分析:(I)利用導(dǎo)數(shù)的運(yùn)算法則可得f′(x)=f′(1)ex-1-f(0)+x,令x=1得f′(1)=f′(1)-f(0)+1,解得f(0).
令x=0,得f′(1)=e,即可得到f(x).
(II)設(shè)g(x)=f′(x)=ex-1+x,則g′(x)=ex+1>0,可得f′(x)在R上單調(diào)遞增.進(jìn)而得到f(x)的單調(diào)性.
解答:解:(I)f′(x)=f′(1)e
x-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.
令x=0,得f′(1)=e,
∴
f(x)=ex-x+x2.
(II)設(shè)g(x)=f′(x)=e
x-1+x,
則g′(x)=e
x+1>0,∴f′(x)在R上單調(diào)遞增.
而f′(0)=0,∴當(dāng)x>0時(shí),f′(x)>0;當(dāng)x<0時(shí),f′(x)<0.
因此f(x)在區(qū)間(-∞,0)上單調(diào)遞減;在區(qū)間(0,+∞)單調(diào)遞增.
點(diǎn)評(píng):熟練掌握利用導(dǎo)數(shù)研究函數(shù)得到單調(diào)性是解題的關(guān)鍵.