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设a>1,f(t)=at-at在(-∞,+∞)内的驻点为t(a).问a为何值时,t(a)最小?并求出最小值.
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设a>1,f(t)=at-at在(-∞,+∞)内的驻点为t(a).问a为何值时,t(a)最小?并求出最小值.
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答案和解析
令f'(t)=atlna-a=0,得唯一驻点t(a)=1-
.
考察函数t(a)=1-
在a>1时的最小值.
令t′(a)=-
=-
=0,
得唯一驻点a=ee.
当a>ee时,t'(a)>0,函数单调递增;当a<ee时,t'(a)<0,函数单调递减,因此t(ee)=1-
为极小值,从而是最小值.
lnlna |
lna |
考察函数t(a)=1-
lnlna |
lna |
令t′(a)=-
| ||||
(lna)2 |
1-lnlna |
a(lna)2 |
得唯一驻点a=ee.
当a>ee时,t'(a)>0,函数单调递增;当a<ee时,t'(a)<0,函数单调递减,因此t(ee)=1-
1 |
e |
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