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[函数] 函数

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hjfmhh Post time 2015-3-11 09:56 |Read mode

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战巡 Post time 2015-3-11 10:36

回复 1# hjfmhh

你这$f(x)$不就一个常函数么...
对任意$a>0$都有$f(x+a)=...$，也就是$f(x+a)=c$
然后可知$f(x)=\frac{1}{4}(2-\sqrt{2})$

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Author| hjfmhh Post time 2015-3-11 11:56

常数a

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tommywong Post time 2015-3-11 13:19

A
$f(x)\le \cfrac{1}{2}$

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战巡 Post time 2015-3-11 14:12

回复 3# hjfmhh

对任意常数$a$

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kuing Post time 2015-3-11 14:27

居然变成语文问题

……

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