Functions: Definition, Domain, Codomain and Range of Function and Number of Functions

A function is basically a special type of Relation.We can visualize a function as a machine, which takes some valid inputs(domain) and generates some outputs(range) corresponding to each valid input.

There are many terms such as ‘map’ or ‘mapping’ used to denote a function.Let us now try to understand “ functions” in detail in this article.

Table of contents

What is a Function?
Domain of a Function
Codomain of a Function
Range of a Function
Domain & Range of some standard functions
Number of Functions
Practice Problems
FAQs

What is a Function?

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image (unique image) in set B and we write f: A B

In other words, 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.

Mathematically, if f: A B and (a, b)f, then f (a) = b, where b is called the image of a under f and a is called the preimage of b under f.

Note: Method to identify if a relation is a function:

Every element in set A should be uniquely associated with some element in set B

Vertical Line Test(Graphical approach),any straight line drawn parallel to y-axis should intersect the graph only once.

Example: Consider these relations represented by the following arrow diagrams.

For R1 the elements 3 & 5 of set A do not have an image in set B. Therefore, the given relation R1 is not a function.For R2 all the elements in A have unique image in B.Therefore, R2 is a function.

Domain of a Function

The set of all the points where the given function is defined is the Domain of the function, denoted by D(f) is the set of all those reals x for which f(x) is meaningful.

Let f:AB, then the set A is known as the domain of f . It is denoted by D(f)=A

Note: For two functions f and g

D(f+g)=D(f)D(g)
D(f-g)=D(f)D(g)
D(fg)=D(f)D(g)
$D (\frac{f}{g})$ =D(f)D(g)-{xD(g) : g(x)=0}

Codomain of a Function

Let f:AB, then the set B is known as the co-domain of f . It is denoted by C(f)=B.

Range of a Function

What actually comes out of a function as an output is known as the range (collection of images).

Let f:AB. Then the set of all the images of elements of A under f is known as the range of f. It is denoted by R(f)={f(x):xA}

Note: Range Codomain

Example: Let f:AB, A= {1, 2, 3} & B={3, 4, 5, 6, 7, 8, 9} such that f(x) = 2x+1, then find the domain, codomain, and range of f.

Solution:

Domain of the function, D(f) =A= {1, 2, 3}

Codomain of the function, C(f) =B= {3, 4, 5, 6, 7, 8, 9}

For Range, we need to find the output corresponding to each element in the Domain of f

f(1) = 2(1)+1 = 3

f(2) = 2(2)+1 = 5

f(3) = 2(3)+1 = 7

So, Range of the function, R(f) = {3, 5, 7}.

Domain & Range of some standard functions

Function	Domain	Range
$a^{x}, (a > 0, a \neq 1)$		$(0, \infty)$
$e^{x}$		$(0, \infty)$
${l o g}_{a} x, a > 0, a \neq 1$	$(0, \infty)$
\|x\|		$(0, \infty)$
[x]
{x}		[0,1)
$\sqrt{x}$	$[0, \infty)$	$[0, \infty)$
sin x		[-1,1]
cos x		[-1,1]
tan x	$<math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">ℝ</mi><mo>-</mo><mfenced open="{" close="}"><mrow><mfenced><mrow><mn>2</mn><mi>n</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>:</mo><mi>n</mi><mo>∈</mo><mi mathvariant="normal">ℤ</mi></mrow></mfenced></math>$
cosec x		-(-1,1)
sec x	$<math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">ℝ</mi><mo>-</mo><mfenced open="{" close="}"><mrow><mfenced><mrow><mn>2</mn><mi>n</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>:</mo><mi>n</mi><mo>∈</mo><mi mathvariant="normal">ℤ</mi></mrow></mfenced></math>$	-(-1,1)
cot x

Number of Functions

Let f:AB such that n(A)=p & n(B)=q.

Then, Total number of possible functions from set A to set B is qp.

Example:Let f:AB where A= {2,5,9} and a set B={3,8}

Here, n(A) =3 & n(B)=2

Therefore, the total number of functions possible from A to B=23=8.

Practice Problems

Example:Consider the following arrow diagrams and check whether they represent a function or not.

(a) (b)

Solution

(a) From the arrow diagram,we have (5,m), (1,l), (6,n) as ordered pairs.

We can clearly observe that the element 2 of set A is not related to any element of B. Hence, it is not a function

(b) From the arrow diagram, we have (a,d), (b,e), (c,e), (d,f) as the ordered pairs.

