Inverse Function

Inversion

求反函数表达式

反函数的求法

1. 识别自变量（输入）和因变量（输出）
2. 用因变量表示自变量

步骤：

1. 求原来函数的值域
2. 解方程y=f(x)，写出x关于y的函数式，并写出y的取值范围（也就是不要忘记写原来函数的值域）
3. 交换x与y。

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$\text{求函数}f(x)=\{\begin{array}{l}x,-\infty <x<1\\ x^2,1\le x\le 4\\ 2^x,4<x<+\infty \end{array}$的反函数

第一段：$x\in (-\infty ,1)$
则$y\in (-\infty ,1)$

$$\begin{array}{r}y=x(-\infty \le y\le 1)\\ x=y(-\infty \le y\le 1)\\ \therefore y=x=f^{-1}(x)(-\infty \le x\le 1)\end{array}$$

第二段：$x\in [1,4]$
$y=x^2,则(1\le y\le 16)$

$$\begin{array}{r}y=x^2(1\le y\le 16)\\ x=\pm \sqrt{y}(1\le y\le 16)\\ \because x\in [1,4]\\ \therefore x=\sqrt{y}(1\le y\le 16)\\ \therefore y=\sqrt{x}(1\le x\le 16)\end{array}$$

第三段：$x\in (4,+\infty )$
则 $y\in (16,+\infty )$

$$\begin{array}{r}y=2^x(16<y<+\infty )\\ x={\log }_{2}y(16<y<+\infty )\\ y={\log }_{2}x(16<x<+\infty )\end{array}$$

综上：

$$f^{-1}(x)=\{\begin{array}{ll}x,& -\infty <x<1\\ & \\ \sqrt{x},& 1\le x\le 16\\ & \\ {\log }_{2}x,& x>16\end{array}$$

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函数$y=x^2(x\le 0)$的反函数
该函数的值域$y\ge 0$
反解

$$\begin{array}{r}y=x^2(y\ge 0)\\ x^2=y(y\ge 0)\\ x=\pm \sqrt{y}(y\ge 0)\\ \because x\le 0\\ \therefore x=-\sqrt{y}(y\ge 0)\\ \therefore y=-\sqrt{x}(x\ge 0)\end{array}$$

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已知函数$f(x)=\sqrt{x+1}$，求$f^{-1}(x)$

$$\begin{array}{r}y=\sqrt{x+1}(y\ge 0)\\ y^2=x+1(y\ge 0)\\ x=y^2-1(y\ge 0)\\ \therefore y=x^2-1=f^{-1}(x)(x\ge 0)\end{array}$$

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求$y=3^{x-1}(x\ge 1)$的反函数

$$\begin{array}{r}x\ge 1\\ x-1\ge 0\\ 当x-1=0时，取得最小值3^0=1\\ \therefore 3^{x-1}\ge 1\end{array}$$ $$\begin{array}{r}y=3^{x-1}(y\ge 1)\\ 3^{x-1}=y(y\ge 1)\\ {\log }_{3}y=x-1(y\ge 1)\\ x={\log }_{3}y+1(y\ge 1)\\ \therefore y={\log }_{3}x+1(x\ge 1)\end{array}$$

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反函数求值

函数$f(x)=\sqrt{x+3}$，求$f^{-1}(0)$
第一种方法，直接求反函数表达式，再代入

$$\begin{array}{r}y=\sqrt{x+3}(y\ge 0)\\ y^2=x+3(y\ge 0)\\ x=y^2-3(y\ge 0)\\ y=x^2-3=f^{-1}(x)(x\ge 0)\\ f^{-1}(0)=0-3=-3\end{array}$$

第二种方法

1. 令右式括号中要求的值，等于左式解析式
2. 解出x，即为所求值

$$\begin{array}{r}即\sqrt{x+3}=0\\ x+3=0^2\\ x=-3\end{array}$$

常见反函数

arc: An arc is a smoothly curving line or movement.
弧形；弧线；

The Aleutian chain is a long arc of islands in the North Pacific.
阿留申群岛是北太平洋上一长串呈弧形排列的小岛。

arcsinx

$$y=\sin x\iff x=\arcsin y$$

最后一步（交换x与y）

$$y=\arcsin x$$

y=\arcsin x

$$\begin{array}{r}定义域：x\in [-1,1]\\ 值域：y\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\end{array}$$

注意定义域

$$\begin{array}{rl}\sin (\arcsin a)& =a(a\in [-1,1])\\ \arcsin (\sin a)& =a\left(a\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]\right)\end{array}$$ $$\begin{array}{r}证明：令y=\arcsin a\Rightarrow a=\sin y\end{array}$$

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arccosx

$$\begin{array}{rl}\cos (\arccos a)& =a(a\in [-1,1])\\ \arccos (\cos a)& =a(a\in [0,\pi ])\end{array}$$

arctan x

$y=\arctan x$

y=\arctan x
y=- \frac{\pi}{2}|dashed|blue
y= \frac{\pi}{2}|dashed|blue

$$\begin{array}{r}定义域：x\in R\\ 值域：y\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\end{array}$$

求反三角函数

求$y=\sin (x-1)$的反函数

$$\begin{array}{r}x-1=\arcsin y\\ x=\arcsin y+1\\ 最后交换\\ y=\arcsin x+1\\ 最后求定义域，也就是求原函数的值域\\ y=\arcsin x+1,x\in [1,-1]\end{array}$$

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求$y=2\sin (3x-1)+4$的反函数

$$\begin{array}{r}y-4=2\sin (3x-1)\\ \frac{y-4}{2}=\sin (3x-1)\\ 3x-1=\arcsin \left(\frac{y-4}{2}\right)\\ 3x=\arcsin \left(\frac{y-4}{2}\right)+1\\ x=\frac{\arcsin \left(\frac{y-4}{2}\right)+1}{3}\\ 互换\\ y=\frac{1}{3}\arcsin \left(\frac{x-4}{2}\right)+\frac{1}{3}\end{array}$$

最后求定义域

$$1\le \frac{x-4}{2}\le 1$$

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