Let a function f : 0 , 5 → R be continuous, f 1 = 3 and F be defined as: F x = ∫ 1 x t 2 g t d t , where g t = ∫ 1 t f u d u . Then for the function F x , the point x = 1 is:

Step-by-step solution

Given, F x = ∫ 1 x t 2 g t d t By Leibnitz rule we get, F ' x = x 2 g x ⇒ F ' 1 = 1 . g 1 = 0 ∵ g 1 = 0 Now F '' x = 2 x g x + x 2 g ' x ⇒ F '' x = 2 x g x + x 2 f x ∵ g ' x = f x ⇒ F '' 1 = 0 + 1 × 3 ⇒ F '' 1 = 3 F x has a local minimum at x = 1 .