Mastering Functions: Domain, Range, and Transformations | 函数精通：定义域、值域与变换

📚 Mastering Functions: Domain, Range, and Transformations | 函数精通：定义域、值域与变换

Functions are one of the most fundamental topics in A-Level Mathematics. A clear understanding of domain, range, and transformations is essential for solving problems across algebra, calculus, and even applied mathematics. This article breaks down the key concepts step by step, ensuring you are fully prepared for your AQA A-Level exam.

函数是 A-Level 数学中最基础也最重要的主题之一。清晰地理解定义域、值域以及图像变换，对于解决代数、微积分乃至应用数学中的各类问题都至关重要。本文将逐步拆解这些核心概念，帮助你为 AQA A-Level 考试做好充分准备。

1. What is a Function? | 什么是函数？

A function is a relation that maps each input value (x) to exactly one output value (y). We often write this as y = f(x). The key property is uniqueness: for a given input, there is only one output. For example, f(x) = x² maps both 2 and -2 to 4, but each input still has exactly one output, so it is a valid function.

函数是一种关系，它将每一个输入值 (x) 唯一地映射到一个输出值 (y)。我们通常写作 y = f(x)。其关键性质是唯一性：对于给定的输入，只有一个输出。例如，f(x) = x² 将 2 和 -2 都映射到 4，但每个输入仍然只有一个输出，因此它是一个合法的函数。

It is important to distinguish functions from general relations. A relation like x² + y² = 1 (a circle) is not a function because a single x value (e.g., x = 0) corresponds to two y values (y = 1 and y = -1). To check if a graph represents a function, we use the vertical line test.

区分函数与一般关系非常重要。像 x² + y² = 1（圆）这样的关系就不是函数，因为一个 x 值（例如 x = 0）对应两个 y 值（y = 1 和 y = -1）。要检查一幅图是否表示函数，我们使用“竖线检验法”。

Key notation: f: x ↦ f(x) means the function f maps x to f(x). The set of all possible inputs is called the domain, and the set of all resulting outputs is called the range.

关键记号： f: x ↦ f(x) 表示函数 f 将 x 映射到 f(x)。所有可能输入的集合称为定义域，所有由此产生的输出的集合称为值域。

2. Domain: The Set of Valid Inputs | 定义域：有效输入的集合

The domain of a function is the complete set of x-values for which the function is defined. In pure mathematics, the domain is often all real numbers, but certain restrictions apply. You must be able to identify these restrictions quickly and accurately.

函数的定义域是函数有意义的全部 x 值的集合。在纯数学中，定义域通常是所有实数，但在某些情况下存在限制。你必须能够快速且准确地识别这些限制。

The two most common restrictions are:

Division by zero: the denominator of a fraction cannot be zero. For f(x) = 1/(x – 3), the domain is all real numbers except x = 3.
分母为零：分数的分母不能为零。对于 f(x) = 1/(x – 3)，定义域是所有实数，除了 x = 3。
Even roots of negative numbers: the square root (and other even roots) of a negative number is not a real number. For f(x) = √(2x + 6), we require 2x + 6 ≥ 0, so x ≥ -3.
偶数次根号下为负数：负数的平方根（以及其他偶数次根）不是实数。对于 f(x) = √(2x + 6)，我们要求 2x + 6 ≥ 0，即 x ≥ -3。

Logarithmic functions also impose a restriction: the argument of a logarithm must be strictly positive. For f(x) = ln(x – 1), the domain is x > 1. Similarly, the domain of g(x) = log₂(x² – 4) requires x² – 4 > 0, giving x < -2 or x > 2.

对数函数也有一个限制：对数的真数必须严格为正。对于 f(x) = ln(x – 1)，定义域是 x > 1。类似地，g(x) = log₂(x² – 4) 的定义域要求 x² – 4 > 0，即 x < -2 或 x > 2。

When a function is modelled on a real-world situation, the domain may be further restricted by context. For instance, if a function represents the height of a ball after t seconds, we only consider t ≥ 0, and up until the ball hits the ground.

