来自李金菊的问题
【已知函数f(x)满足f(x)=f′(1)ex−1−f(0)x+12x2.(Ⅰ)求f(x)的解析式:(Ⅱ)求f(x)的单调区间.】
已知函数f(x)满足f(x)=f′(1)ex−1−f(0)x+12x2.
(Ⅰ)求f(x)的解析式:
(Ⅱ)求f(x)的单调区间.
1回答
2020-12-27 04:03
【已知函数f(x)满足f(x)=f′(1)ex−1−f(0)x+12x2.(Ⅰ)求f(x)的解析式:(Ⅱ)求f(x)的单调区间.】
已知函数f(x)满足f(x)=f′(1)ex−1−f(0)x+12x2.
(Ⅰ)求f(x)的解析式:
(Ⅱ)求f(x)的单调区间.
(I)f′(x)=f′(1)ex-1-f(0)+x,令x=1得f′(1)=f′(1)-f(0)+1,解得f(0)=1.∴f(x)=f′(1)ex−1−x+12x2.令x=0,得f′(1)=e,∴f(x)=ex−x+12x2.(II)设g(x)=f′(x)=ex-1+x,则g′(x)=e...