Function or Mapping

Let {A \ and\  B}   be two non-empty sets. Then, a function or a mapping f from A \ to \  B is a rule which associates to each element x \in A  , a unique f(x) \in B , called the image of x . If f is a function from  A \ to \  B , then we write   f: A \rightarrow B  

 

For  f   to be a function from A \ to \  B :

(i) Every element in A  must have its image in B .

(ii) No element in A  must have more than one image

Example 1:m37x

Let A = \{1,\ 2, \  3\} \ and B\  = \{2,\ 4,\ 6,\ 8\}  

Consider the rule, f(x) = 2x, \ then\ f(1)=2, \ f(2) =4, \ f(3)=6  

Clearly, each x \in A   has a unique image in B . Hence, f   is a function from A \ to \  B .


Representation of a Function

You can represent the function in three different ways:p11

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Arrow Diagram: The function in the above Example can be represented as follows.

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Roster Method: Let  f   be a function between A \ and \  B .  The first thing is to form ordered pairs of all elements in A  that have image in B .  Then the function f is represented as the set of all such ordered pairs.

The function f   in the above example can be written as follows: f=\{(1,\ 2), \ (2,\ 4),\ (3,\ 6)\}  

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Equation Form: Let  f   be a function between A \ and \  B . If f can be represented as a rule of association, then it would take equation for. For example, in the above example, f(x) = 2x

If y \in B \ and\  x \in A, \ then\  y=2x  . Hence,  y=2x   equation represents the function f .

 

Let’s do one example for more clarification.

m38xExample 2:

Let A=\{1,2,3,4,5\} \ and\ B=\{1,4,9,16,20,25\}  

Define f=A \rightarrow B: f(x)=x^2

Represent this function by the above three methods.

Solution:

First find out the following:

f(1)=1,\ f(2)=4,\ f(3)=9, \ f(4)=16,\ f(5)=25
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Arrow Method: Now draw the diagram

Roster Method: In Roster form the function can be represented as:

f=\{(1,\ 1), (2,\ 4), (3,\ 9), (4,\ 16), (5,\ 25)\}

Equation Form: In Equation form the function can be represented as y=x^2  

 

Domain, Co-Domain and Range of a Function

Let f be a function from A \ to \  B . Then, we define:

Domain (f)=A

Co-Domain(f)=B

Range (f)  = Set of all images of A \ in \  B

 

Function as a Relation

Let A and B be two non-empty sets and R be a relation from A \ to \  B . Then R  is called a function from

A \ to \  B , if (i) domain (R) = A   and (ii) no two ordered pairs in R have the same first components.

The following example will make it more clear:

Example 3:

Let A= \{1, 2, 3, 4\} \ and \ B=\{3, 4, 5, 6, 7\}

Let R1= \{(1, 3), (2, 4), (3, 5)\},

R2=\{(1, 3), (1, 7), (2, 4), (3, 5), (4, 6)\},

R3= \{(1, 3), (2, 4), (3, 5), (4, 6)\}

Justify, which of the above relations is a function from A \ to \  B

Solution

  • Domain (R1) = \{1, 2, 3\} \ne A   Hence R1 is not a function of A \ to \  B
  • Two different ordered pairs, namely (1, 3) \ and\  (1, 7)   have the same first co-ordinates. Hence R2   is not a function of A \ to \  B
  • Domain (R3) = \{1, 2, 3, 4\} = A  . Also, no two different ordered pairs in R3 have the same first co-ordinates. Hence R3   is a function of  \ A \ to \ B  

 

Real Valued Functions

A rule f  which associates to each real number  x, a unique real number f(x) , is called a real valued function. Here, f(x)  is an expression in x.

Let’s do an example.

Example 4:

Let f(x)=4x-3,\ x\ belongs\ to\ R. Find the value of f(0), \ f(1), \ f(2),\ f(3)  m39x

 Solution:

Substitute corresponding values of  f(x)   in the function. We get

f(0)=-3,\ f(1)=1,\ f(2)=5\ and\ f(3)=9

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