已知f(x)=f’(1)e(x-1)次幂-f(0)x+(1/2)x² (1)求f(

（1）f'(x)=f'(1)e^(x-1)-f(0)+x f'(1)=f'(1)-f(0)+1,f(0)=1所以f(x)=f'(1)e^(x-1)-x+(1/2)x^2,f(0)=f'(1)/e=1 f'(1)=e,f(x)=e^x-x+x^2/2(2)f(x)≥(1/2)x²+ax+b,e^x-x+x^2/2≥(1/2)x²+ax+...你题目看错了吧。（1）f'(x)=f'(1)e^(x-1)-f(0)+(1/2)x²是（1）f'(x)=f'(1)e^(x-1)-f(0)x +(1/2)x²f'(x)是f(x)的导数，求导你会不会？