已知函數(shù)f(x)=xlnx.
(Ⅰ)求f(x)的最小值;
(Ⅱ)若對(duì)所有x≥1都有f(x)≥ax-1,求實(shí)數(shù)a的取值范圍.
【答案】
分析:(1)先求出函數(shù)的定義域,然后求導(dǎo)數(shù),根據(jù)導(dǎo)函數(shù)的正負(fù)判斷函數(shù)的單調(diào)性進(jìn)而可求出最小值.
(2)將f(x)≥ax-1在[1,+∞)上恒成立轉(zhuǎn)化為不等式
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對(duì)于x∈[1,+∞)恒成立,然后令
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,對(duì)函數(shù)g(x)進(jìn)行求導(dǎo),根據(jù)導(dǎo)函數(shù)的正負(fù)可判斷其單調(diào)性進(jìn)而求出最小值,使得a小于等于這個(gè)最小值即可.
解答:解:(Ⅰ)f(x)的定義域?yàn)椋?,+∞),f(x)的導(dǎo)數(shù)f'(x)=1+lnx.
令f'(x)>0,解得
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;令f'(x)<0,解得
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.
從而f(x)在
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單調(diào)遞減,在
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單調(diào)遞增.
所以,當(dāng)
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時(shí),f(x)取得最小值
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.
(Ⅱ)依題意,得f(x)≥ax-1在[1,+∞)上恒成立,
即不等式
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對(duì)于x∈[1,+∞)恒成立.
令
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,
則
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.
當(dāng)x>1時(shí),
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182153184925098/SYS201310241821531849250019_DA/11.png">,
故g(x)是(1,+∞)上的增函數(shù),
所以g(x)的最小值是g(1)=1,
從而a的取值范圍是(-∞,1].
點(diǎn)評(píng):本題主要考查函數(shù)的單調(diào)性與其導(dǎo)函數(shù)的正負(fù)之間的關(guān)系、根據(jù)導(dǎo)數(shù)求函數(shù)的最值.導(dǎo)數(shù)是高等數(shù)學(xué)下放到高中的內(nèi)容,是每年必考的熱點(diǎn)問(wèn)題,要給予重視.