Let S = { 1 , 2 , 3 } .Determine whether the functions f : S → S defined as below have inverse. Find f - 1 , if it exists. f = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) } .

Given function is f = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) } A function has inverse if it is one-one and onto. Here, f = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) } Since, each element has an unique image, f is one-one. Also, there is a corresponding element for every image, f is onto. ∴ f is inverse.Since, f is both one-one and onto. We know that, inverse of a function exists if and only if a function is bijective, domain of f - 1 = range of f and range of f - 1 = domain of f . ∴ f - 1 = { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) } .

Let S = { 1 , 2 , 3 } .Determine whether the functions f : S → S defined as below have inverse. Find f - 1 , if it exists. f = ( 1 , 2 ) ( 2 , 1 ) , ( 3 , 1 ) .

Given that, function is f = ( 1 , 2 ) ( 2 , 1 ) , ( 3 , 1 ) . We know that, inverse of a function exists if and only if a function is bijective, domain of f - 1 = range of f and range of f - 1 = domain of f . Here, f = ( 1 , 2 ) ( 2 , 1 ) , ( 3 , 1 ) Since, 2 and 3 have the same image 1 . f is not one-one. Since, f is not one-one. f - 1 does not exists.

Let S = { 1 , 2 , 3 } .Determine whether the functions f : S → S defined as below have inverse. Find f - 1 , if it exists. f = { ( 1 , 3 ) , ( 3 , 2 ) , ( 2 , 1 ) } .

Given function is f = { ( 1 , 3 ) , ( 3 , 2 ) , ( 2 , 1 ) } A function has inverse if it is one-one and onto. Here f = { ( 1 , 3 ) , ( 3 , 2 ) , ( 2 , 1 ) } Since, each element has an unique image, f is one-one. Also, there is a corresponding element for every image, f is onto. ∴ f is inverse.Since, f is both one-one and onto. ∴ f - 1 = { ( 3 , 1 ) , ( 2 , 3 ) , ( 1 , 2 ) } .

Let S = { 1 , 2 , 3 } .Determine whether the functions f : S → S defined as below have inverse. Find f - 1 , if it exists. f = { ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 ) } .

Given function is f = { ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 ) } We know that, inverse of a function exists if and only if a function is bijective, domain of f - 1 = range of f and range of f - 1 = domain of f . Here, f = { ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 ) } Since, each element has an unique image, f is one-one. Also, there is a corresponding element for every image, f is onto. ∴ f is inverse.Since, f is both one-one and onto. ∴ f - 1 = { ( 3 , 1 ) , ( 2 , 2 ) , ( 1 , 3 ) } .

Explain why the following functions f : X → Y do not have inverses: X = R - 0 , Y = R and f ( x ) = 1 x for all x ∈ X .

Given that, x = R - { 0 } , y = R f ( x ) = 1 x , ∀ x ∈ X We know that, inverse of a function exists if and only if a function is bijective i.e., one-one and onto function. Suppose f ( x ) = f ( y ) ⇒ 1 x = 1 y ⇒ x = y ∴ f is one-one. Let y = f ( x ) = 1 x ⇒ x = 1 y ⇒ f - 1 ( x ) = 1 x For every y except 0 there exist value of x , therefore the function is onto. Since, the function if one-one and onto function therefore it is bijective. Hence, inverse exist.

Explain why the following functions f : X → Y do not have inverses: x = { - 1 , 0 , 1 , 2 } , y = 2 , 3

Given that, x = { - 1 , 0 , 1 , 2 } y = { 2 , 3 } We know that, inverse of a function exists if and only if a function is bijective i.e., one-one and onto function. f ( x ) = 2 ∀ x = - 1 , 0 , 1 f ( x ) = 3 for x = 2 f - 1 ( 2 ) = - 1 , 1 , 0 Element in domain cannot have more than one range, it is not a one-one function. Hence, the given function is not bijection. ∴ f - 1 cannot be defined.

Explain why the following functions f : x → y do not have inverse: X = R = Y and f ( x ) = x 2 + 1 for all x ∈ R .

Given that, X = R = Y And f ( x ) = x 2 + 1 ∀ x ∈ R Here, 1 , - 1 ∈ R We know that, inverse of a function exists if and only if a function is bijective i.e., one-one and onto function. f ( 1 ) = 1 2 + 1 = 2 And f ( 1 ) = 1 2 + 1 = 2 ∴ Different elements of X have same images in Y ∴ f is not one-one. Thus f is not invertible. Thus f : X → Y do not have inverse.