A-Level Maths: Finding Inverse Functions and Their Graphical Features | A-Level 数学：反函数的求法与图像特征

📚 A-Level Maths: Finding Inverse Functions and Their Graphical Features | A-Level 数学：反函数的求法与图像特征

An inverse function reverses the effect of the original function. If f maps x to y, then its inverse f⁻¹ maps y back to x. Understanding how to find inverses algebraically and interpret their graphs is a core skill in A-Level Mathematics.

反函数的作用是逆转原函数的效果。如果 f 把 x 映射到 y，那么它的反函数 f⁻¹ 就把 y 映射回 x。掌握如何用代数方法求反函数并理解其图像特征，是 A-Level 数学中的核心技能。

1. Definition of an Inverse Function | 反函数的定义

For a function f with domain A and range B, the inverse function f⁻¹ is defined by: f⁻¹(y) = x if and only if f(x) = y. The domain of f⁻¹ is B and the range of f⁻¹ is A.

对于定义域为 A、值域为 B 的函数 f，反函数 f⁻¹ 定义为：当且仅当 f(x) = y 时，f⁻¹(y) = x。f⁻¹ 的定义域是 B，值域是 A。

In A-Level notation, we often write y = f(x) and then solve for x in terms of y to find f⁻¹(y). Finally, replace y with x in the final expression.

在 A-Level 的记号中，我们通常先写出 y = f(x)，然后解出用 y 表示的 x，从而得到 f⁻¹(y)。最后，将表达式中的 y 替换为 x。

2. Conditions for Existence | 反函数存在的条件

A function must be one-to-one (injective) to have an inverse. This means every output value corresponds to exactly one input value. Graphically, any horizontal line must intersect the curve at most once.

函数必须是一一对应（单射）才有反函数。这意味着每个输出值恰好对应一个输入值。从图像上看，任何水平直线与曲线的交点至多只有一个。

If a function is not one-to-one over its natural domain, it can be made invertible by restricting the domain. For example, f(x) = x² is not one-to-one on ℝ, but f(x) = x² with x ≥ 0 is one-to-one.

如果函数在其自然定义域上不是一一对应，可以通过限制定义域使其可逆。例如，f(x) = x² 在 ℝ 上不是一一对应，但限制 x ≥ 0 后 f(x) = x² 是一一对应的。

3. Horizontal Line Test | 水平线测试

The horizontal line test is a visual way to check whether a function has an inverse. If any horizontal line y = c intersects the graph of y = f(x) more than once, then f is not one-to-one and has no inverse over that domain.

水平线测试是一种判断函数是否存在反函数的可视化方法。如果任何水平直线 y = c 与 y = f(x) 的图像相交多于一次，那么 f 就不是一一对应，因而在该定义域上没有反函数。

For example, the graph of y = x³ passes the horizontal line test, so it has an inverse. The graph of y = sin x does not pass it over its whole domain, so we restrict the domain to [-π/2, π/2] to define its inverse.

例如，y = x³ 的图像通过水平线测试，因此它有反函数。y = sin x 的图像在整个定义域上不能通过该测试，因此我们将其定义域限制为 [-π/2, π/2] 来定义其反函数。

4. Algebraic Method: Step-by-Step | 代数求法：逐步步骤

To find the inverse of a function y = f(x):

求函数 y = f(x) 的反函数，步骤如下：

Step 1: Write the function as y = f(x).

第一步：将函数写成 y = f(x) 的形式。
Step 2: Swap x and y in the equation.

第二步：在方程中交换 x 和 y。
Step 3: Solve this new equation for y.

第三步：解这个新方程，求出 y。
Step 4: Replace y by f⁻¹(x) and state the domain of f⁻¹, which is the range of f.

第四步：将 y 替换为 f⁻¹(x)，并写出 f⁻¹ 的定义域，即 f 的值域。

Worked example: Find the inverse of f(x) = (2x + 3)/(x – 1), x ≠ 1.

例题：求 f(x) = (2x + 3)/(x – 1), x ≠ 1 的反函数。

y = (2x + 3)/(x – 1) → x = (2y + 3)/(y – 1)

Multiply both sides by (y – 1): x(y – 1) = 2y + 3, so xy – x = 2y + 3. Rearrange to collect y terms: xy – 2y = x + 3, so y(x – 2) = x + 3. Hence y = (x + 3)/(x – 2).

两边乘以 (y – 1)：x(y – 1) = 2y + 3，即 xy – x = 2y + 3。整理含 y 的项：xy – 2y = x + 3，所以 y(x – 2) = x + 3。于是 y = (x + 3)/(x – 2)。

Thus f⁻¹(x) = (x + 3)/(x – 2), x ≠ 2. The domain x ≠ 2 comes from the range of f, which is all real numbers except 2.

因此 f⁻¹(x) = (x + 3)/(x – 2), x ≠ 2。定义域 x ≠ 2 来自 f 的值域，即除 2 以外的所有实数。

5. Graphical Symmetry about y = x | 关于直线 y = x 的对称性

The graph of y = f⁻¹(x) is a reflection of the graph of y = f(x) in the line y = x. This is because the point (a, b) lies on f if and only if the point (b, a) lies on f⁻¹.

