📚 Inverse Functions, f⁻¹(x) | 反函数，f⁻¹(x)
Inverse functions are a cornerstone of IB Mathematics, enabling us to reverse the mapping of a function. Understanding how to find, verify, and interpret f⁻¹(x) is essential for both SL and HL students, as it connects directly to composite functions, graph transformations, and calculus concepts.
反函数是 IB 数学的基石，使我们能够逆转函数的映射。理解如何求、验证和解释 f⁻¹(x) 对于 SL 和 HL 学生都至关重要，因为它与复合函数、图像变换和微积分概念直接相关。
1. What is an Inverse Function? | 什么是反函数？
An inverse function 'undoes' the inverse function f⁻¹ takes y as its input and gives back x, provided f is one-to-one.
反函数会“撤销”原函数的操作。如果函数 f 接受输入 x 并产生输出 y = f(x)，那么反函数 f⁻¹ 接受 y 作为输入并还原为 x，前提是 f 是一一映射。
and range R, the inverse f⁻¹ has domain R and range D, satisfying f⁻¹(f(x)) = x for all x ∈ D, and f(f⁻¹(x)) = x for all x ∈ R.
记号 f⁻¹(x) 不是指数，不代表 1/f(x)。它只是表示反函数。对于定义域为 D、值域为 R 的函数 f，其反函数 f⁻¹ 的定义域为 R、值域为 D，且满足 f⁻¹(f(x)) = x（对所有 x ∈ D）以及 f(f⁻¹(x)) = x（对所有 x ∈ R）。
2. One-to-one Functions | 一一映射函数
A function must be one-to-one (injective) to possess an inverse. Being one‑to‑one means that no two different inputs map to the same output; if a ≠ b then f(a) ≠ f(b).
函数必须是一一映射（单射）才能有反函数。一一映射意味着没有两个不同的输入对应同一个输出；如果 a ≠ b，则 f(a) ≠ f(b)。
If a function is not one‑to‑one, its 'inverse' would be ambiguous because one y‑value could link back to multiple x‑values, which violates the definition of a function.
如果一个函数不是一一映射，其“反函数”将产生歧义，因为一个 y 值可能对应多个 x 值，这违反了函数的定义。
3. Horizontal Line Test | 水平线检验
A quick way to check whether a function is one‑to‑one is the horizontal line test. If any horizontal line cuts the graph more than once, the function is not one‑to‑one and therefore does not have an inverse over its entire domain.
检验函数是否一一映射的快捷方法是水平线检验。如果任意水平线与图像相交超过一次，该函数就不是一一映射，因而在整个定义域上没有反函数。
comes from both x = 2 and x = −2. However, f(x) = x³ passes the test and has an inverse on ℝ.
例如，f(x) = x² 未通过水平线检验，因为 y = 4 来源于 x = 2 和 x = −2。然而，f(x) = x³ 通过了检验，在 ℝ 上存在反函数。
4. Steps to Find an Inverse Function | 求反函数的步骤
To find the inverse of a function f(x) algebraically, follow these steps:
通过代数方法求函数 f(x) 的反函数，可遵循以下步骤：
Step 1: Write y = f(x).
步骤1：写出 y = f(x)。
Step 2: Swap x and y in the equation. This reflects the idea that inputs and outputs exchange roles.
步骤2：交换方程中的 x 和 y。这体现了输入与输出角色互换的思想。
Step 3: Solve the new equation for y. This may involve rearranging terms, factorising, or applying inverse operations.
步骤3：解出新方程中的 y。这可能涉及移项、因式分解或使用逆运算。
Step 4: Replace y with f⁻¹(x). State the domain of f⁻¹(x) based on the range of the original function.
步骤4：将 y 替换为 f⁻¹(x)。根据原函数的值域给出 f⁻¹(x) 的定义域。
3x − 7, then y = 3x − 7 → swap: x = 3y − 7 → solve: y = (x + 7)/3 → so f⁻¹(x) = (x + 7)/3.
示例：若 f(x) = 3x − 7，则 y = 3x − 7 → 交换：x = 3y − 7 → 求解：y = (x + 7)/3 → 因此 f⁻¹(x) = (x + 7)/3。
5. Domain and Range Swap | 定义域与值域的互换
A fundamental property of inverse functions is that the domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹.
反函数的一个基本性质是：f 的定义域变成 f⁻¹ 的值域，而 f 的值域变成 f⁻¹ 的定义域。
This swap is crucial when working with functions that have restricted domains, such as square‑root, logarithmic or trigonometric functions.
