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 f_{1}, f_{2}, f_{3 }, f_{4 }and f_{5} 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 :
f_{1} = {(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 f_{1} 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 :
f_{2} = {(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 f_{2} is not a function.
Example 3 :
f_{3} = {(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 f_{3} 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 :
f_{4} = {(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 f_{4} 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 :
f_{5} = {(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 f_{5} is not a function.
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