f(x)=[f'(1)/e]*e^x-f(0)x+1/2(x^2)令x=0代入,得f(0)=f'(1)/ef'(1)=ef(0)令x=1,得f(1)=f'(1)-f(0)+1/2=ef(0)-f(0)+1/2原方程两边求导,得f'(x)=[f'(1)/e]*e^x-f(0)+x令x=1,得f'(1)=f'(1)-f(0)+1f(0)=1所以f'(1)=ef(1)...