Let f be a polynomial function such that f 3 x = f ′ x . f ′ ′ x , for all x ∈ R . Then :

Step-by-step solution

Degree of f x will be 3 f ( x ) = a x 3 + b x 2 + c x + d f 3 x = 27 a x 3 + 9 b x 2 + 3 c x + d f ' ( x ) = 3 a x 2 + 2 b x + c f ′ ′ x = 6 a x + 2 b f 3 x = f ′ x   f ′ ′ x Comparing the coefficient, we get 27 a = 18 a 2 ⇒ a = 3 2 Also b = 0 ,   c = 0 ,   d = 0 f x = 3 2 x 3 , f ( 2 ) = 12 ⇒ f ′ x = 9 2 x 2 ,   f ′ ′ x = 9 x Hence, f ' ( 2 ) = 18 ,   f '' ( 2 ) = 18 Hence, f '' 2 − f ' ( 2 ) = 0