Function
Let A and B be any two non - empty sets. Then a
relation form set A to set B is said to be a function if every element of the
first set has a unique element in second set.
Or, if f (x, y): x ϵ A and y ϵ B), f is a function.
Types of Function
One to one Function
A function form set A to set B is said to be one to
one function if every element of the domain (i.e., the first set) has a unique
/different image in the co-domain (second set where it is mapped). Here the
function shown alongside is one to one function.
Many to one Function
A function form set A to set B is said to be many to
one function if at least two elements of the domain have a unique (same) image
in the co-domain. In the diagram, elements 1, 2 and 3 of domain A have a unique
image ‘a’ in co-domain. So, it is a many to one function.
Onto Function
A function form set A to set B is said to be onto
function if the range is equals to the co-domain i.e., no elements are left
over in the second set after mapping.
A function form set A to set B is said to be into
function if the range is not equal to the co-domain. Instead, the range is
proper subset of co-domain. In other words, if all the elements of co-domain
are not mapped with the elements of domain, then it is known as into function.
In the diagram the element ‘c’ of co-domain is left to be mapped. So, it is
into function.
Inverse of
a function
Let f: A → B be one to one onto function. Then the
function defined from B to A on the same time such that f -1 (y) = x
is called inverse of a function where x ϵ A and y ϵ B. it is denoted by f -1
and read as inverse.
In figure,
Here, in figure no. (i), inverse function does not
exist as it is many to one onto function. Similarly, figure no. (ii), inverse
function does not exist as it is one to one into function but the inverse
function exists in case of figure no. (iii) as it is one to one and onto
function.
From figure no. (iii),
f (1)
= a f
-1 (a) = 1
f (2)
= b f
-1 (b) = 2
f (3)
= c f
-1 (c) = 3
Procedural steps in finding inverse of
algebraic function
(i). Put y in
place of f (x) as f (x) = y
(ii). Interchange the roles of x and y.
(iii).Get new
y which is f -1 (x)
Composite function
Let f: A → B be a function and g: B → C be another
function. Then a function defined from A to C is known as composite function of
f and g. It is denoted by gof and read as g knot f, g circle f or f followed by
g.
This also tells us that f is the first function
operated and g is the second.
If f = {(1, 2), (2, 3), (4, 5)} find the inverse of f.
Solution,
Here, f = {(1, 2),
(2, 3), (4, 5)}
f
-1 = {(2, 1), (3, 2), (5, 4)}
Example 2:
a. If f(x) = 7x – 2,
find f -1 (x).
b. If g(x) = 
, find g -1 (2).
a. Solution,
Here, f(x) = 7x – 2
y = 7x
– 2
Interchanging the roles of x and y,
x = 7y
– 2
x + 2 =
7y
y = 
f -1 (x) =
b. Solution,
Here, g(x) =
Or, y =
Interchanging the roles of x and y,
x = 
x (2y + 1) = 5y – 3
2xy + x = 5y – 3
x + 3 = 5y – 2xy
x + 3 = (5 – 2x) y
y = 
g -1(2) = 
= 
= 5
Example 3:
If f(x) = 3x – k, and f -1 (5) = 4, find the
value of k.
Solution,
Here, f(x) = 3x –
k
And f -1 (5)
= 4 ---------------- (i)
Now, for f(x) = 3x
– k,
y = 3x
– k
Interchanging the roles of x and y,
X = 3y
– k
= y or,
f -1(x) =
Therefore, f -1 (5) = 
----------- (ii)
From equation no. (i) and (ii),
= 4
5 + k = 12
K = 12 – 5 = 7
Example 4:
Given a function f (x + 3) = 3x + 5, find f -1
(x)
Solution,
Here, f (x + 3) =
3x + 5
Or, f (x + 3)
= 3x + 9 – 4
Or, f (x + 3)
= 3 (x + 3) – 4
Or, f (a) = 3a
– 4
Now, f (x) = 3x – 4
For f -1
(x)
y = 3x
– 4
Interchanging the role of x and y,
x = 3y
– 4
= y
Therefore, f -1 (x) =
Example 5:
If f (x) = 2x and g (x) = x + 1, find gof (x), gof (-2), fog
(x) and fog (2).
Solution,
Here, f (x) = 2x
g (x) =
x + 1
For, (gof) (x),
gof (x)
= g {f (x)}
=
g {2x}
=
2x + 1
gof
(-2) = 2 (-2) + 1
=
- 4 + 1
=
- 3
Again,
fog (x) = f {g (x)}
=
f {x + 1)
=
2 (x + 1)
=
2x + 2
And, fog (2) = 2.2 + 2
=
4 + 2 = 6
Example 6:
If f (x) = 3x – 7, g (x) = 
and g -1 of (x) = f (x), find the
value of x.
Solution,
Here, f (x) = 3x –
7
g (x) =
for, g -1
g (x) =
y =
interchanging the roles of x and y,
x = 
5x = y
+ 2
Y = g -1
(x) = 5x – 2
Now,
By the question,
g -1
of (x) = f (x)
g -1
{f (x)} = f (x)
g -1
{3x – 7} = 3x – 7
5 (3x –
7) – 2 = 3x – 7
15x –
35 – 2 = 3x – 7
12x = 30
x = 
= 
Therefore, x =
Example 7:
If f (x) = 2x – 4 then prove that fof -1 (x) is
an identity function.
Solution,
Here, f (x) = 2x –
4
For f -1 (x),
f (x) =
2x – 4
y = 2x
– 4
interchanging the roles of x and y,
x = 2y
– 4
y = 
f -1
(x) =
Now, fof -1
(x) = f {f -1 (x)}
= f ( )
=
2 ( ) – 4
=
x + 4 – 4
=
x
Therefore, fof -1 (x) = x
It is an identity function.
0 Comments
Thank you for your feedback.