已知函數(shù)f(x)=x3+ax2+x+1,a∈R.
(1)若函數(shù)f(x)在x=-1處取得極值,求a的值;
(2)在滿足(1)的條件下,探究函數(shù)f(x)零點的個數(shù);如果有零點,請指出每個零點處于哪兩個連續(xù)整數(shù)之間,并說明理由;
(3)討論函數(shù)f(x)的單調(diào)區(qū)間.
【答案】
分析:(1)先求函數(shù)f(x)的導(dǎo)函數(shù),再根據(jù)函數(shù)f(x)在x=-1處取得極值得到f'(-1)=0,解方程即可;
(2)先求出f′(x)=0的值,再討論滿足f′(x)=0的點附近的導(dǎo)數(shù)的符號的變化情況,來確定極值,發(fā)現(xiàn)極值都大于零,從而函數(shù)f(x)有零點且只有一個,又函數(shù)f(x)在[-2,-1]上連續(xù),且f(-1)=1>0,f(-2)=-1<0,所以函數(shù)f(x)的零點介于-2和-1之間.
(3)討論a的值,在函數(shù)的定義域內(nèi)解不等式fˊ(x)>0和fˊ(x)<0,求出單調(diào)區(qū)間即可.
解答:解:(1)f'(x)=3x
2+2ax+1
因為函數(shù)f(x)在x=-1處取得極值所以f'(-1)=0
解得a=2
(2)由(1)知f(x)=x
3+2x
2+x+1f'(x)=3x
2+4x+1
令f'(x)=3x
2+4x+1=0解得
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從上表可以看出
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,
所以函數(shù)f(x)有零點且只有一個
又函數(shù)f(x)在[-2,-1]上連續(xù),且f(-1)=1>0,f(-2)=-1<0,所以函數(shù)f(x)的零點介于-2和-1之間.
(3)f'(x)=3x
2+2ax+1△=4a
2-12=4(a
2-3)
當(dāng)a
2≤3,即
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時,△≤0,f'(x)≥0,所以函數(shù)f(x)在R上是增函數(shù)
當(dāng)a
2>3,即
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時,△>0,解f'(x)=0得兩根為
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,
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(顯然x
1<x
2)
當(dāng)x∈(-∞,x
1)時f'(x)>0;x∈(x
1,x
2)時f'(x)<0;x∈(x
2,+∞)時f'(x)>0
所以函數(shù)f(x)在
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,
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上是增函數(shù);
在
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上是減函數(shù)
綜上:當(dāng)
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時,函數(shù)f(x)在R上是增函數(shù);
當(dāng)
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時,函數(shù)f(x)在
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,
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上是增函數(shù);在
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上是減函數(shù)
點評:本題主要考查了利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,以及函數(shù)的零點和函數(shù)在某點取得極值的條件,屬于基礎(chǔ)題.