Inverse Functions f⁻¹(x): Solving and Applications in IB Maths | IB数学：反函数f⁻¹(x)的求解与应用

📚 Inverse Functions f⁻¹(x): Solving and Applications in IB Maths | IB数学：反函数f⁻¹(x)的求解与应用

The concept of inverse functions is one of the cornerstones of IB Mathematics, appearing across both Analysis and Approaches (AA) and Applications and Interpretation (AI) syllabuses. An inverse function essentially “undoes” what the original function does, providing a powerful tool for solving equations, modelling real-world relationships, and understanding symmetry in mathematics.

反函数是IB数学的核心概念之一，在分析与方法（AA）和应用与解释（AI）两大课程体系中均有涉及。反函数本质上是“撤销”原函数所进行的运算，它为解方程、建模现实世界关系以及理解数学中的对称性提供了强大的工具。

1. Definition and the Horizontal Line Test | 定义与水平线检验

A function f maps each element x from its domain to a unique element y in its range. The inverse function f⁻¹ reverses this mapping: it takes y back to x. However, for an inverse to exist, the original function must be one-to-one (injective), meaning each y-value corresponds to exactly one x-value.

函数 f 将定义域中的每个元素 x 映射到值域中的唯一元素 y。反函数 f⁻¹ 则逆转这一映射：它将 y 映射回 x。然而，要存在反函数，原函数必须是单射（一一对应）的，即每个 y 值恰好对应一个 x 值。

The horizontal line test is a quick graphical method to determine whether a function has an inverse. If any horizontal line intersects the graph of the function at more than one point, the function is not one-to-one, and its inverse is not a function (unless the domain is restricted).

水平线检验是一种快速的图形方法，用于判断函数是否存在反函数。如果任何一条水平线与函数图像相交于多于一个点，则该函数不是一一对应的，其“反函数”就不是函数（除非限制定义域）。

Horizontal Line Test: f is one-to-one ⇔ every horizontal line intersects the graph at most once.

水平线检验：f 是单射 ⇔ 每条水平线与图像至多相交一次。

For example, f(x) = x² fails this test because y = 4 corresponds to both x = 2 and x = -2. Restricting the domain to x ≥ 0 makes it one-to-one, allowing us to define the inverse function f⁻¹(x) = √x.

例如，f(x) = x² 不通过此检验，因为 y = 4 同时对应 x = 2 和 x = -2。将定义域限制为 x ≥ 0 后，它就成为单射，从而可以定义反函数 f⁻¹(x) = √x。

2. Step-by-Step Algebraic Method | 代数求解逐步方法

To find the inverse of a function algebraically, we follow a systematic procedure that works for most IB-level functions.

要求一个函数的反函数，我们遵循一套系统化的步骤，这套方法适用于IB水平的大多数函数。

Step 1: Write y = f(x). 步骤一：写出 y = f(x)。

Step 2: Swap x and y in the equation, so x now becomes a function of y. 步骤二：在方程中交换 x 和 y，使 x 成为 y 的函数。

Step 3: Solve for y in terms of x. 步骤三：解出用 x 表示的 y。

Step 4: Replace y with f⁻¹(x) and state the domain of the inverse function. 步骤四：将 y 替换为 f⁻¹(x)，并注明反函数的定义域。

Let’s illustrate with f(x) = 2x + 3. First, y = 2x + 3. Swapping gives x = 2y + 3. Solving for y: 2y = x − 3, so y = (x − 3)/2. Thus f⁻¹(x) = (x − 3)/2.

我们用 f(x) = 2x + 3 来示范。首先 y = 2x + 3。交换后得到 x = 2y + 3。解出 y：2y = x − 3，所以 y = (x − 3)/2。因此 f⁻¹(x) = (x − 3)/2。

For a rational function such as f(x) = (2x + 1)/(x − 2), the same method applies. Set y = (2x + 1)/(x − 2), swap to get x = (2y + 1)/(y − 2), then multiply both sides by (y − 2): x(y − 2) = 2y + 1. Expanding: xy − 2x = 2y + 1. Rearranging terms with y: xy − 2y = 2x + 1. Factor out y: y(x − 2) = 2x + 1. Therefore y = (2x + 1)/(x − 2), which interestingly is the original function itself — a self-inverse function.

