已知b>-1,c>0,函數(shù)f(x)=x+b的圖象與函數(shù)g(x)=x2+bx+c的圖象相切.
(Ⅰ)求b與c的關(guān)系式(用c表示b);
(Ⅱ)設(shè)函數(shù)F(x)=f(x)g(x)在(-∞,+∞)內(nèi)有極值點,求c的取值范圍.
【答案】
分析:(1)注意把握題目中的信息,f(x)和g(x)在同一點處具有相同的切線斜率.即f′(x
)=g′(x
)
(2)由構(gòu)造的新函數(shù)F(x)在R上有極值點,得到二次函數(shù)F′(x)有兩個零點,再將上題的結(jié)論代入可解.
解答:解:(Ⅰ)依題意,令f'(x)=g'(x),得2x+b=1,
故
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.由于
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,得(b+1)
2=4c.
∵
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,∴
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.
(Ⅱ)F(x)=f(x)g(x)=x
3+2bx
2+(b
2+c)x+bc.
F′(x)=3x
2+4bx+b
2+c.
令F'(x)=0,即3x
2+4bx+b
2+c=0.
則△=16b
2-12(b
2+c)=4(b
2-3c).
若△=0,則F'(x)=0有一個實根x
,且F'(x)的變化如下:
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于是x=x
不是函數(shù)F(x)的極值點.若△>0,
則F′(x)=0有兩個不相等的實根x
1,x
2(x
1<x
2)且F′(x)的變化如下:
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由此,x=x
1是函數(shù)F(x)的極大值點,x=x
2是函數(shù)F(x)的極小值點.
綜上所述,當(dāng)且僅當(dāng)△=0時,函數(shù)F(x)在(-∞,+∞)上有極值點.
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.
∵
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,∴
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.
解之得0<c<7-4
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或c>7+4
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.
故所求c的取值范圍是(0,7-4
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)∪(7+4
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,+∞).
點評:本題考查導(dǎo)數(shù)、切線、極值等知識及綜合運用數(shù)學(xué)知識解決問題的能力.其中三次多項式函數(shù)也是高考中對導(dǎo)數(shù)考查的常見載體.