在处理具有受限定义域的函数（如平方根、对数或三角函数）时，这种互换至关重要。
| Function f(x) | Domain of f | Range of f | Inverse f⁻¹(x) | Domain of f⁻¹ |
|---|---|---|---|---|
| eˣ | ℝ | (0, ∞) | ln x | (0, ∞) |
| x², x ≥ 0 | [0, ∞) | [0, ∞) | √x | [0, ∞) |
| sin x, −π/2 ≤ x ≤ π/2 | [−π/2, π/2] | [−1, 1] | arcsin x | [−1, 1] |
6. Graphs and Symmetry | 图像与对称性
ross the line y = x. If the point (a, b) lies on y = f(x), then (b, a) lies on y = f⁻¹(x).
反函数的图像是原函数图像关于直线 y = x 的对称。如果点 (a, b) 在 y = f(x) 上，那么 (b, a) 就在 y = f⁻¹(x) 上。
This symmetry helps to visualise the domain‑range swap: the x‑intercept of f becomes the y‑intercept of f⁻¹, and vice versa.
这种对称性有助于直观理解定义域与值域的互换：f 的 x 截距变成 f⁻¹ 的 y 截距，反之亦然。
For instance, the graph of f(x) = 2ˣ and its inverse f⁻¹(x) = log₂ x are mirror images over the line y = x.
例如，f(x) = 2ˣ 与其反函数 f⁻¹(x) = log₂ x 的图像关于直线 y = x 互为镜像。
7. Composite Function Property | 复合函数性质
The defining property of an inverse function is captured by compositions:
反函数的定义性质可用复合函数来表示：
f⁻¹(f(x)) = x for all x in the domain of f
f(f⁻¹(x)) = x for all x in the domain of f⁻¹
These equations must hold true. They are often used to verify whether two functions are indeed inverses of each other.
这些等式必须成立，常用于验证两个函数是否互为反函数。
Example: Verify that f(x) = 2x + 3 and g(x) = (x − 3)/2 are inverses. x; g(f(x)) = [(2x + 3) − 3]/2 = 2x/2 = x. Both compositions yield x, so they are inverses.
示例：验证 f(x) = 2x + 3 与 g(x) = (x − 3)/2 互为反函数。f(g(x)) = 2[(x − 3)/2] + 3 = x − 3 + 3 = x；g(f(x)) = [(2x + 3) − 3]/2 = 2x/2 = x。两个复合均得到 x，故互为反函数。
8. Restricting the Domain to Find an Inverse | 限制定义域以获得反函数
Many common functions, such as quadratics, are not one‑to‑one on their natural domain. However, by restricting the domain, we can create a one‑to‑one branch and thus obtain an inverse.
许多常见函数（如二次函数）在其自然定义域上并非一一映射。但通过限制定义域，我们可以创建一个一一映射的分支，从而得到反函数。
The classic example is f(x) = x². For domain {x | x ≥ 0}, the inverse is f⁻¹(x) = √x. For domain {x | x ≤ 0}, the inverse is f⁻¹(x) = −√x. Both are valid inverses, but the restriction must be clearly stated.
经典例子是 f(x) = x²。对于定义域 {x | x ≥ 0}，反函数为 f⁻¹(x) = √x；对于定义域 {x | x ≤ 0}，反函数为 f⁻¹(x) = −√x。两者都是有效的反函数，但必须明确说明限制条件。
In IB exams, you may be asked to find the largest possible domain for which a function is invertible, or to restrict a given domain.
在 IB 考试中，你可能会被要求找出使函数可逆的最大可能定义域，或对给定定义域进行限制。
9. Self-Inverse Functions | 自反函数
A function is called self‑ ts own inverse, that is, f⁻¹(x) = f(x). Equivalently, f(f(x)) = x for all x in its domain.
若一个函数是其自身的反函数，即 f⁻¹(x) = f(x)，则称之为自反函数。等价地，对于其定义域内的所有 x，有 f(f(x)) = x。
Classic examples include:
经典例子包括：
f(x) = x (the identity function)
f(x) = x（恒等函数）
f(x) = −x
f(x) = −x
f(x) = 1/x, x ≠ 0
f(x) = 1/x，x ≠ 0
f(x) = (x + a)/(x − 1) for certain values of a (e.g. a = 2 gives f(x) = (x+2)/(x−1))
f(x) = (x + a)/(x − 1) 对于某些 a 值（例如 a = 2 得 f(x) = (x+2)/(x−1)）
Self‑ ic about the line y = x and also symmetric about the origin in some cases (odd functions).
