📚 Inverse Functions: Domain, Graphs and Exam Skills | 反函数：定义域、图像与应试技巧
In Edexcel A-Level Mathematics, inverse functions appear throughout Pure 1, Pure 2 and Pure 3. They are used to reverse mappings, solve equations, and understand symmetry between functions such as exponentials and logarithms. This article covers the definitions, algebraic methods, graphs, domain and range swaps, common exam traps and worked examples you need to master.
在 Edexcel A-Level 数学中，反函数贯穿 Pure 1、Pure 2 和 Pure 3。它们用于逆转映射、解方程，以及理解指数与对数等函数之间的对称关系。本文涵盖定义、代数方法、图像、定义域与值域的交换、常见考试陷阱和你必须掌握的例题。
1. What Is an Inverse Function? | 什么是反函数？
sed, the reversed rule is called the inverse function and is written f⁻¹. It satisfies two identity conditions, one on the domain of f and one on the domain of f⁻¹.
函数 f 将输入 x 映射为输出 y = f(x)。如果这种映射可以逆转，反转后的规则就称为反函数，记作 f⁻¹。它满足两个恒等条件，一个在 f 的定义域上成立，一个在 f⁻¹ 的定义域上成立。
f⁻¹(f(x)) = x for all x ∈ D(f)
f(f⁻¹(x)) = x for all x ∈ D(f⁻¹)
In simple terms, f⁻¹ undoes what f does. If f turns 3 into 7, then f⁻¹ turns 7 back into 3. The notation f⁻¹ does not mean 1/f(x); it means the inverse function, which is a completely different object.
简单地说，f⁻¹ 会撤销 f 的作用。如果 f 把 3 变成 7，那么 f⁻¹ 就把 7 变回 3。记号 f⁻¹ 并不表示 1/f(x)，它表示反函数，是完全不同的对象。
2. One-to-One Functions and the Horizontal Line Test | 一一函数与水平线检验
no two different inputs produce the same output. Equivalently, if f(a) = f(b), then a = b.
函数只有在一一对应（单射）时才存在反函数。这意味着不存在两个不同输入产生相同输出。等价地说，如果 f(a) = f(b)，那么必有 a = b。
t most once. This is the horizontal line test. It is different from the vertical line test, which checks whether a curve is a function at all.
从图像上看，任意水平线 y = k 与曲线 y = f(x) 至多相交一次，这就是水平线检验。它不同于垂直线检验，后者检验一条曲线是否构成函数。
x ≥ 0, then f becomes one-to-one and an inverse exists. The restricted inverse is f⁻¹(x) = √x for x ≥ 0.
例如，f(x) = x² 定义在全体实数上不是一一函数，因为 f(2) = f(-2) = 4。但如果把定义域限制为 x ≥ 0，则 f 变为一一函数，反函数存在。限制后的反函数为 f⁻¹(x) = √x，x ≥ 0。
3. Domain and Range Swap | 定义域与值域的交换
The domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹. This swap is fundamental when stating an inverse function fully. Marks are often lost when the domain of f⁻¹ is omitted.
f 的定义域成为 f⁻¹ 的值域，f 的值域成为 f⁻¹ 的定义域。在完整写出反函数时，这种交换是核心。遗漏 f⁻¹ 的定义域通常会丢分。
| Function | Domain | Range |
|---|---|---|
| f(x) | D | R |
| f⁻¹(x) | R | D |
For f(x) = 2x + 3 with domain x ∈ ℝ, the range is also ℝ. Therefore f⁻¹(x) = (x - 3)/2 has domain ℝ and range ℝ is invisible here because both sets are the whole real line, but the method still applies.
对于 f(x) = 2x + 3，定义域为 x ∈ ℝ，值域也是 ℝ。因此 f⁻¹(x) = (x - 3)/2 的定义域为 ℝ，值域为 ℝ。这里因为两个集合都是全体实数，交换看起来不明显，但方法仍然适用。
4. Finding an Inverse Algebraically | 用代数方法求反函数
Write y = f(x), swap x and y, solve for y, then replace y by f⁻¹(x) and state the domain.
标准代数方法有四个可靠步骤：写出 y = f(x)，交换 x 与 y，解出 y，然后用 f⁻¹(x) 替换 y，并写出定义域。
make y the subject. Step 4: write y = f⁻¹(x) and give the domain, which is the range of f.
第一步：从 y = f(x) 开始。第二步：交换 x 与 y，得到 x = f(y)。第三步：整理方程，使 y 成为主元。第四步：写出 y = f⁻¹(x)，并给出定义域，即 f 的值域。
- 4. Solve: 5y = x + 4, so y = (x + 4)/5. Therefore f⁻¹(x) = (x + 4)/5. Since f has domain ℝ and range ℝ, f⁻¹ also has domain ℝ and range ℝ.
