解:(1)∵函數(shù)
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,∴f
′(x)=3x
2-t.
1°若t≤0,則f
′(x)≥0(不恒等于0)在[0,1]上恒成立,∴f(x)在[0,1]上單調(diào)遞增;
2°若t≥3時,∵3x
2≤3,∴f
′(x)≤0在[0,1]上恒成立,∴f(x)在[0,1]上單調(diào)遞減;
3°若0<t<3,則
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,令f
′(x)=0,解得
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,
當
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時,f
′(x)<0,∴f(x)在
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上單調(diào)遞減;
當
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時,f
′(x)>0,∴f(x)在
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上單調(diào)遞增.
(2)
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?
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,因此,只需求出當x∈[0,1],t∈R時,
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的最小值即可.
方法一:令g(x)=f(x)+
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,x∈[0,1],
而g
′(x)=f
′(x),由(1)的結論可知:
當t≤0或t≥3時,則g(x)在[0,1]上單調(diào),故g(x)
min=min{g(0),g(1)}=min{
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,
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}=0.
當0<t<3時,則
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=-
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.
∴h(t)=
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.
下面求當t∈R時,關于t的函數(shù)h(t)的最小值.
當t∈(0,1)時,h(t)=
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在(0,1)上單調(diào)遞減;
當1<t<3時,h(t)=
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,
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>0,∴h(t)在(1,3)上單調(diào)遞增.又h(t)在t=1處連續(xù),故h(t)在t∈(0,3)上的最小值是h(1)=-
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.
綜上可知:當t∈[0,1]且t∈R時,
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的最小值為
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,即得h的最小值為-m=
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.
方法2:對于給定的x∈[0,1],求關于t的函數(shù)(t∈R),
g(t)=f(x)+
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=-xt+
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+x
3=
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的最小值.
由于-x≤0,當t∈(-∞,1)時,g
′(t)≤0;由于1-x≥0,故當t∈(1,+∞)時,g
′(t)≥0.
考慮到g(t)在t=1處連續(xù),∴g(t)的最小值h(x)=x
3-x.
下面再求關于x的函數(shù)h(x)=x
3-x在x∈[0,1]時的最小值.
h
′(x)=3x
2-1,令h
′(x)=0,解得
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.
當
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時,h
′(x)<0,函數(shù)h(x)在此區(qū)間上單調(diào)遞減;當
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時,h
′(x)>0,函數(shù)h(x)在此區(qū)間上單調(diào)遞增.
故h(x)的最小值為
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.
綜上可得:當x∈(0,1)時,且t∈R.
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的最小值m=-
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,即得h的最小值為-m=
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.
分析:(1)對t分類討論,利用導數(shù)與單調(diào)性的關系即可得出;
(2)把問題正確等價轉化,通過分類討論,利用導數(shù)研究函數(shù)的單調(diào)性和最值,即可得出.
點評:熟練掌握分類討論的思想方法、利用導數(shù)研究函數(shù)單調(diào)性、極值、最值、及把問題正確等價轉化是解題的關鍵.