Function

Function

 

Consider two sets A and B. Any non-empty subset R of the cartesian product A × B is called a relation from A to B. A special type of relation associates each element of set A with the one and only element of B. This is indeed a refinement of the concept of relation. Such a refinement is known as a function. We may define a function in the following way:


 

A function from a set A to a set B is a relation (or rule) which associates each element of A with a unique element of B.

 

Symbolically, we write

 

f : A → B

 

to mean “f is a function from A to B”. Further, an element y of B associated with an element x of A is denoted by f(x). Equivalently, we write y = f(x) which reads ‘y equals f of x’. Here f(x) is known as the image of f at x or the value of f at x.

 

The letters f, g, h, F, G, H, φ etc. are reserved for denoting functions.


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Representation of Functions

 

The various ways by which a function can be represented are mentioned below:

(i) A table

(ii) A set of ordered pairs

(iii) An arrow diagram

(iv) A graph

(v) A machine

(vi) A formula


 

Example: Let A = {1, 2, 3} and B = {1, 4, 9}. And a function assigns each number of A with its square in B. So, f(1) = 1, f(2) = 4 and f(3) = 9. Here, the domain of the definition of f is A and the range of f is f(A) = B.

 

This function f : A → B can be represented by

 

(a) A table:


(b) A set of ordered pairs:

f = {(1, 1), (2, 4), (3, 9)}

 

(c) An arrow diagram:

Arrow diagram

(d) A graph:

Graph

(e) A machine:

Machine

(f) A formula:

y = f(x) = x2, where x A and y B.

 

 

Testing of Function

 

We can test whether a given relation is a function or not by applying the following tests.


 

(i) If a function is in the form of a set of ordered pairs, examine whether the first element of all the ordered pairs are different or not. If not, it is not a function.

Functions 1. f = {(1, 2), (2, 3), (3, 4)} 2. g = {(4, 5), (6, 7), (8, 9)} 3. h = {(2, 3), (4, 3), (5, 3)} Every x-components are distinct.  Not the functions 1. R1 = {(1, 2), (1, 3), (2, 3)} 2. R2 = {(4, 5), (6, 7), (6, 9)} 3. R3 = {(a, 1), (a, 2), (a, 3)} At least two x-components or the  ordered pairs are the same.

(ii) In the arrow diagram, examine whether each element of the first set has only one image in the second set or not. If not, it is not a function.

Functions: Every element of f has unique image.  Not the functions: At least one element of R does not  have unique image.

(iii) Vertical Line Test: In the graph of a relation, it will represent a function if each vertical line cuts the graph at only one point. The graph will not represent a function if a vertical line cuts the graph at more than one point.

Graphs of vertical line test: Functions      Vertical line cuts the graph at only one  point.  Not the functions      Vertical line cuts the graph at more  than one point.

 

Domain, Range, and Co-domain of a Function

 

If f is a function from set A to set B i.e. f : A → B then set A is said to be the domain of f and B, the co-domain of f. The set of elements of B which are the images of the elements of A is known as the range of f.

 

The image of a function is the set of all output values it may produce. For a given function, the set of all elements of the domain that are mapped into a given subset of co-domain (Range) is known as the pre-images and the elements of the range are called images.


 
 

Types of Functions

 

Onto Function:

A function f is called an onto function if its range and co-domain are equal.

 

Into Function:

A function f is said to be an into function if its range is a proper subset of its co-domain.

 

One to one Function:

A function f is called one to one function if every element in the range has a single pre-image.

 

Many to one Function:

A function f is called many to one function if at least one element of the range has more than one pre-image.


 

Value of a Function

 

If f is a function and (p, q) is in f then we write f(p) = q, where f(p) is called the value of the function at p.

 

 

Worked Out Examples

 

Example 1: Let A = {1, 2, 3}, B = {2, 4, 6} and f : A → B such that f(1) = 2, f(2) = 4 and f(3) = 6. Represent the function f by

(i) Tabular form

(ii) Set of ordered pairs

(iii) Arrow diagram

(iv) Graph

(v) Formula (an equation)

 

Solution: Here,

A = {1, 2, 3}

B = {2, 4, 6}

f : A → B

(i) In a tabular form:

f(1) = 2       i.e. when x = 1 then y = 2

f(2) = 4       i.e. when x = 2 then y = 4

f(3) = 6       i.e. when x = 3 then y = 6

Table

(ii) A set of ordered pairs form:

f = {(1, 2), (2, 4), (3, 6)}

 

(iii) In an arrow diagram:

1 corresponds with f(1) = 2

2 corresponds with f(2) = 4

3 corresponds with f(3) = 6

Arrow diagram

(iv) In a graph:

The points with ordered pairs: (1, 2), (2, 4), (3, 6) are plotted.