例如，设 f(x) = 5x - 4。写出 y = 5x - 4。交换得到 x = 5y - 4。解得 5y = x + 4，所以 y = (x + 4)/5。因此 f⁻¹(x) = (x + 4)/5。由于 f 的定义域为 ℝ，值域为 ℝ，所以 f⁻¹ 的定义域为 ℝ，值域为 ℝ。
5. Graphs of Inverse Functions and Reflection in y = x | 反函数图像与关于 y = x 的对称
ause the point (a, b) lies on y = f(x) if and only if the point (b, a) lies on y = f⁻¹(x).
y = f⁻¹(x) 的图像是 y = f(x) 的图像关于直线 y = x 的反射。这是因为点 (a, b) 在 y = f(x) 上，当且仅当点 (b, a) 在 y = f⁻¹(x) 上。
so reflect: a horizontal asymptote y = c for f becomes a vertical asymptote x = c for f⁻¹.
作图时，用虚线画出 y = x 作为参考。标出关键点并反射它们。渐近线也会反射：f 的水平渐近线 y = c 变成 f⁻¹ 的垂直渐近线 x = c。
tion with its inverse must lie on y = x. This fact is useful when solving equations of the form f(x) = f⁻¹(x).
如果 f 与 f⁻¹ 相交，交点要么落在 y = x 上，要么成对关于 y = x 对称出现。对于连续递增函数，它与反函数的任何交点都必须在 y = x 上。这个事实在解 f(x) = f⁻¹(x) 型方程时很有用。
6. Self-Inverse Functions | 自反函数
ently, f(f(x)) = x for every x in the domain. Its graph is symmetric about the line y = x.
如果一个函数是它自己的反函数，即 f⁻¹(x) = f(x)，则称它为自反函数。等价地，对定义域内每个 x 都有 f(f(x)) = x。它的图像关于直线 y = x 对称。
For f(x) = 3 - x, applying f twice gives f(f(x)) = 3 - (3 - x) = x, so it is self-inverse.
经典例子包括 f(x) = 1/x（x ≠ 0）、f(x) = 3 - x，以及 f(x) = (x + 1)/(x - 1)（x ≠ 1）。对于 f(x) = 3 - x，连续作用两次得 f(f(x)) = 3 - (3 - x) = x，因此它是自反函数。
Exam questions may ask you to prove that a given function is self-inverse. The safest proof is to compute f(f(x)) carefully and show it simplifies to x, stating any excluded values.
考试可能要求你证明某个函数是自反函数。最稳妥的证明方法是仔细计算 f(f(x))，并说明它化简为 x，同时指出所有排除值。
7. Domain Restrictions and Quadratics | 定义域限制与二次函数
Quadratics are not one-to-one on ℝ, so an inverse exists only after restricting the domain. Choose a restriction that makes the function one-to-one, usually to one side of the vertex.
二次函数在 ℝ 上不是一一函数，因此只有在限制定义域之后才存在反函数。选择一个使函数一一对应的限制，通常取顶点的一侧。
- 1. The vertex is at x = -2. If the domain is restricted to x ≥ -2, the range is y ≥ -1.
对于 f(x) = x² + 4x + 3，配方得 f(x) = (x + 2)² - 1。顶点在 x = -2。如果定义域限制为 x ≥ -2，值域为 y ≥ -1。
o find the inverse, write y = (x + 2)² - 1. Swap x and y: x = (y + 2)² - 1. Solve: y + 2 = √x + 1, because the restricted domain gives the positive square root. Hence f⁻¹(x) = √x + 1 - 2 for x ≥ -1.
求反函数时，写出 y = (x + 2)² - 1。交换 x 与 y：x = (y + 2)² - 1。解得 y + 2 = √x + 1，因为限制后的定义域取正平方根。因此 f⁻¹(x) = √x + 1 - 2，x ≥ -1。
8. Exponential and Logarithmic Inverses | 指数函数与对数反函数
The exponential function f(x) = eˣ has domain ℝ nge y > 0. Its inverse is the natural logarithm f⁻¹(x) = ln x, with domain x > 0 and range ℝ. Their graphs are reflections in y = x.
指数函数 f(x) = eˣ 的定义域为 ℝ，值域为 y > 0。它的反函数是自然对数 f⁻¹(x) = ln x，定义域为 x > 0，值域为 ℝ。它们的图像关于 y = x 对称。
More generally, f(x) = aˣ for a > 0 and a ≠ 1 has inverse f⁻¹(x) = logₐ x. The key identities are alogₐ x = x for x > 0, and logₐ(aˣ) = x for all real x.
