HOW TO FIND THE DOMAIN AND RANGE FROM A RELATION

Let A and B be any two non empty sets. A function f from A to B is a rule of correspondence that assigns each element x  A to only one element y  B. We denote y = f(x) to mean y is a function of x.

The set A is called the domain of the function and set B is called the co-domain of the function. Also, y is called the image of x under f and x is called the pre image of y.

The set of all images of elements of A under f is called the range of f .

Note that the range of a function is a subset of its co-domain.

How to check whether a relation is a function? 

Let f be a relation from A to B.  The relation f to be a function, it has to satisfy the following two conditions. 

(i) All the elements of A must have images in B.

(ii) Each element of A must have only one image in B. 

Note :

1. When a relation is defined from A to B, if anyone of the element of A does not have image in B, then the relation is not a function.

2. And also, if anyone of the element of A has more than one image in B, then the relation is not a function.

Let f1, f2, f, fand f5 be the relations defined from A to B. If A = {1, 4, 9, 16} and B = {–1, 2, –3, –4, 5, 6}, Which of the following relations are functions ? In case of a function, write down its domain and range.

Example 1 :

f1 = {(1, –1), (4, 2), (9, –3), (16, –4)}

Solution :

In the above arrow diagram, each element of A has image in B and also each element of A has only one image in A.

So the relation f1 is a function.

Domain is the elements of A. 

Domain = {1, 4, 9, 16}

Range is the elements of B which have preimages in A.

Range = {-1, 2, -3, -4}

Example 2 :

f2 = {(1, –4), (1, –1), (9, –3), (16, 2)}

Solution :

In the above arrow diagram, element 4 in A does not have image in B.

So the relation f2 is not a function.

Example 3 : 

f3 = {(4, 2), (1, 2), (9, 2), (16, 2)}

Solution :

In the above arrow diagram, each element of A has image in B and also each element of A has only one image in A.

So the relation f3 is a function.

Domain is the elements of A. 

Domain = {1, 4, 9, 16}

Range is the elements of B which have preimages in A.

Range = {2, -3}

Example 4 :

f4 = {(1, 2), (4, 5), (9, –4), (16, 5)}

Solution :

In the above arrow diagram, each element of A has image in B and also each element of A has only one image in A.

So the relation f4 is a function.

Domain is the elements of A. 

Domain = {1, 4, 9, 16}

Range is the elements of B which have preimages in A.

Range = {2, -4, 5}

Example 5 :

f5 = {(1, 2), (4, -3), (4, 5), (9, –4), (16, 5)}

Solution :

In the above arrow diagram, each element of A has image in B. But the element 4 in A has more than one image in B (two images -> -3 and 5).

So the relation f5 is not a function.

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