f(x)=f’(1)e^(x-1)-f(0)x+(1/2)x² ①
f'(x)=f'(1)e^(x-1)-f(0)+x
令x=1,得 f'(1)=f'(1)-f(0)+1
得 f(0)=1
在①式中,令x=0,得
f(0)=f'(1)/e
f'(1)=e
故 f(x)=e^x-x+x²/2
由 f'(x)=e^x+x-1
令 f'(x)=0,得 e^x+x-1=0 即x=0
当 x0 f(x)单调增