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:

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:

Arrow Diagram: The function in the above Example can be represented as follows. 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: $f=\{(1,\ 2), \ (2,\ 4),\ (3,\ 6)\}$

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.

Example 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: $f(1)=1,\ f(2)=4,\ f(3)=9, \ f(4)=16,\ f(5)=25$

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 $f(0), \ f(1), \ f(2),\ f(3)$

Solution:

Substitute corresponding values of $f(x)$  in the function. We get $f(0)=-3,\ f(1)=1,\ f(2)=5\ and\ f(3)=9$