If f : R → R is defined by f ( x ) = x 2 , write f - 1 25 .

Given, f : R → R is defined by f x = x 2 We need to find inverse of f Let f x = y , where x , y ∈ R ⇒ x 2 = y ⇒ x = ± y We know that if f : A → B is an invertible function then f - 1 : B → A is such that f ( x ) = y ⇔ f - 1 ( y ) = x . Hence, f - 1 y = ± y Therefore, f - 1 25 = ± 5

If f : C → C is defined by f ( x ) = x 4 . write f - 1 ( 1 )

Given, f : C → C is defined by f x = x 4 We need to first find inverse of f Let f x = y , where x , y ∈ C ⇒ x 4 = y ⇒ x = y 4 We know that if f : A → B is an invertible function then f - 1 : B → A is such that f ( x ) = y ⇔ f - 1 ( y ) = x . Hence, f - 1 y = y 4 Therefore, f - 1 1 = 1 4 Thus, we need to find the fourth roots of unity. Let z = 1 = 1 + 0 i = cos 0 + i sin 0 z 1 4 = cos 0 + i sin 0 1 4 = cos 0 + 2 k π + i sin 0 + 2 k π 1 4 where k = 0 , 1 , 2 , 3 = cos 2 k π 4 + i sin 2 k π 4 , where k = 0 , 1 , 2 , 3 For k = 0 z 1 4 = 1 For k = 1 z 1 4 = cos π 2 + i sin π 2 = i For k = 2 z 1 4 = cosπ + i sinπ = - 1 For k = 3 z = cos 6 π 4 + i sin 6 π 4 = = cos 3 π 2 + i sin 3 π 2 = - cos π 2 - i sin π 2 = 0 - i = - i Hence, f - 1 1 = 1 , - 1 , i , - i

If f : R → R is defined by f ( x ) = x 2 , find f - 1 ( - 25 )

Given, f : R → R is defined by f x = x 2 We need to first find inverse of f Let f x = y , where x , y ∈ R ⇒ x 2 = y ⇒ x = y 2 We know that if f : A → B is an invertible function then f - 1 : B → A is such that f ( x ) = y ⇔ f - 1 ( y ) = x . Hence, f - 1 y = y 2 Therefore, f - 1 - 25 = - 25 2 ∉ R . Hence, f - 1 ( - 25 ) = ϕ