更一般地，f(x) = aˣ（a > 0 且 a ≠ 1）的反函数为 f⁻¹(x) = logₐ x。关键恒等式为 alogₐ x = x（x > 0），以及 logₐ(aˣ) = x（x 为任意实数）。
This pair is a favourite exam theme because it links inverse functions with the laws of logarithms and with graph transformations. Always state the domain of the logarithm when writing an inverse.
这一对函数是考试热门主题，因为它把反函数与对数律、图像变换联系起来。写反函数时务必写出对数的定义域。
9. Composite Functions and the Inverse Relationship | 复合函数与反函数关系
Composition with an inverse gives the identity function. In symbols, f⁻¹ ∘ f = I on the domain of f, and f ∘ f⁻¹ = I on the domain of f⁻¹. This is the formal way to verify an inverse.
与反函数复合得到恒等函数。用符号表示，在 f 的定义域上 f⁻¹ ∘ f = I，在 f⁻¹ 的定义域上 f ∘ f⁻¹ = I。这是验证反函数的正式方法。
Order matters. In general, f⁻¹(g(x)) is not the same as (f(g(x)))⁻¹. The first means apply g, then apply f⁻¹. The second means form the composite f ∘ g and then invert the whole function.
顺序很重要。一般来说，f⁻¹(g(x)) 与 (f(g(x)))⁻¹ 不相同。前者表示先作用 g，再作用 f⁻¹；后者表示先组成复合函数 f ∘ g，再对整个函数求反。
x) = (x - 1)/3, then f(g(x)) = 3((x - 1)/3) + 1 = x, so g = f⁻¹ on the appropriate domain.
验证反函数时，代入复合函数并化简。例如，若 f(x) = 3x + 1，g(x) = (x - 1)/3，则 f(g(x)) = 3((x - 1)/3) + 1 = x，因此在适当定义域上 g = f⁻¹。
10. Common Exam Traps and Mark Scheme Habits | 常见考试陷阱与评分习惯
Many marks are lost for predictable reasons. Knowing these traps in advance is one of the fastest ways to improve your score in inverse function questions.
很多失分都有可预测的原因。提前了解这些陷阱，是提高反函数题目得分最快的方法之一。
This is wrong. The reciprocal and the inverse are different. Second trap: forgetting to swap domain and range, or failing to state the domain of f⁻¹.
陷阱一：写成 f⁻¹(x) = 1/f(x)。这是错误的，倒数和反函数完全不同。陷阱二：忘记交换定义域与值域，或者没有写出 f⁻¹ 的定义域。
g sign when taking a square root. The restricted domain decides whether you need +√ or -√.
陷阱三：假设每个函数都有反函数。必须先检查是否一一对应。陷阱四：开平方时选错符号。限制后的定义域决定你需要 +√ 还是 -√。
he x-axis or y-axis. Sixth trap: mixing up f⁻¹(g(x)) with (f(g(x)))⁻¹, which is a composition-order error.
陷阱五：关于错误的直线反射。反射线是 y = x，不是 x 轴或 y 轴。陷阱六：混淆 f⁻¹(g(x)) 与 (f(g(x)))⁻¹，这是复合顺序错误。
11. Worked Exam-Style Examples | 典型考题精讲
defined by f(x) = (3x + 2)/(x - 1) for x > 1. Find f⁻¹(x) and state its domain and range.
例 1：函数 f 定义为 f(x) = (3x + 2)/(x - 1)，x > 1。求 f⁻¹(x)，并写出它的定义域和值域。
ing gives x(y - 3) = y + 2, hence x = (y + 2)/(y - 3). Therefore f⁻¹(x) = (x + 2)/(x - 3).
设 y = (3x + 2)/(x - 1)。则 y(x - 1) = 3x + 2，所以 xy - y = 3x + 2。整理得 x(y - 3) = y + 2，因此 x = (y + 2)/(y - 3)。所以 f⁻¹(x) = (x + 2)/(x - 3)。
5/(x - 1). Since x > 1, x - 1 > 0, so 5/(x - 1) > 0. Hence f(x) > 3. Therefore f⁻¹ has domain x > 3 and range f⁻¹(x) > 1.
现在求 f 的值域。写出 f(x) = 3 + 5/(x - 1)。因为 x > 1，所以 x - 1 > 0，故 5/(x - 1) > 0。因此 f(x) > 3。所以 f⁻¹ 的定义域为 x > 3，值域为 f⁻¹(x) > 1。
function g is defined by g(x) = x² - 6x + 5 for x ≥ 3. Find g⁻¹(x). Complete the square: g
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