对于有理函数，如 f(x) = (2x + 1)/(x − 2)，同样适用此方法。设 y = (2x + 1)/(x − 2)，交换得 x = (2y + 1)/(y − 2)，两边乘以 (y − 2)：x(y − 2) = 2y + 1。展开：xy − 2x = 2y + 1。整理含 y 的项：xy − 2y = 2x + 1。提取公因子 y：y(x − 2) = 2x + 1。因此 y = (2x + 1)/(x − 2)，有趣的是它恰好是原函数本身——这是一个自反函数。

3. Domain and Range: A Critical Pair | 定义域与值域：关键对应关系

One of the most frequently tested concepts in IB exams is the relationship between the domain and range of a function and its inverse. This relationship is beautifully reciprocal:

IB考试中最常考的概念之一是函数与其反函数在定义域和值域上的对应关系。这种关系具有完美的互逆性：

Domain of f⁻¹ = Range of f, and Range of f⁻¹ = Domain of f

f⁻¹ 的定义域 = f 的值域，f⁻¹ 的值域 = f 的定义域

This means that when we find an inverse, we automatically know its domain: it is simply the range of the original function. This is especially useful for functions with square roots, logarithms, or trigonometric functions where domains must be carefully restricted.

这意味着当我们找到反函数时，就自动知道了它的定义域：就是原函数的值域。这对含有平方根、对数或三角函数的函数尤其有用，因为这类函数的定义域必须仔细限制。

Consider f(x) = √(x − 3), defined on x ≥ 3. Its range is y ≥ 0. Following the steps: y = √(x − 3), swap to get x = √(y − 3), square both sides: x² = y − 3, hence y = x² + 3. The inverse is f⁻¹(x) = x² + 3, but its domain is x ≥ 0, which matches the range of the original function.

考虑 f(x) = √(x − 3)，定义在 x ≥ 3 上。它的值域是 y ≥ 0。按步骤操作：y = √(x − 3)，交换得 x = √(y − 3)，两边平方：x² = y − 3，因此 y = x² + 3。反函数是 f⁻¹(x) = x² + 3，但其定义域为 x ≥ 0，与原函数的值域一致。

Students often make the mistake of stating the domain of the inverse based on its algebraic expression alone, forgetting that the domain must be restricted to the range of the original function. In the example above, x² + 3 is defined for all real x, but as the inverse of √(x − 3), its domain is only x ≥ 0.

学生常犯的错误是仅根据代数表达式来确定反函数的定义域，而忽略了定义域必须限制在原函数的值域内。以上例来说，x² + 3 对所有实数 x 都有定义，但作为 √(x − 3) 的反函数，它的定义域只能是 x ≥ 0。

4. Graphical Symmetry: The Line y = x | 图像对称性：直线 y = x

The graph of an inverse function has a beautiful geometric relationship with the graph of the original function: they are reflections of each other across the line y = x. This symmetry provides both a visual verification tool and a deep insight into the nature of inverse operations.

反函数的图像与原函数的图像之间存在优美的几何关系：它们关于直线 y = x 互为镜像。这种对称性既提供了直观的检验工具，也让我们深入理解逆运算的本质。

If a point (a, b) lies on the graph of f, then the point (b, a) must lie on the graph of f⁻¹. This follows directly from the definition: if f(a) = b, then f⁻¹(b) = a. Every point reflected across y = x swaps its coordinates, and this is precisely the algebraic operation of swapping x and y that we perform when finding an inverse.

如果点 (a, b) 在 f 的图像上，那么点 (b, a) 必然在 f⁻¹ 的图像上。这直接源于定义：如果 f(a) = b，那么 f⁻¹(b) = a。每个点关于 y = x 反射后都会交换其坐标，这正是我们求反函数时交换 x 和 y 的代数操作。

This symmetry has practical applications. When sketching the graph of an inverse function in an exam, you can reflect key points of the original graph across the line y = x. Points of intersection with the line y = x remain fixed during reflection, which is why f(x) = x can only occur at points where f(x) = f⁻¹(x).

这种对称性有实际应用。在考试中画反函数图像时，你可以将原图像的关键点关于直线 y = x 进行反射。与直线 y = x 的交点在反射中保持不变，这就是为什么 f(x) = x 只能出现在 f(x) = f⁻¹(x) 的点上。

5. Composite Functions and Cancellation | 复合函数与抵消性质

Perhaps the most elegant property of inverse functions is expressed through composition. When a function and its inverse are composed in either order, they cancel each other out, yielding the identity function.

