The minimum value of the twice differentiable function f x = ∫ 0 x e x - t f ' t d t - x 2 - x + 1 e x , x ∈ R , is

Step-by-step solution

Given, f x = ∫ 0 x e x - t f ' t d t - x 2 - x + 1 e x ⇒ f x = e x ∫ 0 x e - t f ' t d t - x 2 - x + 1 e x ⇒ e - x f x = ∫ 0 x e - t f ' t d t - x 2 - x + 1 Differentiate on both side w.r.t x we get, e - x f ' x + - f x e - x = e - x f ' x - 2 x + 1 ⇒ f x = e x 2 x - 1 ⇒ f ' x = e x 2 + e x 2 x - 1 ⇒ f ' x = e x 2 x + 1 . . . . . . 1 Now finding critical point we get, 2 x + 1 = 0 ⇒ x = - 1 2 Now differentiating equation 1 to check max