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1.12 互换 \(x\) 与 \(\frac {1}{x}\) 建立方程组求函数
题目
题目：
设函数 \(f(x)\) 的定义域为 \((0,+\infty )\)，且满足
\[ 2f(x)+x^2f\left (\frac {1}{x}\right )=\frac {x^2+2x}{\sqrt {1+x^2}}, \]
则求 \(f(x)\)。
题解
思路入口：
题设中同时出现 \(f(x)\) 与 \(f\left (\frac {1}{x}\right )\)。要求的是 \(f(x)\)，但题设只给出一个关系式，未知量却可看作两个：
\[ f(x),\qquad f\left (\frac {1}{x}\right ). \]
因此自然的入口是把题设中的 \(x\) 换成 \(\frac {1}{x}\)，再得到第二个方程，联立消去 \(f\left (\frac {1}{x}\right )\)。
详细解答：
由题设，对任意 \(x>0\)，有
\[ 2f(x)+x^2f\left (\frac {1}{x}\right )=\frac {x^2+2x}{\sqrt {1+x^2}}. \qquad \text {(1)} \]
因为定义域为 \((0,+\infty )\)，当 \(x>0\) 时也有 \(\frac {1}{x}>0\)，所以可以在题设中把 \(x\) 换为 \(\frac {1}{x}\)，得到
\[ 2f\left (\frac {1}{x}\right )+\frac {1}{x^2}f(x) =\frac {\frac {1}{x^2}+\frac {2}{x}}{\sqrt {1+\frac {1}{x^2}}}. \]
右端需要化简。由于 \(x>0\)，
\[ \sqrt {1+\frac {1}{x^2}} =\sqrt {\frac {x^2+1}{x^2}} =\frac {\sqrt {1+x^2}}{x}. \]
因此
\[ \frac {\frac {1}{x^2}+\frac {2}{x}}{\sqrt {1+\frac {1}{x^2}}} =\frac {\frac {1+2x}{x^2}}{\frac {\sqrt {1+x^2}}{x}} =\frac {1+2x}{x\sqrt {1+x^2}}. \]
于是得到第二个方程
\[ 2f\left (\frac {1}{x}\right )+\frac {1}{x^2}f(x) =\frac {1+2x}{x\sqrt {1+x^2}}. \qquad \text {(2)} \]
为了消去 \(f\left (\frac {1}{x}\right )\)，将 \((2)\) 乘以 \(x^2\)，得
\[ 2x^2f\left (\frac {1}{x}\right )+f(x) =\frac {x(1+2x)}{\sqrt {1+x^2}}. \qquad \text {(3)} \]
再将 \((1)\) 乘以 \(2\)，得
\[ 4f(x)+2x^2f\left (\frac {1}{x}\right ) =\frac {2x^2+4x}{\sqrt {1+x^2}}. \qquad \text {(4)} \]
用 \((4)-(3)\)，左端中 \(f\left (\frac {1}{x}\right )\) 被消去：
\[ 3f(x) =\frac {2x^2+4x-x-2x^2}{\sqrt {1+x^2}} =\frac {3x}{\sqrt {1+x^2}}. \]
所以
\[ f(x)=\frac {x}{\sqrt {1+x^2}}\qquad (x>0). \]
答案：
\[ \boxed {f(x)=\frac {x}{\sqrt {1+x^2}}\quad (x>0)}. \]
知识点
核心知识点：
当函数关系式中同时出现 \(f(x)\) 和 \(f\left (\frac {1}{x}\right )\) 时，常用办法是把 \(x\) 换成 \(\frac {1}{x}\)，得到关于两个未知量的方程组：
\[ f(x),\qquad f\left (\frac {1}{x}\right ). \]
然后通过加减消元求出目标函数。
可迁移方法：
类似地，若关系式中出现 \(f(x)\) 与 \(f(-x)\)，常用 \(x\mapsto -x\)；若出现 \(f(x)\) 与 \(f(a-x)\)，常用 \(x\mapsto a-x\)。核心不是代换本身，而是构造“同一组未知函数值”的方程组。
易错点
易错点：
-
• 忽略定义域：本题能把 \(x\) 换成 \(\frac {1}{x}\)，依赖于 \(x>0\) 时 \(\frac {1}{x}>0\)。
-
• 根式化简漏掉正负：\(\sqrt {x^2}=|x|\)，但本题 \(x>0\)，所以 \(\sqrt {x^2}=x\)。
-
• 消元倍数选错：应让 \(f\left (\frac {1}{x}\right )\) 前的系数相同，再相减。
复盘提醒：
看到 \(f(x)\) 与 \(f\left (\frac {1}{x}\right )\) 同时出现，先写两个未知量，再做 \(x\mapsto \frac {1}{x}\) 的对偶代换，最后联立消元。