A-Level Mathematics: Operations and Graphs of Composite Functions | A-Level 数学：复合函数的运算与图像

📚 A-Level Mathematics: Operations and Graphs of Composite Functions | A-Level 数学：复合函数的运算与图像

Composite functions are one of the most fundamental concepts in A-Level Mathematics. Understanding how two functions combine to form a new function is essential for solving problems in calculus, trigonometry, and even mechanics. This article will guide you through the definition, evaluation, domain considerations, and graphical interpretation of composite functions, with worked examples designed for exam success.

复合函数是 A-Level 数学中最基础的概念之一。理解两个函数如何结合成一个新函数，对于解决微积分、三角学乃至力学中的问题都至关重要。本文将引导你系统地学习复合函数的定义、求值、定义域分析以及图像解读，并配有专为考试设计的例题。

1. What is a Composite Function? | 什么是复合函数？

A composite function is formed when the output of one function becomes the input of another. If f(x) and g(x) are two functions, then the composite function ‘f composed with g’ is written as f(g(x)) and read as ‘f of g of x’. The order of composition matters: f(g(x)) is generally not equal to g(f(x)).

复合函数是指将一个函数的输出作为另一个函数输入所形成的新函数。若 f(x) 和 g(x) 是两个函数，则”f 与 g 复合”的函数记作 f(g(x))，读作”f of g of x”。复合的顺序至关重要：f(g(x)) 通常不等于 g(f(x))。

The notation for composite functions can also be written using a small circle: (f ∘ g)(x) = f(g(x)). Here, g is applied first, then f is applied to the result. In other words, the function on the right (g) is applied first.

复合函数也可用一个小圆圈表示：(f ∘ g)(x) = f(g(x))。这里先应用 g，再将 f 应用于其结果。换句话说，右侧的函数 (g) 先被应用。

g(x) → f(g(x))
Input x → [Inner function g] → [Outer function f] → Output

Example: Let f(x) = x² and g(x) = x + 3. Then (f ∘ g)(x) = f(g(x)) = (x + 3)². And (g ∘ f)(x) = g(f(x)) = x² + 3. Clearly, these two results are different!

例：设 f(x) = x²，g(x) = x + 3。则 (f ∘ g)(x) = f(g(x)) = (x + 3)²；而 (g ∘ f)(x) = g(f(x)) = x² + 3。显然，这两个结果完全不同！

2. Evaluating Composite Functions | 复合函数的求值

To evaluate a composite function at a specific value, work from the inside out. First evaluate the inner function at the given value, then substitute the result into the outer function.

求复合函数在具体数值处的值时，应从内向外逐层计算。先求内层函数在该数值处的值，再将结果代入外层函数。

Worked Example: Given f(x) = 2x + 1 and g(x) = x² − 4, find (f ∘ g)(3).

例题：已知 f(x) = 2x + 1，g(x) = x² − 4，求 (f ∘ g)(3)。

Step 1: Evaluate g(3) = 3² − 4 = 9 − 4 = 5.
Step 1：先求 g(3) = 3² − 4 = 9 − 4 = 5。
Step 2: Substitute into f: f(5) = 2(5) + 1 = 11.
Step 2：将结果代入 f：f(5) = 2(5) + 1 = 11。

Therefore, (f ∘ g)(3) = 11. You can also substitute algebraically first: f(g(x)) = 2(x² − 4) + 1 = 2x² − 7, then evaluate at x = 3: 2(9) − 7 = 11. Both methods are valid; the second is more efficient if you need multiple values.

因此，(f ∘ g)(3) = 11。你也可以先进行代数展开：f(g(x)) = 2(x² − 4) + 1 = 2x² − 7，再代入 x = 3：2(9) − 7 = 11。两种方法均正确；若需要计算多个值，第二种方法更高效。

Pitfall: Always pay attention to the order. (f ∘ g)(3) is not the same as (g ∘ f)(3). In this example, (g ∘ f)(3) = g(7) = 49 − 4 = 45.

易错点：务必注意复合顺序。(f ∘ g)(3) 与 (g ∘ f)(3) 不同。本例中，(g ∘ f)(3) = g(7) = 49 − 4 = 45。

3. Domain of Composite Functions | 复合函数的定义域

The domain of a composite function (f ∘ g)(x) consists of all x in the domain of g for which g(x) lies in the domain of f. In other words, you must exclude any x that makes g(x) undefined, and also exclude any x for which g(x) produces a value outside the domain of f.

复合函数 (f ∘ g)(x) 的定义域包含所有满足以下条件的 x：x 在 g 的定义域内，且 g(x) 落在 f 的定义域内。换言之，必须排除使 g(x) 无定义的 x，以及使 g(x) 超出 f 定义域的 x。

Example: Let f(x) = √x (domain x ≥ 0) and g(x) = x − 5 (domain all real numbers). Then (f ∘ g)(x) = f(g(x)) = √(x − 5). For this to be defined, we need x − 5 ≥ 0, so x ≥ 5. Thus the domain of the composite is [5, ∞).

