分类：教育/科学-理工学科-数学
问题：高三题目17已知函数f(x)满足f(x)＝f'(1)e^(x
已知f(x)满足f(x)＝f'(1)e^(x-1)-f(0)x+(1/2)x^2.(1)求f(x)的解析式及单调区间；(2)若f(x)≥(1/2)x^2+ax+b,求(a+1)b的最大值。
回答：(1)f(x)＝f'(1)e^(x-1)-f(0)x+(1/2)x^2,f'(x)=f'(1)e^(x-1)-f(0)+x,∴f'(1)=f'(1)-f(0)+1,∴f(0)=1,f(0)=f'(1)/e,f'(e)=e,∴f(x)=e^x-x+x^2/2,f'(x)=e^x-1+x,是增，x>0时f'(x)>0,f(x)是增函数；x<0时f'(x)<0,f(x)是减函数。（2）f(x)≥(1/2)x^2+ax+b，<==>g(x)=e^x-(a+1)x-b>=0,g'(x)=e^x-(a+1),是增函数，u=a+1>0时g(x)|min=g[ln(a+1)]=(a+1)[1-ln(a+1)]-b>=0,b<=(a+1)[1-ln(a+1)],(a+1)b<=u^2*[1-lnu],记为h(u),h'(u)=2u(1-lnu)+u^2*(-1/u)=u(1-2lnu),由h'(u)=0得u=√e,h(u)<=h(√e)=e/2,u<=0时b<=1,(a+1)b<=0,∴(a+1)b的最大值=e/2.