定義在R上的函數(shù)f(x)滿足f(x+2)=2f(x),當(dāng)x∈[0,2]時(shí),f(x)=x2-2x,則當(dāng)x∈[-4,-2]時(shí),函數(shù)f(x)的最小值為________.
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分析:定義在R上的函數(shù)f(x)滿足f(x+2)=3f(x),可得出f(x-2)=13f(x),由此關(guān)系求出求出x∈[-4,-2]上的解析式,再配方求其最值.
解答:由題意定義在R上的函數(shù)f(x)滿足f(x+2)=2f(x),
任取x∈[-4,-2],則f(x)=
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f(x+2)=
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f(x+4),
由于x+4∈[0,2],當(dāng)x∈[0,2]時(shí),f(x)=x
2-2x,
故f(x)=
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f(x+2)=
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f(x+4)=
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[(x+4)
2-2(x+4)]=
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(x
2+6x+8)=
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[(x+3)
2-1],x∈[-4,-2]
當(dāng)x=-3時(shí),f(x)的最小值是-
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.
故答案為:-
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.
點(diǎn)評(píng):本題考查函數(shù)的最值及其幾何意義,解題的關(guān)鍵是正確正解定義在R上的函數(shù)f(x)滿足f(x+2)=2f(x),且由此關(guān)系求出x∈[-4,-2]上的解析式,做題時(shí)要善于利用恒恒等式.