反函数最优雅的性质或许体现在复合运算中。当一个函数与其反函数以任意顺序复合时，它们会相互抵消，得到恒等函数。

f⁻¹(f(x)) = x for all x in the domain of f, and f(f⁻¹(x)) = x for all x in the domain of f⁻¹

对 f 定义域中的所有 x，f⁻¹(f(x)) = x；对 f⁻¹ 定义域中的所有 x，f(f⁻¹(x)) = x

These two identities are the algebraic essence of what it means for two functions to be inverses. They provide a powerful verification method: after computing f⁻¹(x), substitute it into f and simplify. If you don’t get x, something has gone wrong in your algebra.

这两个恒等式是两个函数互逆的代数本质。它们提供了一种强大的验证方法：计算出 f⁻¹(x) 之后，将其代入 f 并化简。如果得不到 x，说明代数运算中出了问题。

For example, verify that the inverse of f(x) = eˣ is f⁻¹(x) = ln x. Check: f⁻¹(f(x)) = ln(eˣ) = x ✓, and f(f⁻¹(x)) = e^(ln x) = x (for x > 0) ✓.

例如，验证 f(x) = eˣ 的反函数是 f⁻¹(x) = ln x。检验：f⁻¹(f(x)) = ln(eˣ) = x ✓，且 f(f⁻¹(x)) = e^(ln x) = x（对 x > 0）✓。

In IB exam questions, you may be asked to find f⁻¹ given f, or conversely, you may be given relationships like f(g(x)) = x and asked to identify g as the inverse. Recognizing these patterns quickly is key to efficient problem-solving in timed conditions.

在IB考试题目中，你可能会被要求根据 f 求 f⁻¹，或者反过来，题目给出类似 f(g(x)) = x 的关系，要求你识别出 g 就是反函数。在限时条件下，快速识别这些模式是高效解题的关键。

6. Exponential and Logarithmic Functions | 指数函数与对数函数

The most important pair of inverse functions in the IB syllabus is undoubtedly the exponential and logarithmic functions. The natural exponential function eˣ and the natural logarithm ln x are inverses of each other. More generally, aˣ and logₐx are inverses.

IB课程大纲中最重要的一对反函数无疑是指数函数和对数函数。自然指数函数 eˣ 和自然对数 ln x 互为反函数。更一般地，aˣ 和 logₐx 互为反函数。

This inverse relationship underpins many IB exam problems. When solving eˣ = 5, we apply ln to both sides to “undo” the exponential: x = ln 5. Similarly, when solving ln x = 3, we apply eˣ to both sides: x = e³. This is the essence of logarithmic and exponential equations.

这种互逆关系支撑着许多IB考试题目。解方程 eˣ = 5 时，我们对两边取 ln 来“撤销”指数：x = ln 5。类似地，解 ln x = 3 时，我们对两边取 eˣ：x = e³。这就是对数和指数方程的本质。

For other bases, the relationship is equally important. The equation 2ˣ = 10 can be solved by taking log₂ of both sides: x = log₂10, or by rewriting in terms of natural logs: x = ln 10 / ln 2. Both approaches rely on inverse function properties.

对于其他底数，这种关系同样重要。方程 2ˣ = 10 可以通过两边取 log₂ 来求解：x = log₂10，或者用自然对数改写：x = ln 10 / ln 2。两种方法都依赖于反函数性质。

A common IB question type involves finding the inverse of a composite exponential expression. For example, find the inverse of f(x) = 2e³ˣ − 1. Working step by step: y = 2e³ˣ − 1, swap: x = 2e³ʸ − 1, rearrange: 2e³ʸ = x + 1, so e³ʸ = (x + 1)/2. Taking natural logs: 3y = ln((x + 1)/2), hence f⁻¹(x) = (1/3)ln((x + 1)/2), with domain x > −1.

一种常见的IB题型是求复合指数表达式的反函数。例如，求 f(x) = 2e³ˣ − 1 的反函数。逐步求解：y = 2e³ˣ − 1，交换：x = 2e³ʸ − 1，整理：2e³ʸ = x + 1，所以 e³ʸ = (x + 1)/2。取自然对数：3y = ln((x + 1)/2)，因此 f⁻¹(x) = (1/3)ln((x + 1)/2)，定义域为 x > −1。

7. Trigonometric Inverse Functions | 三角函数的反函数

Inverse trigonometric functions, also known as arc functions, are another important category in IB Mathematics. Since trigonometric functions are periodic and therefore not one-to-one over their entire domains, we must restrict their domains to define inverses.

