Show that the function f : N → N defined by f ( x ) = 3 x - 2 is one-one but not onto.

Given function f : N → N defined by f ( x ) = 3 x - 2 . We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ N , then its a One-one function. Let x 1 , x 2 ∈ N be such that f x 1 = f x 2 ⇒ 3 x 1 - 2 = 3 x 2 - 2 ⇒ 3 x 1 = 3 x 2 ⇒ x 1 = x 2 Hence, f is One-one function. Consider any y ∈ N (co-domain of f ) Let y = f ( x ) = 3 x - 2 ⇒ y + 2 = 3 x ⇒ y + 2 3 = x Thus , for every y ∈ N (Co-domain of f ), there does not exist x ∈ N (Domain of f ) Hence, the f is not Onto.

Show that the function f : N → N defined by f ( x ) = 3 x - 2 is one-one but not onto.

Given function f : N → N defined by f ( x ) = 3 x - 2 . We know that, if f x 1 = f x 2 ⇒ x 1 = x 2 , x 1 , x 2 ∈ N , then its a One-one function. Let x 1 , x 2 ∈ N be such that f x 1 = f x 2 ⇒ 3 x 1 - 2 = 3 x 2 - 2 ⇒ 3 x 1 = 3 x 2 ⇒ x 1 = x 2 Hence, f is One-one function. Consider any y ∈ N (co-domain of f ) Let y = f ( x ) = 3 x - 2 ⇒ y + 2 = 3 x ⇒ y + 2 3 = x Thus , for every y ∈ N (Co-domain of f ), there does not exist x ∈ N (Domain of f ) Hence, the f is not Onto.

Prove that f : N → N , defined by f ( m ) = m 2 + m + 1 for all m ∈ N , is one-one but not onto.

Given that, f : N → N defined by f ( x ) = x 2 + x + 1 We know that, a function f : A → B is said to be a one-one function or an injection, if different elements of A have different images in B . An onto function is such that for every element in the codomain there exists an element in domain which maps to it. f ( x ) = f ( y ) , ∀ x , y ∈ N ⇒ x 2 + x + 1 = y 2 + y + 1 ⇒ x 2 + x - y 2 - y = 0 ⇒ ( x - y ) ( x + y ) + x - y = 0 ⇒ ( x - y ) ( x + y + 1 ) = 0 ⇒ x - y = 0 [ x + y + 1 ≠ 0 ∀ x , y ∈ N ] ⇒ x = y Thus f is one-one. Further f ( x ) = x 2 + x + 1 ≥ 3 ∀ x ∈ N Thus f ( x ) = 1 and 2 and have no pre images in N (domain). ∴ f is not onto. Thus f is one-one but not onto.

Show that the function f : R → R given by f ( x ) = 3 x 3 + 5 is bijective.

Given f : R → R be a function defined by f ( x ) = 3 x 3 + 5 , ∀ x ∈ R f ( x ) = f ( y ) , ∀ x , y ∈ R ⇒ 3 x 3 + 5 = 3 y 3 + 5 ⇒ x 3 = y 3 ⇒ x = y Thus, f is one-one. Let y ∈ R be any arbitrary element.Then f ( x ) = y ∴ 3 x 3 + 5 = y ⇒ x = y - 5 3 1 / 3 Since, y ∈ R ∴ y - 5 3 1 / 3 ∈ R ,such that f ( x ) = f y - 5 3 1 3 = 3 y - 5 3 + 5 - y - 5 + 5 = y Thus f is onto. Hence, f is bijective.

Let f : R → R be defined as f ( x ) = 7 x + 3 . Find the function g : R → R such that g o f = f o g = I R .

Given a function f : R → R defined by f ( x ) = 7 x + 3 ∀ x ∈ R Let y ∈ R be any arbitrary element as y ∈ R ⇒ f x = y ⇒ 7 x + 3 = y ⇒ x = y - 3 7 ∈ R Let us define a function g : R → R by g ( y ) = y - 3 7 ∀ y ∈ R Since f : R → R and g : R → R . Thus both g o f and f o g exists as D f = R f = D g = R g = R Now ( f o g ) ( y ) = f ( g ( y ) ) = f y - 3 7 = 7 y - 3 7 + 3 = y ∀ y ∈ R ∴ f o g = I R And gof ( x ) = g ( f ( x ) ) = g ( 7 x + 3 ) = 7 x + 3 - 3 7 = x ∀ x ∈ R ∴ g o f = I R Thus, g o f = I R = f o g

Let R + be the set of all non-negative real numbers. If f : R + → R + and g : R + → R + are the functions defined by f ( x ) = x 2 and g ( x ) = x . Find g o f and f o g . Are these equal functions.

Given, f : R + → R + defined by f ( x ) = x 2 ∀ x ∈ R + And g : R + → R + defined by g ( x ) = x ∀ x ∈ R + Since, g o f exists as R f ⊂ D g and be a function from R + → R + ∀ x ∈ ℝ + , ( g o f ) ( x ) = g ( f ( x ) ) = g x 2 = x 2 = | x | = x ∵ x ∈ R + ⇒ x = x , ∵ x > 0 ∴ ( gof ) ( x ) = x ∀ x ∈ R + g o f = I R + Now, R g ⊂ D f ∴ f o g exists and be a function from R + to R + . ∀ x ∈ R + , ( f o g ) ( x ) = f ( g ( x ) ) = f ( x ) = ( x ) 2 = x Therefore, f o g = g o f = I R +