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# 通过复合函数验证反函数

## 反函数组合原则

• $f\left(g\left(x\right)\right)=x$ ，在定义域$g$区间内任何$x$ 都成立
• $g\left(f\left(x\right)\right)=x$ ，在定义域$f$区间内任何 $x$ 都成立

### 例1：函数 $f$‍  与 $g$‍  互为反函数

$f\left(g\left(x\right)\right)$$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}g\left(f\left(x\right)\right)$
$\begin{array}{rl}f\left(g\left(x\right)\right)& =\frac{g\left(x\right)+1}{3}\\ \\ & =\frac{3x-1+1}{3}\\ \\ & =\frac{3x}{3}\\ \\ & =x\end{array}$$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\begin{array}{rl}g\left(f\left(x\right)\right)& =3\left(f\left(x\right)\right)-1\\ \\ & =3\left(\frac{x+1}{3}\right)-1\\ \\ & =x+1-1\\ \\ & =x\end{array}$

### 例2：函数 $f$‍  和 $g$‍  不是互为反函数

$f\left(g\left(x\right)\right)$$\phantom{\rule{2em}{0ex}}g\left(f\left(x\right)\right)$
$\begin{array}{rl}f\left(g\left(x\right)\right)& =5\left(g\left(x\right)\right)-7\\ \\ & =5\left(\frac{x}{5}+7\right)-7\\ \\ & =x+35-7\\ \\ & =x+28\end{array}\phantom{\rule{2em}{0ex}}$$\phantom{\rule{2em}{0ex}}\begin{array}{rl}g\left(f\left(x\right)\right)& =\frac{f\left(x\right)}{5}+7\\ \\ & =\frac{5x-7}{5}+7\\ \\ & =x-\frac{7}{5}+7\\ \\ & =x+\frac{28}{5}\end{array}$

（这里要注意，看到 $f\left(g\left(x\right)\right)=x+28$，我们可以得出结论，就是 $f$$g$ 不是互为反函数。）

## 检查你对内容的理解

### 1) $f\left(x\right)=2x+7$‍  且 $h\left(x\right)=\frac{x-7}{2}$‍

$f\left(h\left(x\right)\right)$$h\left(f\left(x\right)\right)$ 表达成变量为 $x$的函数形式。
$f\left(h\left(x\right)\right)=$
$h\left(f\left(x\right)\right)=$

### 2) $f\left(x\right)=4x+10$‍  且 $g\left(x\right)=\frac{1}{4}x-10$‍

$f\left(g\left(x\right)\right)=$
$g\left(f\left(x\right)\right)=$

### 3) $f\left(x\right)=\frac{2}{3}x-8$‍  和 $h\left(x\right)=\frac{3}{2}\left(x+8\right)$‍

$f\left(h\left(x\right)\right)$$h\left(f\left(x\right)\right)$ 表达成变量为 $x$的函数形式。
$f\left(h\left(x\right)\right)=$
$h\left(f\left(x\right)\right)=$