反三角函数（也称为弧函数）是IB数学中的另一重要类别。由于三角函数是周期函数，在其整个定义域上不是单射，因此我们必须限制其定义域来定义反函数。

The standard restricted domains are as follows: for y = sin x, the restricted domain is −π/2 ≤ x ≤ π/2, giving the inverse arcsin x with range −π/2 ≤ arcsin x ≤ π/2. For y = cos x, the restricted domain is 0 ≤ x ≤ π, giving arccos x with range 0 ≤ arccos x ≤ π. For y = tan x, the restricted domain is −π/2 < x < π/2, giving arctan x with range −π/2 < arctan x < π/2.

标准的受限定义域如下：对于 y = sin x，受限定义域为 −π/2 ≤ x ≤ π/2，反函数 arcsin x 的值域为 −π/2 ≤ arcsin x ≤ π/2。对于 y = cos x，受限定义域为 0 ≤ x ≤ π，反函数 arccos x 的值域为 0 ≤ arccos x ≤ π。对于 y = tan x，受限定义域为 −π/2 < x < π/2，反函数 arctan x 的值域为 −π/2 < arctan x < π/2。

IB exam questions often require students to evaluate expressions such as sin(arctan x) or cos(arcsin(3/5)). These can be solved by drawing a right-angled triangle. For sin(arctan x): let θ = arctan x, so tan θ = x = opposite/adjacent. If opposite = x and adjacent = 1, then the hypotenuse is √(x² + 1), and sin(arctan x) = opposite/hypotenuse = x/√(x² + 1).

IB考试题通常要求学生计算如 sin(arctan x) 或 cos(arcsin(3/5)) 这样的表达式。这些可以通过构造直角三角形来求解。对于 sin(arctan x)：设 θ = arctan x，则 tan θ = x = 对边/邻边。如果对边 = x，邻边 = 1，那么斜边为 √(x² + 1)，于是 sin(arctan x) = 对边/斜边 = x/√(x² + 1)。

8. Real-World Applications | 实际应用

Inverse functions are not merely abstract mathematical constructs; they have concrete applications across science, economics, and engineering. Understanding these applications is particularly important for students following the Applications and Interpretation pathway.

反函数不仅仅是抽象的数学构造；它们在科学、经济学和工程学中具有具体的应用。理解这些应用对于学习应用与解释路径的学生尤为重要。

Temperature conversion: The formula F(C) = (9/5)C + 32 converts Celsius to Fahrenheit. Its inverse, C(F) = (5/9)(F − 32), converts Fahrenheit back to Celsius. Both functions are inverses of each other, and the conversion point where F = C = −40 satisfies f(−40) = −40.

温度换算：公式 F(C) = (9/5)C + 32 将摄氏温度转换为华氏温度。其反函数 C(F) = (5/9)(F − 32) 将华氏度转换回摄氏度。两个函数互为反函数，且在 F = C = −40 处满足 f(−40) = −40，即两种温标读数相同的点。

Compound interest: The formula A = P(1 + r)ᵗ gives the amount A after t years with initial principal P and annual interest rate r. The inverse function t = ln(A/P) / ln(1 + r) tells us how many years are needed to reach a target amount. This is a fundamental tool in financial mathematics and appears in IB AI Financial Mathematics topics.

复利计算：公式 A = P(1 + r)ᵗ 给出本金 P 在年利率 r 下经过 t 年后的本息总额 A。反函数 t = ln(A/P) / ln(1 + r) 告诉我们达到目标金额需要多少年。这是金融数学的基础工具，出现在IB AI的金融数学主题中。

Physics and decay: The radioactive decay formula N(t) = N₀e^(−kt) models the amount of radioactive material remaining after time t. Its inverse, t = −(1/k)ln(N/N₀), allows scientists to determine the age of a sample based on the remaining proportion of radioactive isotope — this is the principle behind carbon dating.

物理与衰变：放射性衰变公式 N(t) = N₀e^(−kt) 模拟了 t 时间后剩余的放射性物质数量。其反函数 t = −(1/k)ln(N/N₀) 使科学家能够根据剩余放射性同位素的比例来确定样本年龄——这就是碳定年法的原理。

Economics: The demand function Q = D(P) relates quantity demanded to price. The inverse demand function P = D⁻¹(Q) expresses price as a function of quantity, which is often more useful in microeconomic analysis when drawing supply and demand diagrams.

经济学：需求函数 Q = D(P) 将需求量与价格联系起来。反需求函数 P = D⁻¹(Q) 将价格表示为数量的函数，在绘制供需图时，这种形式在微观经济分析中往往更有用。

9. Common Mistakes and Exam Pitfalls | 常见错误与考试陷阱

Students frequently lose marks on inverse function questions due to several recurring errors. Being aware of these pitfalls is the first step toward avoiding them in the exam.

