A function \(f\) is a relation from a non-empty set \(A\) to a non-empty set \(B\) such that the domain of \(f\) is \(A\) and no
two distinct ordered pairs in \(f\) have the same first element.
For a relation to be a function, there must be only and exactly one \(y\) that corresponds to a given \(x\)
If \(f\) is a function from \(A\) to \(B\) and \(\left( {a,b} \right) \in f\), then \(f\left( a \right) = b\), where \(b\) is called the image of \(a\) under \(f\) and \(a\) is called the pre-image of \(b\) under\(f\).
A function is also termed as a map or a mapping
A function from A to B is denoted as f: A -> B
we often denote function as y=f(x). Here y is the function of x. x is called the independent variable and y is called the dependent variable.
Function Notation
Function are generally denoted by
a. f: A -> B . This tells that ordered pair is formed by elements of A and elements . This is called arrow notation
b. Another popular way to notate function is f(x). Here f(x) is called function(f) of x. We can express like the formula
$f(x) = x^2 +2$
$f(x) = x + 6$
We can also denote this as y=f(x)
Here y is the function of x. x is called the independent variable and y is called the dependent variable
Example 1:
Which of the following relations are functions? Give reasons. If it is a function, determine
its domain and range.
\(\left\{ {\left( {3,1} \right),\left( {5,1} \right),\left( {7,1} \right),\left( {11,1} \right),\left( {14,1} \right),\left( {17,1} \right)} \right\}\)
Since 3, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having
their unique images, this relation is a function.
\({\left\{ {\left( {2,1} \right),\left( {4,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {10,5} \right),\left( {12,6} \right),\left( {14,7} \right)} \right\}}\)
Since the same first element i.e 6 corresponds to two different images 3 and 4, this relation is not a function
\({\left\{ {\left( {1,3} \right),\left( {1,5} \right),\left( {2,5} \right)} \right\}}\)
Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5,
this relation is not a function.
Example 2:
Let P ={1,2,3,4} and Q={11,12,13,14}. Relation are defined from P to Q as R:P-> Q
a. R ={(1,11),(2,12),(3,13),(4,14)}
b. R ={(1,11),(2,12),(3,13)}
c. R ={(1,11),(2,11),(3,11), (4,11)}
d. R ={(1,11),(1,11),(3,11), (4,11),(4,14)}
Which of the following relations are functions? Give reasons. Answer
a. A function as each element in P has corresponding image in Q
b. Not a function because there is no element in Q which corresponds to element 4 in P
c. A function
d. Not a function as element has multiple images in Q
Pictorial representation of Functions
Just like relations, functions are also represented pictorially using arrow diagram
Let P ={1,2,3,4} and Q={a,b,c,d}. function is defined from P to Q as f:P-> Q
f={(1,a),(2,b),(3,c),(4,d)}
Arrow diagram for this function will be
We can identify which relations are function from arrow diagram also. Lets check out some example
The above is not a function
The above is a function
The above is not a function
Real valued function and Real Functions
A function which has either R or one of its subsets as its range is called a real valued function. Further, if its domain is also either R or a subset of R, it is called a real function.
Example
1. Function f: R -> R defined by f(x) =$2x+1$ for each $x \in R$ is a real function
2. Function f:N -> N defined by f (x) = $3x + 1$ for each $x \in N$ is a real function
Examples of Functions in Maths
1. Let A ={1,2,3,4}. A relation R is defined from X to N
$R={(x,y),y=x^3 + 2, x \in A ,y \in N }$
Determine if this relation is a function? Solution
Here $R={(x,y),y=x^3 + 2, x \in A ,y \in N }$
Now x takes the values as 1,2,3,4
Therefore for x= 1 y = 1 +2=3
for x=2 , y=8+2 =10
for x=3,y=27+2=29
for x=4, y=64+2=66
So relation can be written as
R={(1,3),(2,10),(3,29),(4,66)}
Here We can see that each element of A is the first element of some ordered pair in the relation and no element in A has multiple images in R
Hence it is a function
2.Let N be the set of natural numbers and the relation R be defined on N
$R = {(x, y) : y = 2x, x, y \in N}$
Is this relation a function? Solution
Here The domain of R is the set of natural numbers N. The co-domain is also N.
The range is the set of even natural numbers.
Since every natural number n has one and only one image, this relation is a function.
