Mathematics Composition of Functions and Invertible Function For CBSE-NCERT

### Topics Covered

star Composition of Functions
star Invertible Function

### Composition of Functions

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"FUNCTION"

f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A → C given by

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ gof (x) = g(f (x)), ∀ x ∈ A.

\color{green}"Properties of composition functions:"

\color{blue}ul"Commutativity:" Generally, composition of functions is not commutative. So, we can write for two functions f and g:

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ gof != fog

Does f (g(x)) = g (f(x)) ?

Let f(x) = 2x+3 let g(x) = 3x+2

f (g(x)) = 2 (3x+2) +3 = 6x + 7
g(f(x)) = 3 (2x+3) +2 = 6x +11 Not equal

Function composition is not commutative !

\color{blue}ul"Associativity:" Composition is always associative. If we have three functions f, g and h, then

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (f\circ g)\circ h=f\circ (g\circ h)=f\circ g\circ h\quad "for" \ \"all" \ \ f,g,h

Associative Property :

 ( (fog) oh ) (x) = (fo (goh) ) (x)

= f (g (h (x) ))

f(x) = x fog (x) = x^2

g(x) = x^2 ( ( fog) oh) (x) = (x-1)^2

goh (x) = (x-1)^2 = fo (goh)

\color { maroon} ® \color{maroon} ul (" REMEMBER")

color { blue} ("if"\ \ f : A → B \ \ "and"\ \ g : B → C \ \ "are onto, then" \ \ gof : A → C\ \ "is also onto.")
Q 3164267155

Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be
functions defined as f (2) = 3, f (3) = 4, f(4) = f (5) = 5 and g (3) = g (4) = 7 and
g (5) = g (9) = 11. Find gof.

Solution:

We have gof (2) = g (f (2)) = g (3) = 7, gof (3) = g (f (3)) = g (4) = 7,
gof (4) = g (f (4)) = g (5) = 11 and gof (5) = g (5) = 11.
Q 3174267156

Find gof and fog, if f : R → R and g : R → R are given by f (x) = cos x
and g (x) = 3x^2 . Show that gof ≠ fog.

Solution:

We have gof (x) = g (f (x)) = g (cos x) = 3 (cos x)^2 = 3 cos^2 x. Similarly,
fog(x) = f (g (x)) = f (3x^2) = cos (3x^2). Note that 3cos^2 x ≠ cos 3x^2, for x = 0. Hence,
gof ≠ fog.
Q 3104367258

Show that if  f : R - { 7/5} -> R - {3/5 }  is defined by f(x) = (3x+4)/(5x -7)  and

g : R- {3/5 } -> R - {7/5 } is defined by g(x) = (7x+4)/( 5x-3) , then fog = I_A and gof =I_B where ,

A= R- { 3/5} , B =R - { 7/5} ; I_A(x) = x, AA x in A , I_B (x) = x , AA x in B are called identity

functions on sets A and B, respectively.

Solution:

We have

gof (x) = g ( (3x+4)/(5x - 7 ) ) = ( 7 ( (3x+4)/( 5x-7) ) +4 )/( 5 ( ( 3x+ 4)/(5x - 7 ) ) -3) = (21 x +28 + 20 x -28 )/(15 x +20 -15x +21) = (41x)/(41) = x

Similarly , fog (x) = f ( (7x+4)/( 5x-3) ) = ( 3 ( (7x+4)/(5x-3) ) +4)/( 5 ( ( 7x +4)/( 5x-3) ) - 7 ) = (21 x +12 +20 x -12)/(35 x +20 -35 x + 21) = (41 x)/(41) = x

Thus, gof (x) = x, ∀ x ∈ B and fog (x) = x, ∀x ∈ A, which implies that gof = I_B
and fog = I_A.
Q 3164467355

Show that if f : A → B and g : B → C are one-one, then gof : A → C is
also one-one.

Solution:

Suppose gof (x_1) = gof (x_2)

⇒ g (f (x_1)) = g(f (x_2))

⇒ f (x_1) = f (x_2), as g is one-one

⇒ x_1 = x_2, as f is one-one
Hence, gof is one-one.
Q 3154567454

Show that if f : A → B and g : B → C are onto, then gof : A → C is
also onto

Solution:

Given an arbitrary element z ∈ C, there exists a pre-image y of z under g
such that g (y) = z, since g is onto. Further, for y ∈ B, there exists an element x in A

with f (x) = y, since f is onto. Therefore, gof (x) = g (f (x)) = g (y) = z, showing that gof
is onto

### Invertible Function

\color{purple} "★ DEFINITION ALERT"

\color{fuchsia} \mathbf\ul"INVERSE FUNCTION"

\color{fuchsia} ✍️ A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that gof = I_x and fog = I_y. The function g is called the inverse of f and is denoted by f^( –1).

\color{green} ★ \color{green} \mathbf(KEY \ CONCEPT)

\color{green} ✍️ if f is invertible, then f must be one-one and onto and conversely, if f is one-one and onto, then f must be invertible.

\color{green} ✍️ This fact significantly helps for proving a function f to be invertible by showing that f is one-one and onto, specially when the actual inverse of f is not to be determined.
Q 3124667551

Let f : N → Y be a function defined as f (x) = 4x + 3, where,
Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.

Solution:

Consider an arbitrary element y of Y. By the definition of Y, y = 4x + 3,

for some x in the domain N. This shows that  x= (y-3)/4 . Define g : Y → N by

g(y) = ( y-3)/4 . Now gof = g (f(x) ) = g (4x+3) = ( 4x +3-3)/4 = x and

fog (y) = f (g (y)) = f ( ( y-3)/4) = (4 (y-3) )/4 +3 = y-3 + 3 = y . This shows that gof = I_N

and fog = I_Y, which implies that f is invertible and g is the inverse of f.
Q 3154667554

Let Y = {n^2 : n ∈ N} ⊂ N. Consider f : N → Y as f (n) = n^2. Show that
f is invertible. Find the inverse of f.

Solution:

An arbitrary element y in Y is of the form n^2, for some n ∈ N. This

implies that n = sqrt y This gives a function g : Y → N, defined by g (y) = sqrt y . Now

gof (n) = g (n^2) = sqrt (n^2) = n and  fog ( y ) = f (sqrt y ) = (sqrt y )^2 = y , which shows that

gof = I_N and fog = I_Y. Hence, f is invertible with f^(- 1 ) = g .
Q 3104767658

Let S = {1, 2, 3}. Determine whether the functions f : S → S defined as
below have inverses. Find f( –1), if it exists.
(a) f = {(1, 1), (2, 2), (3, 3)}
(b) f = {(1, 2), (2, 1), (3, 1)}
(c) f = {(1, 3), (3, 2), (2, 1)}

Solution:

(a) It is easy to see that f is one-one and onto, so that f is invertible with the inverse
f^(-1 )of f given by f^(-1) = {(1, 1), (2, 2), (3, 3)} = f.

(b) Since f (2) = f (3) = 1, f is not one-one, so that f is not invertible.

(c) It is easy to see that f is one-one and onto, so that f is invertible with
f^(-1) = {(3, 1), (2, 3), (1, 2)}.