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函数$f(x)$的定义域为$D$，若对于任意的$x_1，x_2∈D$，当$x_1<x_2$时，都有$f(x_1)≤f(x_2)$，则称函数$f(x)$为定义域$D$上的非减函数．设函数$f(x)$在$[0,1]$上为非减函数，且满足以下三个条件：$①f(0)=0,②f(\frac{x}{3})=\frac{f(x)}{2},③f(1-x)=1-f(x)$,则$f(\frac{1}{x})=?$
$f(1)=1-f(0)=1$
$f(\frac{1}{2})=1-f(\frac{1}{2})=\frac{1}{2}$
$f(\frac{1}{3})=\frac{f(1)}{2}=\frac{1}{2}$
$f(\frac{2}{3})=1-f(\frac{1}{3})=\frac{1}{2}$
$f(\frac{2}{9})=\frac{f(\frac{2}{3})}{2}=\frac{1}{4}$
$f(\frac{1}{6})=\frac{f(\frac{1}{2})}{2}=\frac{1}{4}$
$f(\frac{1}{5})=f(\frac{2}{9})=f(\frac{1}{6})=\frac{1}{4}$
$f(\frac{1}{4})=\frac{f(\frac{3}{4})}{2}=\frac{1-f(\frac{1}{4})}{2}=\frac{1}{3}$
$f(\frac{1}{9})=\frac{f(\frac{1}{3})}{2}=\frac{1}{4}$
$f(\frac{1}{7})=f(\frac{1}{8})=f(\frac{1}{9})=\frac{1}{4}$
$f(\frac{1}{10})=\frac{f(\frac{3}{10})}{2}=\frac{f(\frac{9}{10})}{4}=\frac{1-f(\frac{1}{10})}{4}=\frac{1}{5}$
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kuing
发表于 2013-11-16 01:40