当函数用于模拟现实世界情境时，定义域可能受实际背景的进一步限制。例如，如果一个函数表示球在 t 秒后的高度，我们只考虑 t ≥ 0，并且直到球落地为止。

3. Range: The Set of Possible Outputs | 值域：可能输出的集合

The range is the set of all y-values that the function can actually produce. Finding the range often involves understanding the shape of the graph, especially the vertex of a parabola or the asymptotes of a rational function.

值域是函数实际上能产生的所有 y 值的集合。求值域通常需要理解图像的形状，特别是抛物线的顶点或有理函数的渐近线。

For a quadratic function in completed square form, f(x) = a(x – h)² + k, the turning point is (h, k). If a > 0, the parabola opens upwards and the range is y ≥ k. If a < 0, the parabola opens downwards and the range is y ≤ k.

对于配成完全平方形式的二次函数 f(x) = a(x – h)² + k，其顶点是 (h, k)。如果 a > 0，抛物线开口向上，值域为 y ≥ k。如果 a < 0，抛物线开口向下，值域为 y ≤ k。

Example: For f(x) = 2(x – 1)² + 3, the vertex is (1, 3), and since a = 2 > 0, the range is f(x) ≥ 3. Therefore the range is written as [3, ∞).

例如：对于 f(x) = 2(x – 1)² + 3，顶点为 (1, 3)，且由于 a = 2 > 0，值域为 f(x) ≥ 3。因此值域写作 [3, ∞)。

For rational functions such as f(x) = 1/x, the range excludes y = 0 because no finite x makes 1/x equal to 0. The range is (-∞, 0) ∪ (0, ∞). For exponential functions like f(x) = eˣ, the range is (0, ∞), never reaching zero.

对于像 f(x) = 1/x 这样的有理函数，值域不包括 y = 0，因为没有有限的 x 能让 1/x 等于 0。值域是 (-∞, 0) ∪ (0, ∞)。对于像 f(x) = eˣ 这样的指数函数，值域是 (0, ∞)，永远不会到达零。

4. Composite Functions | 复合函数

A composite function is formed when the output of one function is used as the input of another. The notation is (f ∘ g)(x) = f(g(x)). This means: first apply g to x, then apply f to the result. The order matters greatly — f(g(x)) is generally not the same as g(f(x)).

复合函数是指将一个函数的输出用作另一个函数的输入。记号 (f ∘ g)(x) = f(g(x)) 表示：先将 g 作用于 x，然后将 f 作用于其结果。顺序非常重要——f(g(x)) 通常不等于 g(f(x))。

Example: If f(x) = 2x + 1 and g(x) = x², then:

f(g(x)) = f(x²) = 2x² + 1

g(f(x)) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1

Observe that the two results are very different. When evaluating composite functions with numbers, work from the inside out. For example, if f(x) = x² and g(x) = x + 2, then f(g(3)) = f(5) = 25.

观察可知，两个结果截然不同。当用数值计算复合函数时，从内向外运算。例如，如果 f(x) = x² 且 g(x) = x + 2，那么 f(g(3)) = f(5) = 25。

The domain of f(g(x)) is restricted to those x for which g(x) is in the domain of f. If g(x) = x – 1 and f(x) = √x, then we need x – 1 ≥ 0, so the domain is x ≥ 1.

f(g(x)) 的定义域受到限制：只有当 g(x) 属于 f 的定义域时，x 才是有效的。如果 g(x) = x – 1 且 f(x) = √x，那么我们需要 x – 1 ≥ 0，所以定义域为 x ≥ 1。

5. Inverse Functions | 反函数

The inverse of a function f, denoted f⁻¹, reverses the effect of f. If f(a) = b, then f⁻¹(b) = a. The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f. Not every function has an inverse — it must be one-to-one.

函数 f 的反函数，记作 f⁻¹，是逆转 f 效果的函数。如果 f(a) = b，那么 f⁻¹(b) = a。f⁻¹ 的定义域就是 f 的值域，f⁻¹ 的值域就是 f 的定义域。并非每个函数都有反函数——它必须是一一对应的。

A function is one-to-one (injective) if every output corresponds to exactly one input. Graphically, it must pass the horizontal line test. For example, f(x) = x² is not one-to-one over all real numbers because f(2) = f(-2) = 4, so it has no inverse over its full domain.

