Let f x , be a polynomial of degree 3 , such that f - 1 = 10 ,   f 1 = - 6 ,   f x , has a critical point at x = - 1 and f ' x , has a critical point at x = 1 . Then f x , has local minima at x =

Step-by-step solution

f " x = λ x - 1 Integrating both the side with respect to x ⇒ f ' x = λ x - 1 2 2 + c As f ' 1 = 0 ⇒ c = - 2 λ f ' x = λ x - 1 2 2 - 2 λ Integrating both the side with respect to x f x = λ x - 1 3 6 - 2 λ x + u f 1 = - 6 ⟹ - 2 λ + u = - 6 f - 1 = 10 ⟹ 2 λ 3 + u = 10 ∴ λ = 6 and u = 6 f ' x = 0 ⟹ x - 1 or x = 3 f ' ' 3 > 0 ∴ x = 3 , is a point of local minima.