Question

已知函数f(x)=f’(1)e^(x-1)- f(0)x+1/2x^2,(1)求f(x)的解析式及单调区间.（2）若f(

Answer 1

f(0)=f'(1)/e

f'(x)=f'(1)e^(x-1)-f(0)+x

f'(1)=f'(1)-f(0)+1=f'(1)-f'(1)/e+1

解得

f'(1)=e

f(0)=1

f(x)=e^x-x+1/2 x^2

令 f'(x)=e^x+x-1=0 解得 x=0

f''(x)=e^x+1>0,f'(x)单调递增

x>0 f'(x)>0 f(x)单调递增

x