（1）f (x)=f'(1)e^(x-1)-f(0)x+1/2x^2令x=0,f(0)=f'(1)/ef'(x)=f'(1)e^(x-1)-f'(1)/e+x令x=1,则得f'(1)=e,f(0)=1所以f (x)=e^x-x+1/2x^2（2）令g(x)=f(x)-1/2x^2+ax+b=e^x-(a+1)x-b>=0 所以e^x >=(a+1)x+b令x=1,...