令 g(x) = e^x f(x)g'(x) = e^x [ f(x) + f'(x) ]由拉格朗日中值定理存在 m属于(0,1) ,使[ g(1) - g(0) ] / (1 - 0) = g'(m)即 e*f(1) - f(0) = e^m [ f(m) +f'(m) ]即 f(m) + f'(m) = e^(-m) * [ e*f(1) - f(0) ]证...