A = {2,5,7,8}, B = {-3,0,1,2,3,4} and f = {(2,-3),(5,0),(7,2),(8,3)} then as f in the set builder form can be written as
f = {(): x∈A,y∈B, x is greater than y}
3. By formula
In this form , a formula can be used to represent a function.
For example, the equation y = 3x + 1 represents a function where x takes all values on the set of natural numbers N and the values of y are obtained by using the above equation.
4. By table
In this form, a table can be used to represent a function.
For example, the table given below represents a function:
 x 1 2 3 4 5 6 7 y 1 4 9 16 25 36 49
5. By arrow diagram
In this form, the function is represented by drawing arrows from first components to the second components of all ordered pairs which belong to the given function.
For example, the function
f = {(-1,2),(2,0),(3,2),(4,-1),(5,0)} can be represented by the following arrow diagram.
6. By graph
A function can be represented by a graph.
For example, consider the function f(x) = 2x + 1, x is real number.
Here, someorderedpairsof f are:
 x 0 1 2 f(x) 1 3 5
##### Things to remember
• To be a relation a function all the elements of the domain must have 9image in co-domain. Otherwise, it will not be the function.
• Also, the condition for uniqueness should be a function exist. Instead of these, the function can also be tested by using a test known as vertical line test. For this test, a vertical line is drawn in the graph at any point. If the vertical line cuts the graph at only one point it is a function and if it cuts at more than one point then it is not a function.
• Main features of a function

1. to every x∈A,there exists an element y∈B such that (x,y)∈f i.e. y is an image of x under f i.e. y = f(x).
2. no element of A can have more than one image in B.
3. there may be elements of B which are not associated with any element of A.
4. distinct elements of A may have the same image in B.

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Function ##### Questions and Answers

Solution:

Here, f = {(1,-1),(2,1)(3,3)(4,5)}

Now, Domain = set of all elements of the first set.

$\therefore$ Domain = {1,2,3,4,5}

Range = set of all elements of the set that are mapped with first set.

$\therefore$ Range = {-1,1,3,5}

Solution:

Here, the function f(x) = 3x + 5.

Image = 8

So, f(x) = 3x + 5

or, 8 = 3x + 5

or, 3x = 8 - 5or, 3x = 3

$\therefore$ x = 1

So, the element 1 in the domain has the image 8 under the function.

f(x) = 3x + 5.

Solution

Here, f(x) = x + 3

So f(-1) = -1 +3 = 2

f(2) = 2 + 3 = 5

Here, 3 does not belong to the closed interval of -2 ≤ x ≤ 2. So, f(3) is not defined.

Solution:

Here, f(x) = 2x + 3

And Range = {1,3,5}

Now, f(x) = 1

or, 2x + 3 = 1

or, x = -1

Again f(x) = 3

or, 2x + 3 = 3

or, x = 0

Also, f(x) = 5

or, 2x + 3 = 5

or, x = 1

So, domain of f = {-1,0,1}

Solution:

Here f(x+3) = f(x) + f(3)

Putting x = 0, we get,

f(0+3) = f(0) + f(3)

or, f(3) = f(0) + f(3)

or, f(3) - f(3) = f(0)

or, f(0) = 0 proved.

Again , putting x = -3, we get

f(-3+3) = f(-3) + f(3)

or, f(0) = f(-3) + f(3)

or, 0 = f(-3) + f(3)

or, f(-3) = -f(3) proved.

Solution:

Here, f(x+1) = 2x - 3

or, f(x+1) = 2(x+1) - 3 - 2

= 2(x+1) - 5

So, f(x) = 2x - 5

Now, f(2) = 2×2 - 5

= 4 - 5

= -1

Solution

Given that f(x) = ax + b .

f(-3) = a(-3) + b

= -3a + b
But, f(-3) = -4

So, -3a + b = -4........(i)

Also, f(-3) = a(3) + b

= 3a + b

But f(3) = 2

$\therefore$ 3a + b = 2.........(ii)

Solving equation (i) and (ii), we get,

2b = -2

$\therefore$ b = -1

Putting the value of b in equation (i), we get

-3a + (-1) = -4

or, -3a = -4 + 1

or, -3a = -3

a = 1

$\therefore$ The value of a is 1 and b = -1.

Solution

Let x be the element in domain which has the image 0 under the function f(x) = x2 + 5x + 6.

Then, f(x) = 0

or, x2 + 5x + 6 = 0

or, x2 + 3x + 2x + 6 = 0

or, x(x+3) + 2(x+3) = 0

or, (x+3) (x+2) = 0

$\therefore$ Possible values of x are -3 and -2.

Solution

Here, f(x) = g(x)

$\therefore$ x2 + 2x + -1 = 5x + 3

or, x2 + 2x - 1 = 5x + 3

or, x2 - 3x - 4 = 0

or, x2 - 4x + x - 4 = 0

or, x(x-4) + x(x-4) =0

or, (x-4) (x+1) = 0

$\therefore$ Possible values x are 4 and -1.