﻿ 嚴格遞增函數(頁 1) - 高中的數學 - I：數與函數 - Math Pro 數學補給站

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### 回復 1# tsyr 的帖子

\begin{align} & f\left( 1 \right)\times f\left( f\left( 1 \right)+1 \right)=1 \\ & k\times f\left( k+1 \right)=1 \\ & f\left( k+1 \right)=\frac{1}{k} \\ & \\ & f\left( k+1 \right)\times f\left( f\left( k+1 \right)+\frac{1}{k+1} \right)=1 \\ & f\left( \frac{1}{k}+\frac{1}{k+1} \right)=\frac{1}{f\left( k+1 \right)}=k=f\left( 1 \right) \\ & \frac{1}{k}+\frac{1}{k+1}=1 \\ & k=\frac{1\pm \sqrt{5}}{2} \\ \end{align}
$$k=\frac{1+\sqrt{5}}{2}$$不合，為什麼不合，就留給您了

### 回復 1# tsyr 的帖子

$$f\left( f\left( x \right)+\frac{1}{x} \right)f\left( f\left( f\left( x \right)+\frac{1}{x} \right)+\frac{1}{f\left( x \right)+\frac{1}{x}} \right)=\frac{1}{f\left( x \right)}\cdot f\left( \frac{1}{f\left( x \right)}+\frac{1}{f\left( x \right)+\frac{1}{x}} \right)=1$$

$$f\left( \frac{1}{f\left( x \right)}+\frac{1}{f\left( x \right)+\frac{1}{x}} \right)=f\left( x \right)$$

, 整理成$$x{{\left( f\left( x \right) \right)}^{2}}-f\left( x \right)-\frac{1}{x}=0\Rightarrow f\left( x \right)=\frac{1\pm \sqrt{5}}{2x}$$
(取正號時不合)，故$$f\left( 1 \right)=\frac{1-\sqrt{5}}{2}$$

[[i] 本帖最後由 hua0127 於 2014-7-1 09:58 PM 編輯 [/i]] wow!