Mathematics Composition of Functions and Invertible Function For CBSE-NCERT
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Class 12 Chapter 1 - RELATIONS AND FUNCTIONS

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)}`.
 
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