已知函數(shù)f(x)=xlnx
(Ⅰ)求f(x)的單調區(qū)間;
(Ⅱ)設函數(shù)f(x)的最小值為M,求與曲線y=f(x)相切且斜率為e•M(其中e為常數(shù))的切線方程.
解:(I)函數(shù)的定義域為:(0,+∞)
對函數(shù)求導可得f′(x)=lnx+1
令f′(x)>0可得
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f′(x)<0可得
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則函數(shù)的單調增區(qū)間為(
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),單調減區(qū)間為(0,
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)
(II)由(I)可知函數(shù)x=
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取得最小值,故M=f(
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)=
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,e•M=-1
設滿足條件的切點為(x
0,y
0),則根據(jù)導數(shù)的幾何意義有l(wèi)nx
0+1=-1即
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切點坐標為(
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切線方程為
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分析:(I)先求函數(shù)的定義域,然后對函數(shù)求導可得f′(x)=lnx+1分別令f′(x)>0f′(x)<0可求函數(shù)的單調增區(qū)間,單調減區(qū)間
(II)由(I)可知函數(shù)x=
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取得最小值,從而可求故M=f(
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),e•M=-1
設滿足條件的切點為(x
0,y
0),則根據(jù)導數(shù)的幾何意義可求切點坐標為(
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,進一步可得切線方程
點評:(1)想要求函數(shù)的單調區(qū)間,可先求函數(shù)的定義域,然后結合導數(shù)的符號進行求解,此類問題容易忽略對定義域的判斷
(2)利用導數(shù)的幾何意義設出切點坐標是解決該問題的關鍵