## Domain Co-domain and Range of Function

Here we will discuss about domain, co-domain and range of function.
Let : A → B (f be function from A to B), then

● Set A is known as the domain of the function ‘f’

● Set B is known as the co-domain of the function ‘f’

● Set of all f-images of all the elements of A is known as the range of f. Thus, range of f is denoted by
f(A).

**Note: **

Range ∈ co-domain

**Example on Domain, co-domain and range of function: **

**1. ** Which of the arrow diagrams given below represents a mapping? Give reasons to support your answer.

**Solution:**

(a) a has unique image p.

(b) has unique image q.

(c) has unique image q.

(d) has unique image r.

Thus, each element of A has a unique image in B.

Therefore, the given arrow diagram represents a mapping.

(b) In the given arrow diagram, the element ‘a’ of set A is associated with two elements, i.e., q and r of set B. So, each element of set A does not have a unique image in B.

Therefore, the given arrow diagram does not represent a mapping.

(c) The element ‘b’ of set A is not associated with any element of set B. So b ∈ A does not have any image. For a mapping from A to B, every element of set A must have a unique image in set B which is not represented by this arrow diagram. So, the given arrow diagram does not represent a mapping.

(d) a has a unique image p. b has a unique image q. c has a unique image r. Thus, each element in set A has a unique image in set B.

Therefore, the given arrow diagram represents a mapping.

**2. ** Find out if R is a mapping from A to B.

(i) Let A = {3, 4, 5} and B= {6, 7, 8, 9} and R = {(3, 6) (4, 7) (5, 8)}

**Solution:**

Since, R = {(3, 6); (4, 7); (5, 8)} then Domain (R) = {3, 4, 5} = A

We observe that no two ordered pairs in R have the same first component.

Therefore, R is a mapping from A to B.

(ii) Let A = {1, 2, 3} and B= {7, 11} and R = {(1, 7); (1, 11); (2, 11); (3, 11)}

**Solution:**

Since, R = {(1, 7); (1, 11); (2, 11); (3, 11)} then Domain (R) = {1, 2, 3} = A

But the ordered pairs (1, 7) (1, 11) have the same first component.

Therefore, R is not a mapping from A to B.

**3. ** Let A = {1, 2, 3, 4} and B = {0, 3, 6, 8, 12, 15}

Consider a rule f (x) = x^{2} - 1, x∈A, then

(a) show that f is a mapping from A to B.

(b) draw the arrow diagram to represent the mapping.

(c) represent the mapping in the roster form.

(d) write the domain and range of the mapping.

**Solution:**

Using f (x) = x^{2} - 1, x ∈ A we have

f(1) = 0,

f(2) = 3,

f(3) = 8,

f(4) = 15

We observe that every element in set A has unique image in set B.

Therefore, f is a mapping from A to B.

(b) Arrow diagram which represents the mapping is given below.

(c) Mapping can be represented in the roster form as

f = {(1, 0); (2, 3); (3, 8); (4, 15)}

(d) Domain (f) = {1, 2, 3, 4} Range (f) = {0, 3, 8, 15}

**Representation of a function by an arrow diagram: **

In this, we represent the sets by closed figures and the elements are represented by points in the closed figure.

The mapping f : A → B is represented by arrow which originates from elements of A and terminates at the elements of B.

**Some examples of functions:**

**figure (i)**
Each element of A has a unique image in B

**figure (ii)**
Two elements of A are associated with same element in B

**figure (iii)**
Each element of A has a unique image in B

**figure (iv)**
Every element of A has a unique image in B

**Note: **
• Observe in figure (i) and figure (ii), there are some elements in B which are not f-images of any elements of A.

• In figure (iii), figure (iv), two elements of A have the same image in B.

**Function as a special type of relation: **
If A and B are two non-empty sets, A relation f from A to B is called a function from A to B if every element of A (say x) has one and only one image (say y) in B. The f-image of x is denoted by f (x) and so we write y = f (x). The element x is called the pre-image of y under ‘f’.

**Real valued function of a real variable: : **
If the domain and range of a function ‘f' are subsets of R (set of real numbers), then f is said to be the real valued function of real variable or simply a real function. It may be defined as

A function f A → B is called a real valued function if B is a subset of R. If A and B are subsets of R then f is called a real function.

**More examples on domain, co-domain and range of function: **
**1.** Let N be the set of natural number if f: N → N by f (x) = 3x +2, then find f (1), f (2), f (-3), f (-4).

**Solution:**
Since for f(x) = 3x + 2

then f(1) = 3 × 1 + 2 = 3 + 2 = 5

f(2) = 3 × 2 + 2 = 6 + 2 = 8

there for f(-3) = 3 × (-3) + 2 = -9 + 2 = -7

f(-4) = 3 × -4 + 2 = -12 + 2 = -10

**2.** Let A = {a, b, c, d} and B= {c, d, e, f, g}

Let R

_{1} = {(a, c) (b, d) (c, e)}

R

_{2} = {(a, c) (a, g) (b, d) (c, e) (d, f)}

R

_{3} = {(a, c) (b, d) (c, e) (d, f)}

Justify which of the given relation is a function from A to B.

**Solution:**
We have,

(i) Domain R

_{1} {a, b, c} ≠ A

Therefore, R

_{1} is not a function from A to B.

(ii) Two different ordered pairs (a, c) (a, g) have the same first component.

Therefore, R

_{2} is not a function from A → B.

(iii) Domain R

_{3} = {a, b, c, d} = A and not two different ordered pair have same first component.

Therefore, R

_{3} is a function from A to B.

**Related Concepts**
●

Ordered Pair
●

Cartesian Product of Two Sets
●

Relation
●

Domain and Range of a Relation
●

Functions or Mapping
7th Grade Math Problems

8th Grade Math Practice

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