📚 Inverse Functions: A Complete AQA A-Level Maths Guide | 反函数：AQA A-Level 数学完整指南
Understanding inverse functions is essential for success in AQA A-Level Mathematics. An inverse function reverses the effect of the original function, linking inputs and outputs in the opposite direction. This guide covers notation, conditions for existence, algebraic methods, domain and range, graph relationships, self-inverse functions, key inverse pairs such as exponential and logarithmic functions, restricted trigonometric inverses, differentiation of inverses, and common exam pitfalls.
理解反函数对于在 AQA A-Level 数学中取得成功至关重要。反函数会反转原函数的作用，以相反方向联系输入与输出。本指南涵盖记号、存在条件、代数求法、定义域与值域、图像关系、自反函数、指数函数与对数函数等关键互逆对、限定定义域的反三角函数、反函数求导以及常见考试易错点。
1. What is an Inverse Function? | 什么是反函数？
mbols, if f(a) = b then f⁻¹(b) = a. Every input of f becomes an output of f⁻¹, and every output of f becomes an input of f⁻¹.
反函数写作 f⁻¹(x)，它撤销原函数 f(x) 的作用。如果 f 把输入 a 映射为输出 b，那么 f⁻¹ 就把 b 映回 a。用符号表示：若 f(a) = b，则 f⁻¹(b) = a。f 的每一个输入都成为 f⁻¹ 的输出，f 的每一个输出都成为 f⁻¹ 的输入。
example, if f(x) = x + 5, then f(2) = 7, so f⁻¹(7) must be 2. The operation that reverses "add 5" is "subtract 5" can always check your answer by composing the two functions: f(f⁻¹(x)) = (x - 5) + 5 = x.
例如，若 f(x) = x + 5，则 f(2) = 7，因此 f⁻¹(7) 必为 2。逆转 "加 5" 的运算就是 "减 5"，所以反函数为 f⁻¹(x) = x - 5。你可以通过复合两个函数来检验答案：f(f⁻¹(x)) = (x - 5) + 5 = x。
n slip is to read f⁻¹(x) as the reciprocal 1/f(x). The superscript ⁻¹ here is notation for "inverse function" power. For f(x) = x + 5, f⁻¹(x) = x - 5, which is definitely not 1/(x + 5). Keep the words "inverse" and "reciprocal" strictly separate in your written answers.
一个非常常见的失误是把 f⁻¹(x) 读作倒数 1/f(x)。这里的上标 ⁻¹ 是 "反函数" 的记号，而不是幂。对于 f(x) = x + 5，f⁻¹(x) = x - 5，绝不是 1/(x + 5)。在书面作答时，务必把 "反函数" 与 "倒数" 严格区分开。
2. Notation and the Two Core Identities | 记号与两个核心恒等式
The inverse of f is written f⁻¹, and its rule is f⁻¹(x). The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f. Two composition identities summarise the whole idea and are frequently quoted in mark schemes:
f 的反函数写作 f⁻¹，其对应法则写作 f⁻¹(x)。f⁻¹ 的定义域是 f 的值域，f⁻¹ 的值域是 f 的定义域。下面两个复合恒等式概括了整个概念，也常常出现在评分标准中：
f⁻¹(f(x)) = x for every x in the domain of f
f(f⁻¹(x)) = x for every x in the domain of f⁻¹
hese identities only hold on the correct domains: if f is defined on x ≥ 2, then f⁻¹(f(x)) = x only makes sense for x ≥ 2.
第一个恒等式表示先作用 f 再作用 f⁻¹ 会回到出发点；第二个恒等式表示反过来的顺序也一样。注意这些恒等式只在正确的定义域上成立：若 f 的定义域为 x ≥ 2，那么 f⁻¹(f(x)) = x 也只在 x ≥ 2 时有意义。
Remember that an inverse function is unique. If a function has an inverse, there is exactly one such function, so you never need to worry about two different correct answers. What you must always state is the domain on which the inverse is defined, because that is where most marks are lost.
