【答案】
分析:(1)把點(diǎn)P的坐標(biāo)代入f(x)中,得到a,b及c的關(guān)系式,記作①,求出f(x)的導(dǎo)函數(shù),又函數(shù)在P處的切線與直線x-3y=0垂直,得到切線的斜率為-3,所以把x=1代入導(dǎo)函數(shù)中得到導(dǎo)函數(shù)值等于-3,列出關(guān)于a,b及c的另一關(guān)系式,記作②,聯(lián)立①②,利用c表示出a與b,代入導(dǎo)函數(shù)中得到導(dǎo)函數(shù)的系數(shù)與c有關(guān),然后根據(jù)c的范圍,分c大于等于0小于
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和c大于等于
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小于1兩種情況,討論導(dǎo)函數(shù)的正負(fù)進(jìn)而得到相應(yīng)的函數(shù)的單調(diào)區(qū)間;
(2)當(dāng)a與b群毆大于0時(shí),得到導(dǎo)函數(shù)等于0時(shí)x的兩個(gè)值,根據(jù)(-∞,m),(n,+∞)是f(x)的單調(diào)遞增區(qū)間,得到m與n關(guān)于a與b的關(guān)系式,根據(jù)(1)中用c表示的a與b代入所求的式子中,得到關(guān)于c的關(guān)系式,化簡(jiǎn)后,由a與b都大于0解出c的取值范圍,利用基本不等式即可求出所求式子的范圍.
解答:解:由f(x)=ax
3+bx
2+c的圖象過(guò)點(diǎn)P(-1,2)可知:-a+b+c=2①,
又f′(x)=3ax
2+2bx,因?yàn)閒(x)點(diǎn)P處的切線與直線x-3y=0垂直,
所以f′(-1)=3a-2b=-3②,
聯(lián)立①②解得:a=1-2c,b=3-3c,
則f′(x)=3(1-2c)x
2+6(1-c)x,
(i)當(dāng)c∈[0,
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)時(shí),1-2c>0,
令f′(x)=0,解得x
1=0,x
2=-
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<0,
顯然,當(dāng)x>0或x<-
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時(shí),f′(x)>0;當(dāng)-
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<x<0時(shí),f′(x)<0,
所以當(dāng)c∈[0,
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)時(shí),f(x)的單調(diào)遞增區(qū)間是(-∞,-
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)和(0,+∞),
f(x)的單調(diào)遞減區(qū)間是(0,-
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);
(ii)當(dāng)c∈[
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,1)時(shí),f(x)的單調(diào)遞減區(qū)間是(-
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,+∞)和(-∞,0),
f(x)的單調(diào)遞增區(qū)間是(0,-
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);
(2)當(dāng)a>0,b>0時(shí),令f′(x)=3ax
2+2bx=x(3ax+2b)=0,解得:x=0或x=-
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<0,
由(-∞,m),(n,+∞)是f(x)的單調(diào)遞增區(qū)間,得到m=-
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,n=0,
又a=1-2c>0,b=3-3c>0,得到c<
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,即1-2c>0,
則n-m-2c=
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-2c=
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-2c=
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+(1-2c)≥2,當(dāng)且僅當(dāng)1-2c=
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即c=0或1時(shí)取等號(hào),
所以n-m-2c的范圍是[2,+∞).
點(diǎn)評(píng):此題考查學(xué)生會(huì)利用導(dǎo)函數(shù)的正負(fù)確定函數(shù)的單調(diào)區(qū)間,考查分類討論的數(shù)學(xué)思想,會(huì)利用基本不等式求函數(shù)的最小值,是一道中檔題.