问题
解答题
已知函数f(x)满足f(x)=f′(1)ex-1-f(0)x+
(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+
x2.1 2
令x=0,得f′(1)=e,
∴f(x)=ex-x+
x2.1 2
(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.