记住反函数是唯一的。若一个函数存在反函数，那么它只有一个，因此不必担心会出现两个不同的正确答案。你真正必须写明的是反函数的定义域，因为丢分大多发生在这里。
3. One-to-One Functions and the Horizontal Line Test | 一一函数与水平线检验
ut value: different inputs always give different outputs. A many-to-one function, such as f(x) = x² on the whole real line, cannot be reversed, because f(2) = 4 and f(-2) = 4 both lead back to an impossible choice between 2 and -2.
一个函数只有在其整个定义域上是一一函数（单射）时，才存在反函数。一一函数意味着每个输出值只由一个输入值产生：不同的输入总是给出不同的输出。多对一的函数，例如在整个实数集上的 f(x) = x²，无法被反转，因为 f(2) = 4 与 f(-2) = 4 会让我们无法在 2 和 -2 之间做出选择。
The horizontal line test gives a quick graphical check. Draw any horizontal line across the graph of the function. If every such line meets the curve at most once, the function is one-to-one and an inverse exists. If some horizontal line meets the curve twice or more, the function is not one-to-one.
水平线检验提供了快速的图像判断方法。在函数图像上画任意一条水平线。如果每条这样的直线与曲线至多相交一次，则函数是一一的，反函数存在；如果某条水平线与曲线相交两次或更多，则函数不是一一的。
ally, suppose f(a) = f(b) and show that this forces a = b. This is the method expected for "show that f is one-to-one" h. For f(x) = 3x - 4, if 3a - 4 = 3b - 4 then 3a = 3b and hence a = b, so f is one-to-one.
要用代数方法证明函数是一一的，可假设 f(a) = f(b)，并证明这必然推出 a = b。这是 "证明 f 是一一函数" 这类题目的标准方法，比单纯引用图像更严谨。对于 f(x) = 3x - 4，若 3a - 4 = 3b - 4，则 3a = 3b，故 a = b，所以 f 是一一的。
4. Domain Restrictions and When No Inverse Exists | 定义域限制与不存在反函数的情形
When a natural domain makes a function many-to-one, you can often restrict the domain to a smaller interval on which it becomes one-to-one. Restricting the domain changes the function, so the restricted domain must be stated clearly in every step of your working.
当自然定义域使函数成为多对一时，你通常可以把定义域限制在较小的区间上，使它成为一一函数。限制定义域会改变函数本身，因此在每一步运算中都必须清楚写出限制后的定义域。
-one. Restrict the domain to x ≥ 2: now the right-hand branch is strictly increasing, so the function is one-to-one and has an inverse.
以 f(x) = x² - 4x + 7 为例。配方得 f(x) = (x - 2)² + 3，因此图像是顶点为 (2, 3) 的抛物线。在全体实数上它不是一一函数。把定义域限制为 x ≥ 2：此时右半支严格递增，函数成为一一函数，于是存在反函数。
To find that inverse, write y = (x - 2)² + 3, so y - 3 = (x - 2)² and x - 2 = ±√y - 3. Because x ≥ 2, the positive root is correct, giving x = 2 + √y - 3. Swapping the letters gives the inverse rule:
求该反函数时，写 y = (x - 2)² + 3，于是 y - 3 = (x - 2)²，x - 2 = ±√y - 3。由于 x ≥ 2，应取正根，得 x = 2 + √y - 3。交换字母后得到反函数的法则：
f⁻¹(x) = 2 + √x - 3, x ≥ 3
r leave it out. If instead you had chosen x ≤ 2, the correct inverse would be f⁻¹(x) = 2 - √x - 3, again with x ≥ 3.
限制条件 x ≥ 3 来自 f 的值域 f(x) ≥ 3。这一步本身就有分，所以绝不能省略。如果你选择的是 x ≤ 2，那么正确的反函数就是 f⁻¹(x) = 2 - √x - 3，定义域同样是 x ≥ 3。
5. Domain and Range of an Inverse Function | 反函数的定义域与值域
The domain and range of an inverse are swapped relative to the original function. Formally, the domain of f⁻¹ equals the range of f, and the range of f⁻¹ equals the domain of f. This single fact answers most "state the domain of f⁻¹" questions without any algebra.
