解:(1)f(x)=-x
3-x
2+x+1,f′(x)=-3x
2-2x+1=-(3x-1)(x+1).
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f(x)的極大值為
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,極小值為0.
f(x)的單調(diào)增區(qū)間為(-1,
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),單調(diào)減區(qū)間為(-∞,-1),(
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).
(2)∵f(x)=-x
3-ax
2+b
2x+1,
∴f′(x)=-3x
2-2ax+b
2,又x
1,x
2為f(x)的極值點,
∴x
1,x
2為方程-3x
2-2ax+b
2=0的兩根,
x
1+x
2=-
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,x
1x
2=-
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,
∵|f(x
1)-f(x
2)|=
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|x
1-x
2|,
∴|-x
13-ax
12+b
2x
1+1+x
23+ax
23-b
2x
2-1|=
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|x
1-x
2|,
整理得|x
12+x
1x
2+x
22+a(x
1+x
2)-b
2|=
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,
即|9+
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-
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-b
2|=
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,
∴a
2+3b
2=1,∴a
2≤1.
∵k=f′(x)=-3x
2-2ax+b
2=-3x
2-2ax+
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,
f′(x)
max=f′
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=
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,
∴m>
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.
分析:(1)把a=1,b=1代入函數(shù)f(x)=-x
3-ax
2+b
2x+1,求導,分析導函數(shù)的符號,可得f(x)的單調(diào)性、極值;
(2)根據(jù)x
1,x
2為f(x)的極值點,得到x
1,x
2為方程-3x
2-2ax+b
2=0的兩根,利用韋達定理得到x
1+x
2=-
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,x
1x
2=-
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,并把|f(x
1)-f(x
2)|=
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|x
1-x
2|代入化簡得到|9+
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-
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-b
2|=
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,利用導數(shù)的幾何意義得到k=f′(x)=-3x
2-2ax+b
2=-3x
2-2ax+
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,要求函數(shù)y=f(x)的圖象上任意一點的切線斜率恒小于m,實際上是求k=f′(x)的最大值.
點評:此題是個難題.考查利用導數(shù)研究函數(shù)的單調(diào)性和極值、最值問題以及導數(shù)的幾何意義.考查了同學們觀察、推理以及創(chuàng)造性地分析問題、解決問題的能力.