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 f 上当且仅当点 (b, a) 在 f⁻¹ 上。

When drawing an inverse function, always sketch the line y = x first, then reflect the original curve across it. Pay attention to intercepts: the x-intercept of f becomes the y-intercept of f⁻¹, and vice versa.

绘制反函数图像时，先画出直线 y = x，然后将原曲线关于它反射。注意截距：f 的 x 截距变为 f⁻¹ 的 y 截距，反之亦然。

6. Domain and Range Interchange | 定义域与值域互换

If f has domain A and range B, then f⁻¹ has domain B and range A. This property is essential for determining the valid domain of an inverse function and for solving equations involving inverse trigonometric functions.

如果 f 的定义域为 A，值域为 B，那么 f⁻¹ 的定义域为 B，值域为 A。这一性质对于确定反函数的有效定义域以及求解涉及反三角函数的方程至关重要。

For example, f(x) = eˣ has domain ℝ and range (0, ∞). Therefore f⁻¹(x) = ln x has domain (0, ∞) and range ℝ.

例如，f(x) = eˣ 的定义域为 ℝ，值域为 (0, ∞)。因此 f⁻¹(x) = ln x 的定义域为 (0, ∞)，值域为 ℝ。

7. Composite Functions and Inverses | 复合函数与反函数

If f and g are inverse functions, then f(g(x)) = x and g(f(x)) = x for all x in the appropriate domains. This provides a powerful verification method.

如果 f 和 g 互为反函数，那么在相应的定义域内，f(g(x)) = x 且 g(f(x)) = x。这提供了一种强大的验证方法。

In general, (f⁻¹)⁻¹ = f. Also, if f and g are both invertible, then (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹. Note the order of composition reverses.

一般地，(f⁻¹)⁻¹ = f。另外，如果 f 和 g 都可逆，那么 (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹。注意复合的顺序会反转。

8. Inverse Trigonometric Functions | 反三角函数

A-Level exams often require inverse trigonometric functions. Since sin x, cos x and tan x are periodic, we restrict their domains to make them one-to-one:

A-Level 考试常涉及反三角函数。由于 sin x、cos x 和 tan x 是周期函数，我们限制其定义域使其一一对应：

Function	Restricted Domain	Range of Inverse
sin⁻¹ x	x ∈ [-1, 1]	[-π/2, π/2]
cos⁻¹ x	x ∈ [-1, 1]	[0, π]
tan⁻¹ x	x ∈ ℝ	(-π/2, π/2)

The graph of y = sin⁻¹ x is the reflection of y = sin x with domain [-π/2, π/2] about the line y = x. Memorise the standard shapes and ranges to avoid domain mistakes.

y = sin⁻¹ x 的图像是 y = sin x（定义域为 [-π/2, π/2]）关于直线 y = x 的反射。牢记标准图形和值域，避免定义域错误。

9. Transforming f(x) to f⁻¹(x) Graphically | 从 f(x) 的图像得到 f⁻¹(x) 的图像

To sketch y = f⁻¹(x) from y = f(x):

要从 y = f(x) 的图像画出 y = f⁻¹(x) 的图像：

Draw the line y = x.

画出直线 y = x。
Reflect the graph of y = f(x) in this line.

将 y = f(x) 的图像关于这条直线反射。
Mark the new intercepts and asymptotes: vertical asymptotes become horizontal asymptotes, and horizontal asymptotes become vertical ones.

标记新的截距和渐近线：垂直渐近线变为水平渐近线，水平渐近线变为垂直渐近线。

For example, f(x) = ln(x – 2) has a vertical asymptote x = 2 and x-intercept at (3, 0). Its inverse f⁻¹(x) = eˣ + 2 has a horizontal asymptote y = 2 and y-intercept at (0, 3).

例如，f(x) = ln(x – 2) 有垂直渐近线 x = 2，x 截距为 (3, 0)。其反函数 f⁻¹(x) = eˣ + 2 有水平渐近线 y = 2，y 截距为 (0, 3)。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Mistake 1: Forgetting to state the domain of f⁻¹. Always write the domain explicitly; it is often tested.

错误一：忘记写出 f⁻¹ 的定义域。务必显式写出定义域，这经常是考点。

Mistake 2: Trying to find the inverse of a function that is not one-to-one without restricting the domain. Always check the horizontal line test first.

错误二：未限制定义域就去求非一一对应函数的反函数。先检查水平线测试。

Mistake 3: Incorrectly swapping variables in composite inverse problems. Remember: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹, not g⁻¹ ∘ f⁻¹.

错误三：在复合反函数问题中错误地交换变量。记住：(g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹，而不是 g⁻¹ ∘ f⁻¹。

Exam Tip: After finding f⁻¹, verify by computing f(f⁻¹(x)) and simplifying to x. This catches algebra errors.

考试技巧：求出 f⁻¹ 后，通过计算 f(f⁻¹(x)) 并化简为 x 来验证。这能发现代数错误。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

A-Level Maths: Finding Inverse Functions and Their Graphical Features | A-Level 数学：反函数的求法与图像特征