反函数的定义域与值域相对于原函数是互换的。严格地说，f⁻¹ 的定义域等于 f 的值域，f⁻¹ 的值域等于 f 的定义域。仅凭这一条事实，就能在没有代数运算的情况下回答大多数 "写出 f⁻¹ 的定义域" 的题目。
For example, if f(x) = eˣ with domain x ∈ ℝ, then the range of f is y > 0. Therefore f⁻¹(x) = ln x has domain x > 0 and range y ∈ ℝ. Similarly, if f(x) = ln x for x > 0, its range is ℝ, so f⁻¹(x) = eˣ has domain ℝ and range y > 0.
例如，若 f(x) = eˣ，定义域为 x ∈ ℝ，则 f 的值域为 y > 0。因此 f⁻¹(x) = ln x 的定义域为 x > 0，值域为 y ∈ ℝ。类似地，若 f(x) = ln x（x > 0），其值域为 ℝ，所以 f⁻¹(x) = eˣ 的定义域为 ℝ，值域为 y > 0。
Use this reversal as a checking tool. Once you have found a formula for f⁻¹, ask yourself: does its domain match the range I calculated for f? If not, either your inverse formula or your range is wrong. Examiners reward candidates who explicitly state both the rule and the domain.
把这种互换当作检验工具。一旦求出 f⁻¹ 的表达式，就问自己：它的定义域与我算出的 f 的值域一致吗？如果不一致，那么要么反函数表达式错了，要么值域错了。考官会奖励那些明确指出法则与定义域的考生。
6. Finding Inverse Functions Algebraically | 代数求反函数的方法
Most AQA questions ask you to find an inverse function by rearranging. The routine has four steps, and following them in order prevents careless errors.
大多数 AQA 题目要求你通过变形求出反函数。这个过程有四个步骤，按顺序执行可以避免粗心错误。
Step 1: Write the function as y = f(x). | 第一步：把函数写成 y = f(x)。
Step 2: Swap x and y to get x = f(y). | 第二步：交换 x 与 y，得到 x = f(y)。
Step 3: Rearrange to make y the subject. | 第三步：变形，使 y 成为主元。
Step 4: Replace y with f⁻¹(x) and state the domain. | 第四步：把 y 换成 f⁻¹(x)，并写出定义域。
Worked example: find the inverse of f(x) = (3x + 1)/2 for x ∈ ℝ ng gives x = (3y + 1)/2. Multiply by 2: 2x = 3y + 1. Subtract 1: 2x - 1 = 3y. Divide by 3:
例题：求 f(x) = (3x + 1)/2（x ∈ ℝ）的反函数。写 y = (3x + 1)/2。交换后得 x = (3y + 1)/2。两边乘 2：2x = 3y + 1。减 1：2x - 1 = 3y。除以 3：
f⁻¹(x) = (2x - 1)/3, x ∈ ℝ
quick and gives confidence, especially in an exam where a single sign error can cost several marks.
用数值检验：f(1) = 2，而 f⁻¹(2) = (4 - 1)/3 = 1。这种检验很快，而且能增强信心，尤其是在考试中，一个符号错误就可能丢掉好几分。
7. Graphs: Reflection in the Line y = x | 图像：关于直线 y = x 的对称
y = x does. A point (a, b) on the graph of f becomes the point (b, a) on the graph of f⁻¹.
y = f⁻¹(x) 的图像是 y = f(x) 的图像关于直线 y = x 的镜像，因为交换 x 与 y 正是关于 y = x 反射的效果。f 图像上的点 (a, b) 变成 f⁻¹ 图像上的点 (b, a)。
else, the intersections must occur in symmetric pairs across y = x. Sketching both curves on the same axes is a fast way to check that your algebra is sensible.
f 的图像与直线 y = x 的交点在这个反射下保持不变，因此这些点也在 f⁻¹ 的图像上。如果 f 与 f⁻¹ 还在别处相交，交点必须关于 y = x 成对对称出现。把两条曲线画在同一坐标系中是检验代数结果是否合理的快捷方法。
mptote at x = 0 and a horizontal asymptote at y = 0; since f is self-inverse, its graph is symmetric about y = x and the two asymptotes simply swap roles.
渐近线也遵循同样的规律。f 的竖直渐近线 x = c 变成 f⁻¹ 的水平渐近线 y = c。例如 f(x) = 1/x 有竖直渐近线 x = 0 与水平渐近线 y = 0；由于 f 是自反函数，其图像关于 y = x 对称，两条渐近线只是互换角色。
8. Self-Inverse Functions | 自反函数
unctions appear regularly in AQA papers because they combine graph reasoning with algebra.