学生在反函数题目中经常因几个反复出现的错误而失分。了解这些陷阱是在考试中避免它们的第一步。

Forgetting the horizontal line test: Attempting to find an inverse for a function that is not one-to-one without restricting the domain. For instance, writing f⁻¹(x) = ±√x for f(x) = x² without stating x ≥ 0.
忽略水平线检验：在未限制定义域的情况下，试图对不是单射的函数求反函数。例如，在未声明 x ≥ 0 的情况下，写出 f(x) = x² 的“反函数” f⁻¹(x) = ±√x。
Incorrect domain of the inverse: Stating the domain of f⁻¹ based on its formula rather than the range of f. Always check that the domain of f⁻¹ equals the range of f.
反函数定义域错误：根据公式而非 f 的值域来确定 f⁻¹ 的定义域。务必检查 f⁻¹ 的定义域是否等于 f 的值域。
Algebraic errors when swapping: Making mistakes when rearranging equations after swapping x and y, especially in rational or square root functions. Always verify your answer using the identity f⁻¹(f(x)) = x.
交换后的代数错误：在交换 x 和 y 后整理方程时出错，尤其是在有理函数或平方根函数中。始终使用恒等式 f⁻¹(f(x)) = x 验证答案。
Confusing reciprocal with inverse: Writing f⁻¹(x) = 1/f(x). This is one of the most common misconceptions. The notation f⁻¹ means inverse function, not reciprocal.
将倒数与反函数混淆：写成 f⁻¹(x) = 1/f(x)。这是最常见的误解之一。记号 f⁻¹ 表示反函数，而不是倒数。
Forgetting self-inverse functions: Some functions like f(x) = 1/x and f(x) = (ax + b)/(cx − a) satisfy f⁻¹(x) = f(x). Not recognising these can lead to unnecessary computation errors.
忘记自反函数：某些函数如 f(x) = 1/x 和 f(x) = (ax + b)/(cx − a) 满足 f⁻¹(x) = f(x)。未能识别这些会导致不必要的计算错误。

10. Worked Examples from IB-Style Questions | IB风格例题精解

Let us work through a few representative problems that mirror the style and difficulty of IB exam questions involving inverse functions.

让我们演练几道具有代表性的题目，这些题目反映了IB考试中反函数问题的风格和难度。

Example 1: The function f is defined by f(x) = (x − 2)/3 for x ∈ ℝ. Find f⁻¹(x) and verify your answer.

例题1：函数 f 定义为 f(x) = (x − 2)/3，x ∈ ℝ。求 f⁻¹(x) 并验证答案。

Solution: Let y = (x − 2)/3. Swap x and y: x = (y − 2)/3. Multiply by 3: 3x = y − 2. Hence y = 3x + 2, so f⁻¹(x) = 3x + 2. Verification: f⁻¹(f(x)) = 3((x − 2)/3) + 2 = (x − 2) + 2 = x ✓.

解答：设 y = (x − 2)/3。交换 x 和 y：x = (y − 2)/3。两边乘以 3：3x = y − 2。因此 y = 3x + 2，所以 f⁻¹(x) = 3x + 2。验证：f⁻¹(f(x)) = 3((x − 2)/3) + 2 = (x − 2) + 2 = x ✓。

Example 2: Let f(x) = 4ˣ + 1. Find f⁻¹(17).

例题2：设 f(x) = 4ˣ + 1。求 f⁻¹(17)。

Solution: We need to find x such that f(x) = 17. Solve 4ˣ + 1 = 17, so 4ˣ = 16. Since 16 = 4², we have x = 2. Therefore f⁻¹(17) = 2. No need to derive the full inverse function — simply solve the equation f(x) = 17.

解答：我们需要找到 x 使得 f(x) = 17。解方程 4ˣ + 1 = 17，得 4ˣ = 16。由于 16 = 4²，所以 x = 2。因此 f⁻¹(17) = 2。无需推导完整反函数——只需解方程 f(x) = 17 即可。

Example 3: A function is defined by f(x) = √(2x − 1) for x ≥ 1/2. Find f⁻¹(x) and state its domain.

例题3：函数定义为 f(x) = √(2x − 1)，其中 x ≥ 1/2。求 f⁻¹(x) 并指出其定义域。

Solution: Let y = √(2x − 1). Swap: x = √(2y − 1). Square: x² = 2y − 1. Thus 2y = x² + 1, so y = (x² + 1)/2. The range of the original function is y ≥ 0, so the domain of f⁻¹ is x ≥ 0. Hence f⁻¹(x) = (x² + 1)/2, for x ≥ 0.

