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Functions in Maths





3. What is Function

  • A function is a "well-behaved" relation
  • 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
Example 1:
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
  1. \(\left\{ {\left( {3,1} \right),\left( {5,1} \right),\left( {7,1} \right),\left( {11,1} \right),\left( {14,1} \right),\left( {17,1} \right)} \right\}\)
  2. \({\left\{ {\left( {2,1} \right),\left( {4,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {10,5} \right),\left( {12,6} \right),\left( {14,7} \right)} \right\}}\)
  3. \({\left\{ {\left( {1,3} \right),\left( {1,5} \right),\left( {2,5} \right)} \right\}}\)
Answer
  1. \(\left\{ {\left( {3,1} \right),\left( {5,1} \right),\left( {7,1} \right),\left( {11,1} \right),\left( {14,1} \right),\left( {17,1} \right)} \right\}\)
    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.
  2. \({\left\{ {\left( {2,1} \right),\left( {4,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {10,5} \right),\left( {12,6} \right),\left( {14,7} \right)} \right\}}\)
    Since the same first element i.e 6 corresponds to two different images 3 and 4, this relation is not a function
  3. \({\left\{ {\left( {1,3} \right),\left( {1,5} \right),\left( {2,5} \right)} \right\}}\)
    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
Arrow diagram for functions in Maths
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.

Quiz Time

Question 1 which is of the below relation is not a function.
A. R = {(2,2),(2,4),(3,3), (4,4)}
B. R = {(2,1),(3,1), (4,2)}
C. R = {(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)}
D. None of the above
Question 2Let A = {1, 2, 3,4} and B = {5, 7}. Then possible number of relation from A to B ?
A. 64
B. 256
C. 16
D. 32
Question 3 if function f : X -> R, $f (x) = x^3 $, where X = {-1, 0, 3, 9, 7}, the function expressed in ordered pair will be
A. {(-1,-1),(1,1),(3,27), (7,343),(9,729)}
B. {(-1,-1),(0,0),(3,27)}
C. {(-1,1),(0,0),(3,9), (7,343),(9,729)}
D. {(-1,-1),(0,0),(3,27), (7,343),(9,729)}
Question 4 Let f(x) =x2 ,find the value of $\frac {f(2.1) -f(2)}{2.1 -2}$
A. 4
B. .41
C. 4.1
D. .14
Question 5 Find the range of the function defined as $f(x)=\sqrt {9-x^2}$
A. [-3,3]
B. [0,3]
C.(0,3)
D.[-3,0)
Question 6Find the domain of the function $f(x) =\frac {x^2+1}{x^2-5x+4}$
A. R
B. R - {4}
C. R - {1}
D. R -{1,4}



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