如果一个函数是自身的反函数，即 f⁻¹(x) = f(x)，则称它为自反函数。从几何上看，其图像关于直线 y = x 对称。自反函数经常出现在 AQA 试卷中，因为它把图像推理与代数结合了起来。
fy it, write y = x/(x - 1), swap to get x = y/(y - 1), then rearrange: x(y - 1) = y, so xy - x = y, giving y(x - 1) = x and finally y = x/(x - 1). The function is unchanged.
简单的例子包括 f(x) = 1/x（x ≠ 0），以及 f(x) = a - x（a 为常数），例如 f(x) = 6 - x。一个不那么明显的例子是 f(x) = x/(x - 1)（x ≠ 1）。验证如下：写 y = x/(x - 1)，交换得 x = y/(y - 1)，再变形：x(y - 1) = y，故 xy - x = y，得 y(x - 1) = x，最终 y = x/(x - 1)。函数没有改变。
ry x in the domain. If composing f with itself returns the identity, then f must be its own inverse. This argument is short, rigorous and often earns full marks.
若不想通过公式证明自反性，可以证明对定义域内每个 x 都有 f(f(x)) = x。如果 f 与自身复合得到恒等映射，那么 f 必为自身的反函数。这种论证简短、严谨，往往能拿满分。
9. Exponential and Logarithmic Functions as Inverses | 指数函数与对数函数互为反函数
QA A-Level Mathematics is the exponential function and the natural logarithm. For f(x) = eˣ with domain ℝ ln x with domain x > 0. The two functions are reflections of each other in the line y = x.
AQA A-Level 数学中最重要的互逆函数对是指数函数与自然对数。对于定义域为 ℝ 的 f(x) = eˣ，其反函数为定义域 x > 0 的 f⁻¹(x) = ln x。这两个函数关于直线 y = x 互为镜像。
The defining identities follow directly from the composition rules:
由复合恒等式可直接得到定义性关系：
ln(eˣ) = x for all x ∈ ℝ and eln x = x for all x > 0
The general case is the same idea with a different base. If f(x) = aˣ 0 and a ≠ 1, then f⁻¹(x) = log_a x for x > 0. In particular, log₁₀ x is the inverse of 10ˣ, and ln x is the inverse of eˣ. Always check the base before writing your answer.
一般情形只是换了一个底数，思想完全相同。若 f(x) = aˣ（a > 0 且 a ≠ 1），则 f⁻¹(x) = log_a x（x > 0）。特别地，log₁₀ x 是 10ˣ 的反函数，ln x 是 eˣ 的反函数。写答案前务必确认底数。
| f(x) | f⁻¹(x) | Domain of f⁻¹ |
| eˣ | ln x | x > 0 |
| 10ˣ | log₁₀ x | x > 0 |
| aˣ | log_a x | x > 0 |
| ln x | eˣ | x ∈ ℝ |
A typical exam task is to invert a composed exponential, such as f(x) = e2x + 1. Write y = e2x + 1, swap to x = e2y + 1 ithms: ln x = 2y + 1, so y = (ln x - 1)/2. The inverse is f⁻¹(x) = (ln x - 1)/2 for x > 0.
典型考题是求复合指数函数的反函数，例如 f(x) = e2x + 1。写 y = e2x + 1，交换得 x = e2y + 1，取对数：ln x = 2y + 1，故 y = (ln x - 1)/2。反函数为 f⁻¹(x) = (ln x - 1)/2（x > 0）。
10. Inverse Trigonometric Functions with Restricted Domains | 限定定义域的反三角函数
The trigonometric functions are periodic, so they are many-to-one on their natural domains and have no inverse there. To create an inverse, we restrict each function to a principal interval on which it is strictly increasing or strictly decreasing.
三角函数是周期函数，因此在自然定义域上是多对一的，没有反函数。为了构造反函数，我们把每个函数限制在一个主区间上，使其在该区间上严格递增或严格递减。
| Function | Restricted domain | Inverse | style="border:1px solid #ccc;padding:
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