- 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.

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

Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

- \(\left\{ {\left( {3,1} \right),\left( {5,1} \right),\left( {7,1} \right),\left( {11,1} \right),\left( {14,1} \right),\left( {17,1} \right)} \right\}\)
- \({\left\{ {\left( {2,1} \right),\left( {4,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {10,5} \right),\left( {12,6} \right),\left( {14,7} \right)} \right\}}\)
- \({\left\{ {\left( {1,3} \right),\left( {1,5} \right),\left( {2,5} \right)} \right\}}\)

- \(\left\{ {\left( {3,1} \right),\left( {5,1} \right),\left( {7,1} \right),\left( {11,1} \right),\left( {14,1} \right),\left( {17,1} \right)} \right\}\)

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.

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.

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

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

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

$R={(x,y),y=x^3 + 2, x \in A ,y \in N }$

Determine if this relation is a function?

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?

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.

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