F (0) = f (1) = 0, f (1 / 2) = 1. It is proved that there is at least one point in (0,1) such that f '(x) = 1 F (x) is continuous on [0,1] and differentiable in (0,1).

F (0) = f (1) = 0, f (1 / 2) = 1. It is proved that there is at least one point in (0,1) such that f '(x) = 1 F (x) is continuous on [0,1] and differentiable in (0,1).

According to the Lagrange theorem, according to the Lagrange theorem, the existence of the theorem of lagranglagrange theorem, the existence of the existence of the existence of the existence of the following theorem, the existence of the existence of the following theorem, the existence of the existence of the existence of the existence of the existence of the existence of the existence of the theorem, the existence of the existence of the existence of the existence of the existence of the theorem, the existence of the existence of the existence of the existence of the theorem, the existence of the existence of the existence of the theorem, the existence of the existence of the existence of the existence of the existence of the existence of the existence of the existence of the existence of the existence of the existence of the existence, the existence of LIM (△ x → x → 0) f '(0) f' (0,0) in (0,0,0) f '(0) f' (0,0) f '(0) (x) (x (x (x (x + (x + (x + (x) f' (0) f '(0) (x) (x (x + (x + \1) continuous, So there is at least one point x ∈ (ξ 1, ξ 2), that is, X ∈ (0,1), satisfying f '(ξ 1) = 2 ＞ f' (x) = 1 ＞ f '(ξ 2) = - 2