# Class 9 – Functions

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Functions
• Introduction to functions
• Types of functions
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## Study Material

 Introduction: Let A and B be two non-empty sets. Then a function ‘f ’ from set A to set B is a rule or method or correspondence which associates elements of set A to elements of set B such that. i)      All elements of set A are associated to elements in set B. ii)     An element of set A is associated to a unique element in set B. In other words, a function ‘ f ’ from a set A to set B associates  each element of set A to a unique element of set B. Terms such as “map” (or “mapping”), “correspondence” are used as synonyms for “Function”. If f  is a function from a set A to a set B, then we write or, which is read as f is a function from A to B or f maps A to B. If an element a ∈ A is associated to an element , then b is called ‘the f-image of a (or) image of a under f (or) the value of the function f at a. Also a is called the pre-image of b under the function f. We write it as: b = f (a). Example: Let  and  be two sets and let be rules associating elements A to elements of B as shown in the following figures.

We observe that

i)      f1 is not a function from set A to set B,since there is an element  whichis not associated to any element of B.

ii)     f2 is not a function from A to B because an element 4 ∈ A   is associated to two elements cand e in B.

iii)   f3  and f4 are functions from A to B, because under f3 and f4  each element in A is associated to a unique element in B.

Domain,co- domain and range of a function:
Let f : A⟶B.Then, the set A is known as the domain of f and the set B is known as the

Co-domain of f. The set of all f-imagesof elements of A is known as the range of f or image set of A under f and is denoted by f (A).

Thus, = Range of f.
Clearly,

Example: Let A = {-2, -1, 0, 1, 2} and B = { 0, 1,2, 3, 4, 5, 6 }. Consider a rule.

Then, Clearly, each element of A is associated to unique element of B. So, f : A⟶Bgiven by is a function.

Clearly, domain of f  = A = {-2,-1, 0, 1, 2} and range of f  = {0, 1, 4}.

Example: Consider a function f : N ®N defined by the rule f(x) = 2x – 3. This rule does not define a function from N to itself, because   i.e.,  is not associated to any element of N.

Number of Functions: Let A and B be two finite sets having m and n elements respectively.  Then each element of set A can be associated to any one of n elements of set B. So, total number of functions from set A to set B is equal to the number of ways of doing m jobs where each job can be done in n ways. The total number of such ways is

Example: Given A = {-1, 0, 2, 5, 6, 11}, B = {-2,-1, 0, 18, 28, 108} and  Find. Is
Solution:We have:

Weobserve that. So,.

Example: Let be given by .
Find (i)         (ii)The pre-images of 39 and 2 under f.

Solution: (i) We have

(ii)Let x be the pre-image of 39. Then,

So,pre-images of 39 are -6 and 6.

Letx be the pre-image of 2. Then,

Since,no real value of x satisfies the equation ,so, 2 does not have any pre-image in f.

The total number of functions from A to B is
The total number of relations from a set A having m elements to a set Bhaving n elements is .So, the number of relations from A to B which are not functions is

i.e.,
Example: The total number of functions from a set A = {a, b, c, d} to set B = {1, 2, 3} is

Function as a relation: Let A and B be two non-empty sets. A relation f  from A to B i.e. a sub-set of , is called a function (or a mapping ora map) from A to B if
(i) for each there exists a unique such that .
(ii)
Thus, a non-void subset of  is a function from A to B if each element of A appears in some ordered pair in f and no two ordered pairs in f  have the same first element.
If , then b is called the image of a under f  and a is called the pre-image of b under f  .

Example: Let A = {1,2,3}, B = {2,3,4} and  be three subsets of  as given below.
, ,

Then

i)       is a function from A to B.

ii)     is not a function from A to B, because has two images .

iii)   is not a function from A to B because  has no image in B.

If a function f is expressed as the set of ordered pair, the domain of f is the set of all first components of the ordered pairs of f  and the range is the set of second components of ordered pairs of f i.e.,

Domain of
Range of

Example: Expressthe following functions as sets of ordered pairs and determine their ranges:

(i)

(ii)

Solution:

i)

So, . Range of.

ii)    We have A = {1, 2,3, ……. 10}. So,

Equal Functions: Two functions f and g are said to be equal if

i)the domain of f = domain of g.

ii)the co-domain of =the co-domain of g, and

iii) for every x belonging to their common domain.

If two function f and g are equal, then we write .

Example: Let  given by  and  given by . Then, we observe that are having same domain and co-domain i.e.,  Hence,

Algebra of real valued functions: If f and are real valued functions with domains A and B respectively, then both f and are defined on  when

(i) Let   and  suppose that , we define  and  on   as  and .

Let  and c be aconstant and a function is defined on A then from the above definition
and

Thefunction  isdenoted by

(ii) Let , we define  on E by

If ,then  is not defined.

(iii) Let  and . We define  and and A by  and
for all

(iv) If , then we define  on E by for all .

Example: If and then find

(i)        (ii)        (iii)   (iv)        (v )          (vi)

(vii)       (viii)         (ix)            (x)

Solutions: Domain of f = A = {4, 5, 6}, Domain of g = B = {4, 6, 8}

Domainof

(i)

(ii)

(iii) Domain  of ,Domain of

(iv)Domain of f + 4 = A = {4,5,6}

(v) Domain of

(vi)Domain of

(vii) Domain of

(viii) Domainof

(ix) Domain of

(x) Domain of

Functional Equations: Functional equations are equations for unknown functions instead of unknown numbers. In this Synopsis, we shall try to explore how we can find the unknown function when we know that the condition is satisfied.

Functional equations in one variable: Functional equations in one variable are usually easierto solve. Although there is no definite method to solve functional equations,there are some tips.
Example: If , then find
Solution:Let . On direct substitution of y,we get. Thus

Example: If   , find

Solution:   Let , then . On substitution weget
Thus

Example: If  then find

Solution: It is equivalent to.Simplifying it,

wehave