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__:

__(d) A graph__:

__(e) A machine__:

__(f) A formula__:

y = f(x) = x^{2}, 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.

(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.

(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.

**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

__(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

__(iv) In a graph__:

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

__(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) = x^{2} + 4x + 5 then find
the value of

(i) f(5)

(ii) f(-2)

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

**Solution:** Here,

f(x) = x^{2} + 4x + 5

(i) When x = 5 then, f(5) = 5^{2} + 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) = 8^{2} + 4×8 + 5 = 64 + 32 +
5 = 101

When x = 9 then f(9) = 9^{2} + 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) = 4x^{2}
– 3x + 4 and g(x) = 3x^{2} – 7x + 9 such that f(x) = g(x). Find the
value of x.

**Solution:** Here,

f(x) = g(x)

or, 4x^{2} – 3x + 4 = 3x^{2} – 7x + 9

or, 4x^{2} – 3x + 4 – 3x^{2} + 7x – 9 = 0

or, x^{2} + 4x – 5 = 0

or, x^{2} + 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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