如果一个函数每个输出都恰对应一个输入，那么它是一一对应的（单射）。从图像上看，它必须通过水平线检验。例如，f(x) = x² 在整个实数域上不是一一对应的，因为 f(2) = f(-2) = 4，所以它在完整定义域上没有反函数。

However, if we restrict the domain to x ≥ 0, then f(x) = x² becomes one-to-one and its inverse is f⁻¹(x) = √x. In AQA examinations, you will frequently encounter functions whose domains have been restricted precisely to make inverses possible.

然而，如果我们将定义域限制为 x ≥ 0，那么 f(x) = x² 就变为一一对应，其反函数是 f⁻¹(x) = √x。在 AQA 考试中，你经常会遇到通过限制定义域来使反函数存在的题目。

To find an inverse algebraically: Write y = f(x). Swap x and y. Rearrange to make y the subject. Replace y with f⁻¹(x). Verifying, the graph of f⁻¹ is the reflection of f in the line y = x.

代数方法求反函数： 写出 y = f(x)。交换 x 和 y。重新整理以 y 为对象。将 y 替换为 f⁻¹(x)。验证时，f⁻¹ 的图像是 f 关于直线 y = x 的反射。

6. Graph Transformations: Translations | 图像变换：平移

Graph transformations allow you to sketch new functions starting from a known base graph. The two main types are translations and stretches (including reflections). Understanding the notation is crucial to avoid common mistakes.

图像变换允许你从已知的基本图像出发画出新函数的草图。两种主要类型是平移和伸缩（包括反射）。理解记号对于避免常见错误至关重要。

Horizontal translation: f(x + a) shifts the graph of f(x) to the left by a units. f(x – a) shifts it to the right by a units. This is counterintuitive: the sign inside the brackets appears reversed. The transformation is along the x-axis, so it affects the domain.

水平平移：f(x + a) 将 f(x) 的图像向左移动 a 个单位。f(x – a) 将其向右移动 a 个单位。这有点反直觉：括号内的符号看起来是反的。变换沿 x 轴进行，因此影响定义域。

Vertical translation: f(x) + a shifts the graph upwards by a units. f(x) – a shifts it downwards. This transformation is more intuitive, as the addition is applied directly to the output values, affecting the range.

垂直平移：f(x) + a 将图像向上移动 a 个单位。f(x) – a 将其向下移动。这个变换更直观，因为加法直接作用于输出值，影响值域。

For example, starting from y = x²:

y = (x – 3)²: shift 3 units to the right
y = (x – 3)²：向右平移 3 个单位
y = x² + 2: shift 2 units up
y = x² + 2：向上平移 2 个单位
y = (x – 3)² + 2: shift right 3 and up 2
y = (x – 3)² + 2：向右平移 3 个单位并向上平移 2 个单位

7. Graph Transformations: Stretches | 图像变换：伸缩

A vertical stretch multiplies all y-values by a constant factor. The graph of a·f(x) is stretched or compressed vertically by factor a. If a > 1, the graph stretches upward; if 0 < a < 1, it compresses vertically; if a is negative, there is also a reflection in the x-axis.

垂直伸缩将所有 y 值乘以一个常数因子。a·f(x) 的图像在垂直方向按因子 a 拉伸或压缩。如果 a > 1，图像向上拉伸；如果 0 < a < 1，则在垂直方向压缩；如果 a 为负，则还伴随关于 x 轴的反射。

A horizontal stretch multiplies all x-values by a factor. The graph of f(bx) is stretched or compressed horizontally. If b > 1, the graph compresses toward the y-axis; if 0 < b < 1, it stretches away. Careful: f(bx) compresses by a factor of 1/b.

水平伸缩将所有 x 值乘以一个因子。f(bx) 的图像在水平方向拉伸或压缩。如果 b > 1，图像向 y 轴压缩；如果 0 < b < 1，则远离 y 轴拉伸。注意：f(bx) 按 1/b 的因子压缩。

Example: For y = sin(x), the function y = 3 sin(2x) has amplitude 3 and period π (since the period of sin(x) is 2π divided by 2). The 3 stretches the graph vertically, and the 2 compresses it horizontally.

