Here we will discuss about domain, codomain 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 codomain of the function ‘f’
● Set of all fimages of all the elements of A is known as the range of f. Thus, range of f is denoted by
f(A).
Note:
Range ∈ codomain
Example on Domain, codomain 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) = x²  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) = x²  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)
figure (ii)
figure (iii)
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 fimages 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 nonempty 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 fimage of x is denoted by f (x) and so we write y = f (x). The element x is called the preimage 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, codomain 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 R₁ = {(a, c) (b, d) (c, e)}
R₂ = {(a, c) (a, g) (b, d) (c, e) (d, f)}
R₃ = {(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 R₁ {a, b, c} ≠ A
Therefore, R₁ is not a function from A to B.
(ii) Two different ordered pairs (a, c) (a, g) have the same first component.
Therefore, R₂ is not a function from A → B.
(iii) Domain R₃ = {a, b, c, d} = A and not two different ordered pair have same first component.
Therefore, R₃ is a function from A to B.
● Relations and Mapping
Domain and Range of a Relation
Domain Codomain and Range of Function
● Relations and Mapping  Worksheets
Worksheet on Functions or Mapping
8th Grade Math Practice
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