例：设 f(x) = √x（定义域 x ≥ 0），g(x) = x − 5（定义域为全体实数）。则 (f ∘ g)(x) = f(g(x)) = √(x − 5)。要使该式有意义，需 x − 5 ≥ 0，即 x ≥ 5。因此复合函数的定义域为 [5, ∞)。

More complex example: Let f(x) = 1/x (domain x ≠ 0) and g(x) = x² − 1. Then (f ∘ g)(x) = 1/(x² − 1). This is undefined when x² − 1 = 0, i.e., x = ±1. So the domain of the composite is all real numbers except x = 1 and x = −1.

更复杂的例子：设 f(x) = 1/x（定义域 x ≠ 0），g(x) = x² − 1。则 (f ∘ g)(x) = 1/(x² − 1)。当 x² − 1 = 0，即 x = ±1 时该式无定义。因此复合函数的定义域为除去 x = 1 和 x = −1 之外的所有实数。

Rule of thumb: When finding the domain of a composite function, always consider two restrictions: (1) the inner function must be defined, and (2) the output of the inner function must be a valid input for the outer function.

经验法则：求复合函数定义域时，始终考虑两个限制条件：(1) 内层函数必须有定义；(2) 内层函数的输出必须是外层函数的合法输入。

4. Range of Composite Functions | 复合函数的值域

The range of (f ∘ g)(x) is the set of all possible outputs after applying f to the range of g (restricted to the domain found above). A practical method is to first determine the range of the inner function, then input that interval into the outer function.

复合函数 (f ∘ g)(x) 的值域是 f 作用于 g 的值域（并限于上述定义域）后得到的所有可能输出值的集合。一种实用方法是先确定内层函数的值域，再将这个区间代入外层函数。

Example: Let f(x) = x² and g(x) = x − 1, with x ∈ [0, 3]. First, as x ranges from 0 to 3, g(x) = x − 1 ranges from −1 to 2. Then f(g) = (x − 1)². When g ranges over [−1, 2], the square ranges from 0 (at g = 0) to 4 (at g = −1 or g = 2? Actually (−1)² = 1, 2² = 4, so max is 4). Hence the range of (f ∘ g) is [0, 4].

例：设 f(x) = x²，g(x) = x − 1，其中 x ∈ [0, 3]。首先，当 x 从 0 变到 3 时，g(x) = x − 1 的取值范围为 −1 到 2。然后 f(g) = (x − 1)²。当 g 在 [−1, 2] 内变化时，其平方在 0（g = 0 处）到 4（g = 2 处）之间变化（注意 (−1)² = 1）。因此 (f ∘ g) 的值域为 [0, 4]。

Notice that the range is not always simply ‘square every endpoint’. You must consider whether the inner function’s range covers a turning point of the outer function. Here the turning point of x² at g = 0 lies inside the interval [−1, 2], so the minimum is 0.

请注意，值域并不总是简单地将端点代入外层函数。你必须考虑内层函数的值域是否包含外层函数的转折点。本例中 x² 在 g = 0 处的转折点落在区间 [−1, 2] 内，因此最小值为 0。

5. Algebraic Operations: Finding the Composite Formula | 代数运算：求复合函数的解析式

To find the algebraic expression for a composite function, substitute the entire inner function into the outer function wherever ‘x’ appears, then simplify.

要求复合函数的代数解析式，需将整个内层函数代入外层函数中所有出现 ‘x’ 的位置，然后化简。

Example: f(x) = x² + 2x, g(x) = x − 3. Find (f ∘ g)(x).

例：f(x) = x² + 2x，g(x) = x − 3。求 (f ∘ g)(x)。

(f ∘ g)(x) = f(x − 3) = (x − 3)² + 2(x − 3) = x² − 6x + 9 + 2x − 6 = x² − 4x + 3

Alternatively, find (g ∘ f)(x) = g(x² + 2x) = (x² + 2x) − 3 = x² + 2x − 3. Compare the two results: they are clearly different expressions, reinforcing that composition is not commutative.

再求 (g ∘ f)(x) = g(x² + 2x) = (x² + 2x) − 3 = x² + 2x − 3。比较两个结果：它们显然不同，再次印证复合运算不满足交换律。

Special case: When f and g are inverse functions, their composite simplifies to x. For example, if f(x) = 2x + 3 and f⁻¹(x) = (x − 3)/2, then (f ∘ f⁻¹)(x) = x and (f⁻¹ ∘ f)(x) = x.