例如：对于 y = sin(x)，函数 y = 3 sin(2x) 的振幅为 3，周期为 π（因为 sin(x) 的周期 2π 除以 2）。3 在垂直方向拉伸图像，2 在水平方向压缩图像。

8. Reflections | 反射

Reflections are a special case of stretches with a scale factor of -1. There are two key reflections you must know:

反射是伸缩因子为 -1 的特殊情况。你必须掌握两种关键反射：

First, y = -f(x) reflects the graph of f(x) in the x-axis. Every point (x, y) maps to (x, -y). This affects signs of the outputs, so the range is negated. For example, the range of f(x) = x² + 1 is y ≥ 1, and the range of -f(x) becomes y ≤ -1.

第一，y = -f(x) 将 f(x) 的图像关于 x 轴反射。每个点 (x, y) 映射到 (x, -y)。这改变了输出的符号，因此值域会取相反数。例如，f(x) = x² + 1 的值域为 y ≥ 1，而 -f(x) 的值域变为 y ≤ -1。

Second, y = f(-x) reflects the graph in the y-axis. Every point (x, y) maps to (-x, y). This affects the domain. For example, if the domain of f is [-2, 3], then the domain of f(-x) is [-3, 2].

第二，y = f(-x) 将图像关于 y 轴反射。每个点 (x, y) 映射到 (-x, y)。这影响了定义域。例如，如果 f 的定义域为 [-2, 3]，则 f(-x) 的定义域为 [-3, 2]。

A combination like y = -f(-x) reflects in both axes, equivalent to a rotation of 180° about the origin. Combining multiple transformations requires careful order of operations. In exams, always describe the transformations in the order they are applied.

像 y = -f(-x) 这样的组合在两个轴上反射，等效于绕原点旋转 180°。组合多个变换时需要仔细考虑运算顺序。在考试中，务必按应用顺序描述变换。

9. Modelling with Transformations | 变换建模

Transformations are not just abstract — they are widely used to model real-world phenomena. For example, the height of a Ferris wheel passenger over time can be modelled using a sine wave with vertical and horizontal shifts:

变换不仅仅是抽象的，它们被广泛用于模拟现实世界中的现象。例如，摩天轮乘客的高度随时间的变化可以用带有垂直和水平位移的正弦波来建模：

h(t) = A sin(B(t – C)) + D

Here, A represents the amplitude (half the range of heights), D is the vertical shift (the central height), C is the horizontal shift (the starting angle), and B determines the period (time for one complete revolution).

这里，A 代表振幅（高度范围的一半），D 是垂直位移（中心高度），C 是水平位移（起始角度），B 决定周期（完成一整圈的时间）。

In AQA exam questions, you may be given a graph and asked to find the equation by identifying the transformations. Key steps: identify the amplitude, find the period, locate the principal axis (vertical shift), and determine any phase shift.

在 AQA 考试题中，你可能会被给出一幅图，要求通过识别变换来求方程。关键步骤：确定振幅，求周期，定位中心轴（垂直位移），并确定任何相位移动。

10. Common Exam Pitfalls | 常见考试陷阱

Being aware of common mistakes can significantly improve your score. Here are the most frequent errors students make in this topic:

了解常见错误可以显著提高你的分数。以下是学生在本主题中最常犯的错误：

Common Mistake \| 常见错误	Correction \| 正确做法
Forgetting the domain restriction when finding an inverse	Always check whether the function is one-to-one over its domain
求反函数时忘记定义域限制	始终检查函数在其定义域上是否一一对应
Confusing f(x + a) as a shift to the right	f(x + a) shifts LEFT; f(x – a) shifts right
误将 f(x + a) 当作向右平移	f(x + a) 向左平移；f(x – a) 向右平移
Applying horizontal stretches in the wrong direction	f(bx) compresses by factor 1/b when b > 1
水平伸缩方向搞反	当 b > 1 时，f(bx) 按 1/b 的因子压缩
Ignoring the domain of composite functions	Check that the inner function’s output is in the outer function’s domain
忽略复合函数的定义域	检查内层函数的输出是否在外层函数的定义域内
Saying the range of a quadratic is all real numbers	Use the vertex to determine the minimum or maximum of the range
认为二次函数的值域是全体实数	利用顶点确定值域的最小值或最大值

11. Exam-Style Question Walkthrough | 考试题型演练

Let’s walk through a typical AQA-style question to see how these concepts come together. This will help you apply the theory to actual exam conditions.

