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[函数] 又是零点问题

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aishuxue Posted 2020-9-26 20:42 |Read mode
Last edited by aishuxue 2020-9-26 20:50已知$a>0$, 求证: 函数$f(x)=a\ln x+\dfrac{2}{x}-a-2$在$(0,+\infty)$上有两个零点.
我的困惑是: 在区间$(\dfrac{2}{a},+\infty)$上比较好找,在$(0,\dfrac{2}{a})$上不知道怎么找.

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kuing Posted 2020-9-26 22:07
`(0,2/a)` 上的话,分个类吧:

若 `a\leqslant1`,取 `x_1=1/e\in(0,2/a)`,有 `f(x_1)=2(e-a-1)>0`;

若 `a>1`,取 `x_2=1/(ea^2)\in(0,2/a)`,有 `f(x_2)=-2a\ln a+2ea^2-2a-2>-2a(a-1)+4a^2-2a-2=2(a^2-1)>0`。

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isee Posted 2020-9-26 23:19
回复 2# kuing

不可思议,竟然没用极限

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色k Posted 2020-9-26 23:56
Last edited by hbghlyj 2025-3-21 23:53回复 3# isee

面对现实,明知扯极限也是白扯啊……

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 Author| aishuxue Posted 2020-9-27 07:59
回复 2# kuing
取这样的数,是如何想到的

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kuing Posted 2020-9-27 14:10
回复 5# aishuxue

尝试呗,实际上我还先尝试过 1/a, 1/(ea) 之类,都不够小,再升次就够了,如果还不够,再换更高阶形式……

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 Author| aishuxue Posted 2020-9-27 22:05
回复 6# kuing


谢谢,辛苦了!

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