解答：设 y = √(2x − 1)。交换：x = √(2y − 1)。两边平方：x² = 2y − 1。因此 2y = x² + 1，即 y = (x² + 1)/2。原函数的值域为 y ≥ 0，所以 f⁻¹ 的定义域为 x ≥ 0。因此 f⁻¹(x) = (x² + 1)/2，x ≥ 0。

11. Inverse Functions in Calculus | 微积分中的反函数

For students taking Mathematics: Analysis and Approaches at Higher Level, the relationship between a function and its inverse extends into differential calculus through an important theorem.

对于学习数学：分析与方法高级水平的学生来说，函数与其反函数之间的关系通过一个重要定理延伸到了微分学中。

If g = f⁻¹, then g′(x) = 1 / f′(g(x)), provided f′(g(x)) ≠ 0

如果 g = f⁻¹，则 g′(x) = 1 / f′(g(x))，前提是 f′(g(x)) ≠ 0

This derivative formula for inverse functions is derived from the chain rule applied to the identity f(f⁻¹(x)) = x. Differentiating both sides gives f′(f⁻¹(x)) · (f⁻¹)′(x) = 1, from which the result follows immediately.

这个反函数求导公式是对恒等式 f(f⁻¹(x)) = x 应用链式法则推导出来的。两边同时求导得到 f′(f⁻¹(x)) · (f⁻¹)′(x) = 1，结果立即可得。

For example, to find the derivative of g(x) = ln x, which is the inverse of f(x) = eˣ, we use the formula: g′(x) = 1/f′(g(x)) = 1/e^(ln x) = 1/x. This elegantly recovers the well-known derivative of the natural logarithm.

例如，求 g(x) = ln x（即 f(x) = eˣ 的反函数）的导数时，使用公式：g′(x) = 1/f′(g(x)) = 1/e^(ln x) = 1/x。这就优雅地恢复了自然对数的著名导数公式。

Similarly, the derivative of arcsin x can be derived: if f(x) = sin x on [−π/2, π/2], then (arcsin x)′ = 1/cos(arcsin x) = 1/√(1 − x²). This formula appears frequently in AA HL integration problems involving trigonometric substitutions.

类似地，可以推导 arcsin x 的导数：如果 f(x) = sin x 在 [−π/2, π/2] 上，则 (arcsin x)′ = 1/cos(arcsin x) = 1/√(1 − x²)。这个公式在AA高级水平涉及三角换元的积分问题中频繁出现。

12. Summary and Exam Strategy | 总结与考试策略

Inverse functions are a unifying theme in IB Mathematics, connecting algebra, geometry, and calculus. Mastery of this topic requires both procedural fluency and conceptual understanding.

反函数是IB数学中一个统一的主题，将代数、几何与微积分联系在一起。掌握这一主题既需要程序性熟练度，也需要概念性理解。

Here is a concise checklist to guide your exam preparation and problem-solving:

以下是一份简洁的清单，用于指导你的备考和解题：

Always check one-to-one property using the horizontal line test before finding an inverse.
在求反函数之前，始终使用水平线检验来判断单射性质。
Follow the four-step algebraic procedure systematically: substitute, swap, solve, standardise.
系统性地遵循四步代数流程：代入、交换、求解、规范化。
State the domain of f⁻¹ as the range of f, never as an afterthought.
将 f⁻¹ 的定义域表述为 f 的值域，不要事后才想起。
Use composition to verify your answer quickly and reliably.
使用复合运算快速可靠地验证答案。
Remember the graphical symmetry across y = x and use it for sketching.
记住关于 y = x 的图形对称性，并使用它来画图。
For trigonometric inverses, memorise the restricted domains and ranges.
对于反三角函数，牢记受限定义域和值域。
In calculus contexts, apply the inverse function derivative formula correctly.
在微积分情境中，正确应用反函数求导公式。

With consistent practice on past paper questions and a solid grasp of the underlying principles, you will find inverse function problems to be some of the most rewarding marks available in the IB Mathematics exam.

通过在真题上的持续练习以及对基本原理的扎实掌握，你会发现反函数题目是IB数学考试中最值得拿分的一类题目。

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Inverse Functions f⁻¹(x): Solving and Applications in IB Maths | IB数学：反函数f⁻¹(x)的求解与应用