设f(x)在[0,1]上可导,且f(0)＝1,f(1)＝0.证明：至少存在一点η∈（0,1）,使

考察 g(x) = x^3 f(x)因为 g(0) = g(1) = 0,所以存在 η∈(0,1),使得：g'(η) = (g(1) - g(0)) / (1 - 0) = 0而 g'(η) = η^3 f'(η) + 3η^2 f(η) = η^2 (η f'(η) + 3 f(η))因为 η^2 ≠ 0,所以 η f'(η) + ...