Therefore, it represents a function as all the elements of set A are uniquely related to the elements of set B.(image of two or more elements can be the same).

Example:Which of the following graphs represents a function?

(a) (b)

Solution:

If we draw vertical lines parallel to y-axis (Vertical line test), the graph in part (c) will have more than one intersection point and (a), (b), (d) will have only one point of intersection. So, (a), (b), (d) represents a function, while (c) is not a function.

Example: Find the domain of $f (x) = \frac{2}{\sqrt{9 - x^{2}}} l o g (x^{3} - x) .$ .

Solution

For domain of numerator i.e. $2 l o g (x^{3} - x)$ we have,

$x^{3} - x > 0 \Rightarrow x (x^{2} - 1) > 0 \Rightarrow x (x - 1) (x + 1) > 0 \Rightarrow x \in (- 1, 0) \cup (1, \infty)$

For domain of denominator i.e. $\sqrt{9 - x^{2}}$ we have,

$9 - x^{2} \geq 0$

$\Rightarrow (x - 3) (x + 3) \leq 0$

$\Rightarrow x \in [- 3, 3]$

For values of x where the denominator becomes zero we have,

$9 - x^{2} = 0$

$\Rightarrow x = \pm 3$

Domain (f)= {Domain of Numerator Domain of Denominator }- {Values of x where Denominator is zero}

Therefore, Domain $(f) = (- 1, 0) \cup (1, \infty) \cap (- 3, 3)$

$= (- 1, 0) \cup (1, 3) .$

Example: Find the domain and range of the function $f (x) = \frac{2}{x + 1}$

Solution:

f(x) is defined for all xR except for the values of x where the denominator becomes zero. Hence, the domain of f is -{-1}.

For range, let $y = f (x) = \frac{2}{x + 1}$

$\Rightarrow y (x + 1) = 2 \Rightarrow (x + 1) = \frac{2}{y} \Rightarrow x = \frac{2}{y} - 1$

For x to be defined, clearly y0. Hence,range of f is -{0}.

Example: Find the domain of the function $f (x) = \frac{1}{| x^{2} - 4 |} .$

Solution:

f(x) is defined for all xR except for the values of x where the denominator becomes zero.

Now for obtaining the values of x where the denominator becomes zero, we have

$(x^{2} - 4) = 0, w e g e t x = - 2, 2 .$

So, the domain of f will be -{-2,2}.

Example:Find the range of the function $f (x) = | x + 3 | - 2 .$

Solution:

We know that modulus functions always yields non negative values

$\Rightarrow | x + 3 | \geq 0 f o r a l l x \in$

$\Rightarrow | x + 3 | - 2 \geq - 2$

$\Rightarrow f (x) \geq - 2$

Therefore, the range of the function is [-2,).

Example:Find the domain and range of the function $f (x) = \sqrt{- x} .$

Solution

Let us try to get the domain and range of $f (x) = \sqrt{- x} .$ with the help of graph.

Clearly, the given function is defined for all non positive real numbers. Hence,domain is (-,0] .

Since, the graph of the function -x lies above x-axis, therefore the range is [0,)

Example: Find the Range of $y = 6^{x} + 3^{x} + 6^{- x} + 3^{- x} + 2$

Solution:

$y = 6^{x} + 3^{x} + 6^{- x} + 3^{- x} + 2$

We know A.M≥G.M

$\Rightarrow 6^{x} + 6^{- x} \geq 2 \sqrt{6^{x} . 6^{- x}} \Rightarrow 6^{x} + 6^{- x} \geq 2$ .....(i)

And similarly $3^{x} + 3^{- x} \geq 2$ .....(ii)

Now, $y = (6^{x} + 6^{- x}) + {(3}^{x} + 3^{- x}) + 2$

$\Rightarrow y_{m i n} = 2 + 2 + 2 = 6$ ( using (i) & (ii) )

Hence, y6. Therefore, the range is [6,)

FAQs

1.Is there any difference between Relation and Function?
Answer: Yes, every function is a relation but every relation is not a function.

2.How many elements can be present in the Domain and Range of a function?
Answer: There can be infinite number of elements in the domain and range of a function

3.Is there any Horizontal line Test also just like the Vertical line test?
Answer: Yes, Horizontal line test is there to check whether a function is one one or not.

4.Can we represent functions using Capital letters?
Answer: Yes, functions can be represented using capital letters.