让我们逐步讲解一道典型的 AQA 风格题目，看看这些概念如何综合运用。这将帮助你将理论应用到实际考试情境中。

Question: The function f is defined by f(x) = 2x² – 8x + 11 for x ∈ ℝ. (a) Express f(x) in the form a(x – h)² + k. (b) State the range of f. (c) Find the inverse of f, stating the domain of f⁻¹.

题目： 函数 f 定义为 f(x) = 2x² – 8x + 11，其中 x ∈ ℝ。(a) 将 f(x) 表示为 a(x – h)² + k 的形式。(b) 写出 f 的值域。(c) 求 f 的反函数，并写出 f⁻¹ 的定义域。

Solution (a): Completing the square:

f(x) = 2(x² – 4x) + 11

f(x) = 2[(x – 2)² – 4] + 11

f(x) = 2(x – 2)² – 8 + 11

f(x) = 2(x – 2)² + 3

Solution (b): Since 2(x – 2)² ≥ 0, the minimum value of f is 3. The range is therefore f(x) ≥ 3, or [3, ∞). Because the graph is a U-shape (a > 0), the range extends upward without bound.

解答 (b)：由于 2(x – 2)² ≥ 0，f 的最小值为 3。因此值域为 f(x) ≥ 3，即 [3, ∞)。由于图像是 U 形（a > 0），值域向上无限延伸。

Solution (c): For the inverse to exist, we must restrict the domain so that f is one-to-one. The natural restriction is x ≥ 2, making the function strictly increasing. Now we find the inverse:

解答 (c)：要使反函数存在，我们必须限制定义域，使 f 一一对应。自然的限制是 x ≥ 2，使函数严格递增。现在求反函数：

y = 2x² – 8x + 11

x = 2y² – 8y + 11

x = 2(y – 2)² + 3

(x – 3)/2 = (y – 2)²

y = 2 + √((x – 3)/2)

Since the restricted domain is x ≥ 2, we take the positive square root. The domain of f⁻¹ is the range of f on this restricted domain, so f⁻¹ is defined for x ≥ 3.

由于限制定义域为 x ≥ 2，我们取正平方根。f⁻¹ 的定义域是 f 在这个限制定义域上的值域，因此 f⁻¹ 的定义域为 x ≥ 3。

12. Final Revision Tips | 最终复习建议

To master this topic for your AQA A-Level exam, incorporate these strategies into your revision routine. Consistency and practice are the keys to success.

为了在 AQA A-Level 考试中掌握这一主题，请将这些策略纳入你的复习计划。持续性和练习是成功的关键。

First, draw graphs for every function you encounter. Even a quick sketch helps you understand the domain, range, and the effect of transformations. The graphical representation reinforces your algebraic understanding and builds intuition.

第一，为你遇到的每个函数画图。即使是一个简略的草图也能帮助你理解定义域、值域以及变换的效果。图像表示能强化你的代数理解并建立直觉。

Second, practise past papers under timed conditions. This topic appears frequently in Section A of the paper as short-response questions, and in Section B as part of multi-stage problems on graph sketching or function modelling.

第二，在限时条件下练习历年真题。这一主题在试卷的 A 部分经常以短答题形式出现，在 B 部分则作为多阶段问题的一部分，涉及图像描绘或函数建模。

Finally, review your mistakes systematically. Keep a log of errors you make — especially around notation, sign errors, and domain restrictions — and revisit them in the final week before the exam.

最后，系统地回顾你的错误。记录你犯过的错误——特别是关于记号、符号错误和定义域限制的——并在考试前最后一周重新复习它们。

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Mastering Functions: Domain, Range, and Transformations | 函数精通：定义域、值域与变换