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[函数] 求$f(x)解析式$

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reny Posted 2013-11-5 08:10 |Read mode
已知$f(x)+f\left(\dfrac{x+2016}{x+1}\right)=x$,求$f(x)$的解析式.

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longzaifei Posted 2013-11-5 09:34
已经有个帖子问这个问题。

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战巡 Posted 2013-11-5 09:39
回复 1# reny

这个烂题到底哪里来的?怎么好几个人都在问?
感觉存在严重问题,如果这不是连续函数麻烦就大了

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爪机专用 Posted 2013-11-5 11:22
就往下两个帖子就是同一问题

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 Author| reny Posted 2013-11-5 19:04
已知$f(x)+f\left(\dfrac{x+2016}{x+1}\right)=x$,求$f(x)$的解析式.
别人问的,说是一个自主招生试题,怎么做啊?

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其妙 Posted 2013-11-5 22:29
已知$f(x)+f\left(\dfrac{x+2016}{x+1}\right)=x$,求$f(x)$的解析式.
reny 发表于 2013-11-5 08:10
建议改为$f(x)+2f(\dfrac{x+2016}{x-1})=x$,可以不?

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realnumber Posted 2013-11-5 22:32
回复 5# reny

确定是原题,还是他的改编题?比如求f($\sqrt{2006}$)与f(x)解析式是两回事.

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 Author| reny Posted 2013-11-6 22:57
回复 6# 其妙
改了之后的,又怎么做呢?要一定的条件才好求?

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kuing Posted 2013-11-6 23:12
回复 8# reny

g(x)=(x+2016)/(x-1), g(g(x))=...

迭代后会有循环的理论上能做(或者无解),不循环的就说不准了……建议还是核实一下准确的原题先……

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Tesla35 Posted 2018-4-23 12:41
来源
(复旦大学自招)定义在$\mathbf{R}$上的函数$f(x)$满足$f(x)+2f\left(\frac{x+2002}{x-1}\right)=4015-x(x\neq 1)$,
则$f(2004)=$

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游客 Posted 2018-4-23 14:05
Last edited by 游客 2018-4-23 14:16f(x)+mf(y)=g(x),y=u(x)的反函数是自己,就象6楼那样,属于课本习题;
不然就象3楼那样,不知道谁乱改的。

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