自反函数既关于直线 y = x 对称，有时也关于原点对称（奇函数）。
10. Worked Examples | 计算举例
Example 1 (linear): Find f⁻¹(x) for f(x) = 4x − 9.
例1（线性）：求 f(x) = 4x − 9 的反函数。
y = 4x − 9 → swap: x = 4y − 9 → 4y = x + 9 → y = (x + 9)/4, so f⁻¹(x) = (x + 9)/4.
y = 4x − 9 → 交换：x = 4y − 9 → 4y = x + 9 → y = (x + 9)/4，故 f⁻¹(x) = (x + 9)/4。
Example 2 (rational): Find the inverse of f(x) = (3x − 2) / (x + 4), x ≠ −4.
例2（有理函数）：求 f(x) = (3x − 2)/(x + 4)，x ≠ −4 的反函数。
) = −(4x + 2) → y = (4x + 2) / (3 − x). Thus f⁻¹(x) = (4x + 2)/(3 − x), with domain x ≠ 3.
设 y = (3x − 2)/(x + 4)。交换：x = (3y − 2)/(y + 4)。两边乘分母：x(y + 4) = 3y − 2 → xy + 4x = 3y − 2。合并 y 项：xy − 3y = −2 − 4x → y(x − 3) = −(4x + 2) → y = (4x + 2)/(3 − x)。因此 f⁻¹(x) = (4x + 2)/(3 − x)，定义域 x ≠ 3。
matches the range of f, which is all real numbers except y = 3 (the horizontal asymptote).
注意 f⁻¹ 的定义域与 f 的值域一致，即除 y = 3（水平渐近线）外的所有实数。
11. Inverse Trigonometric Functions – IB HL | 反三角函数 – IB HL 内容
In HL Mathematics, students work with the inverse trig functions arcsin x, arccos x and arctan x. Because sine, cosine and tangent are periodic, their domains must be restricted to produce one‑to‑one branches.
在 HL 数学中，学生会学习反三角函数 arcsin x、arccos x 和 arctan x。由于正弦、余弦和正切是周期函数，必须限制定义域以得到一一映射的分支。
The standard restrictions are:
标准限制为：
cted to [−π/2, π/2] gives inverse y = arcsin x, with domain [−1, 1] and range [−π/2, π/2].
y = sin x 限制于 [−π/2, π/2] 得反函数 y = arcsin x，定义域 [−1, 1]，值域 [−π/2, π/2]。
y = cos x restricted to [0, π] gives inverse y = arccos x, domain [−1, 1], range [0, π].
y = cos x 限制于 [0, π] 得反函数 y = arccos x，定义域 [−1, 1]，值域 [0, π]。
y = tan x restricted to (−π/2, π/2) gives inverse y = arctan x, domain ℝ, range (−π/2, π/2).
y = tan x 限制于 (−π/2, π/2) 得反函数 y = arctan x，定义域 ℝ，值域 (−π/2, π/2)。
These inverses are essential in calculus (differentiation and integration) and solving trigonometric equations analytically.
这些反函数在微积分（求导和积分）以及解析求解三角方程中不可或缺。
12. Common Mistakes and Key Tips | 常见错误与重要提示
Mistake 1: Assuming every function has an inverse without checking one‑to‑one condition. Always apply the horizontal line test or algebraic reasoning.
错误1：未检验一一映射条件就假设每个函数都有反函数。务必使用水平线检验或代数推理。
Mistake 2: Interpreting f⁻¹(x) as 1/f(x). The notation is not an exponent; it denotes the inverse mapping.
错误2：将 f⁻¹(x) 解读为 1/f(x)。该记号不是指数，而是表示逆映射。
Mistake 3: Forgetting to swap domain and range when writing the final answer. The domain of f⁻¹ is the range of f.
错误3：在写最终答案时忘记交换定义域和值域。f⁻¹ 的定义域是 f 的值域。
Mistake 4: Algebraic slips when rearranging, especially with rational functions. Double‑check your work by verifying that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
错误4：在移项时出现代数错误，尤其是有理函数。通过验证 f(f⁻¹(x)) = x 和 f⁻¹(f(x)) = x 来复查你的运算。
Key tip: When a function is given with a restricted domain, always write the inverse’s domain explicitly. In IB exams, marks are
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