Graph

(v) In formula (an equation):

When x = 1 then y = f(1) = 2 = 2 × 1

When x = 2 then y = f(2) = 4 = 2 × 2

When x = 3 then y = f(3) = 6 = 2 × 3

The function f can be expressed as y = 2x or f(x) = 2x.


 

Example 2: What element in the domain has image 9 under the function f(x) = 4x + 5?

 

Solution: Here,

f(x) = 4x + 5 and image of f = 9

So, f(x) = 9

or, 4x + 5 = 9

or, 4x = 9 – 5

or, 4x = 4

or, x = 4/4

or, x = 1

Thus, the required element of the domain is 1.

 

 

Example 3: If a function f is such that f(x + 5) = f(x) + f(5), x R, show that f(0) = 0 and f(-5) = - f(5).

 

Solution: Here,

f(x + 5) = f(x) + f(5)

at x = 0

f(0 + 5) = f(0) + f(5)

or, f(5) = f(0) + f(5)

or, f(5) – f(5) = f(0)

or, 0 = f(0)

f(0) = 0

 

Again, at x = -5

f(-5 + 5) = f(-5) + f(5)

or, f(0) = f(-5) + f(5)

or, 0 = f(-5) + f(5)

or, -f(5) = f(-5)

f(-5) = -f(5) proved.

 

 

Example 4: A function f is defined by f(x) = x2 + 4x + 5 then find the value of

(i) f(5)

(ii) f(-2)

(iii) f(8) + f(9)

 

Solution: Here,

f(x) = x2 + 4x + 5

(i) When x = 5 then, f(5) = 52 + 4×5 + 5 = 25 + 20 + 5 = 50

(ii) When x = -2 then f(-2) = (-2)2 + 4 × (-2) + 5 = 4 – 8 + 5 = 1

(iii) When x = 8 then f(8) = 82 + 4×8 + 5 = 64 + 32 + 5 = 101

When x = 9 then f(9) = 92 + 4×9 + 5 = 81 + 36 + 5 = 122

Now, f(8) + f(9) = 101 + 122 = 223

 

 

Example 5: Let A = {a, b, c}, B = {5, 7, 10, 12} and define f(a) = 5, f(b) = 7, f(c) = 10. Identify whether function f : A → B is into or onto.

 

Solution: Here,

f(a) = 5, f(b) = 7 and f(c) = 10

Range = {5, 7, 10}

Co-domain = {5, 7, 10, 12}

The range is a proper subset of the co-domain. Thus the given function is an into function.


 

Example 6: Let the function f : N → N be defined by f(x) = 2x + 1

(i) Is the function f onto?

(ii) What is the range of 8?

(iii) What is the pre-image of 11?

(iv) If (k, 4k – 5) lies in the function, find the value of k.

 

Solution: Here,

(i)

Domain = {1, 2, 3, …………..}

Range = {1, 2, 3, …………….}

y = f(x) = 2x + 1

When x = 1 then y = 2×1 + 1 = 3

When x = 2 then y = 2×2 + 1 = 5

When x = 3 then y = 2×3 + 1 = 7

Range = {3, 5, 7, ………….} ≠ Co-domain of f

Thus, f is not an onto function.

 

(ii)

When x = 8 then y = 2×8 + 1 = 17

Thus, the image of 8 is 17.

 

(iii)

For pre-image of 11, f(x) = 11

or, 2x + 1 = 11

or, 2x = 11 – 1

or, x = 10/2

or, x = 5

The pre-image of 11 is 5.

 

(iv)

(k, 4k – 5) lies on f(x) = 2x + 1

So, x = k and f(x) = 4k – 5

Now, f(x) = 2x + 1

or, 4k – 5 = 2k + 1

or, 4k – 2k = 1 + 5

or, 2k = 6

or, k = 3

The value of k is 3.

 

 

Example 7: If f(x) and g(x) be two functions defined by f(x) = 4x2 – 3x + 4 and g(x) = 3x2 – 7x + 9 such that f(x) = g(x). Find the value of x.

 

Solution: Here,

f(x) = g(x)

or, 4x2 – 3x + 4 = 3x2 – 7x + 9

or, 4x2 – 3x + 4 – 3x2 + 7x – 9 = 0

or, x2 + 4x – 5 = 0

or, x2 + 5x – x – 5 = 0

or, x(x + 5) – 1(x + 5) = 0

or, (x + 5)(x – 1) = 0

Either: x + 5 = 0 or x = -5

Or: x – 1 = 0 or, x = 1

x = -5, 1

Thus, the value of x are -5 and 1.



 

 

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