问题 解答题
已知函数f(x)满足f(x)=f′(1)ex-1-f(0)x+
1
2
x2

(Ⅰ)求f(x)的解析式:
(Ⅱ)求f(x)的单调区间.
答案

(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+

1
2
x2

令x=0,得f′(1)=e,

f(x)=ex-x+

1
2
x2

(II)设g(x)=f′(x)=ex-1+x,

则g′(x)=ex+1>0,∴f′(x)在R上单调递增.

而f′(0)=0,∴当x>0时,f′(x)>0;当x<0时,f′(x)<0.

因此f(x)在区间(-∞,0)上单调递减;在区间(0,+∞)单调递增.

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