Question

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

Answer 1

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