特殊情形：当 f 与 g 互为反函数时，它们的复合函数简化为 x。例如，若 f(x) = 2x + 3，f⁻¹(x) = (x − 3)/2，则 (f ∘ f⁻¹)(x) = x 且 (f⁻¹ ∘ f)(x) = x。

6. Decomposing Composite Functions | 复合函数的分解

In many exam problems, you are given a composite function and asked to identify its component functions. This requires spotting the ‘outer’ and ‘inner’ structure.

许多考试题目会给出一个复合函数，要求你识别其组成部分。这需要你辨认出”外层”和”内层”的结构。

Example: Express h(x) = (3x + 2)⁵ as a composite of two simpler functions.

例：将 h(x) = (3x + 2)⁵ 表示为两个更简单函数的复合。

We see that the operation applied is ‘raise to the 5th power’ applied to the expression ‘3x + 2’. Hence we can set the outer function f(u) = u⁵, and the inner function g(x) = 3x + 2. Then h(x) = f(g(x)) = (3x + 2)⁵.

我们可以看出，运算方式是将表达式 ‘3x + 2’ 整体进行五次方运算。因此设外层函数 f(u) = u⁵，内层函数 g(x) = 3x + 2。则 h(x) = f(g(x)) = (3x + 2)⁵。

Another example: Express h(x) = sin(2x + 1) as a composite. Choose f(u) = sin u and g(x) = 2x + 1. Then h(x) = f(g(x)).

另一个例子：将 h(x) = sin(2x + 1) 表示为复合函数。选择 f(u) = sin u，g(x) = 2x + 1。则 h(x) = f(g(x))。

This skill is particularly important for the chain rule in differentiation: recognising that y = (3x + 2)⁵ can be differentiated by letting u = 3x + 2 and using dy/dx = dy/du × du/dx.

这种分解技巧在微分的链式法则中尤为重要：识别 y = (3x + 2)⁵ 可以通过令 u = 3x + 2，并利用 dy/dx = dy/du × du/dx 来求导。

7. Graphs of Composite Functions | 复合函数的图像

Sketching the graph of a composite function can be challenging, but a systematic approach helps. You can build the graph of (f ∘ g)(x) by considering how the inner function maps x to g(x), and then how the outer function maps g(x) to f(g(x)).

绘制复合函数的图像可能具有挑战性，但系统性的方法会有所帮助。你可以通过先考虑内层函数如何将 x 映射到 g(x)，再考虑外层函数如何将 g(x) 映射到 f(g(x))，来构建 (f ∘ g)(x) 的图像。

Key observations:

关键观察：

Zeros of (f ∘ g)(x) occur where f(g(x)) = 0, i.e., where g(x) equals a root of f.
(f ∘ g)(x) 的零点出现在 f(g(x)) = 0 处，即 g(x) 等于 f 的某个根时。
If f is an increasing function on the range of g, then the composite preserves the monotonicity of g. If f is decreasing, the direction reverses.
若 f 在 g 的值域上是增函数，则复合函数保持 g 的单调性。若 f 是减函数，则单调方向反转。
Points where g(x) = a (where a is a turning point of f) become turning points of the composite, with the same x-coordinate.
当 g(x) = a（a 为 f 的转折点）时，对应的 x 坐标成为复合函数的转折点横坐标。

Example: Sketch the graph of y = (x − 2)². Here g(x) = x − 2 and f(u) = u². The graph of g is a line with slope 1 crossing the x-axis at x = 2. Applying f = squaring gives a parabola that touches the x-axis at x = 2 (because g = 0 there) and opens upwards. The vertex is at (2, 0).

例：绘制 y = (x − 2)² 的图像。这里 g(x) = x − 2，f(u) = u²。g 的图像是一条斜率为 1 的直线，与 x 轴交于点 x = 2。应用 f 即平方运算后，得到一条开口向上的抛物线，在 x = 2 处与 x 轴相切（因为此时 g = 0）。顶点坐标为 (2, 0)。

When dealing with more complicated composite functions, it is often useful to sketch the intermediate function g(x) on the same diagram, then ‘transfer’ values onto the outer function’s graph. This method, sometimes called the ‘chain of mappings’, is especially useful for functions like y = sin(2x) or y = e^{x²}.

处理更复杂的复合函数时，在同一坐标系中先画出中间函数 g(x) 的图像，再将值”传递”到外层函数的图像上往往很有用。这种方法有时被称为”映射链法”，特别适用于 y = sin(2x) 或 y = e^{x²} 等函数。

8. Transformations and Composite Functions | 图像变换与复合函数

Graphical transformations are closely related to composite functions. For example, y = f(x + a) is a horizontal translation of y = f(x) by −a units. This can be written as (f ∘ g)(x) where g(x) = x + a.

图像变换与复合函数密切相关。例如，y = f(x + a) 是 y = f(x) 水平平移 −a 个单位的结果。这可写成 (f ∘ g)(x)，其中 g(x) = x + a。

Similarly, y = f(kx) represents a horizontal stretch by factor 1/k. Here g(x) = kx. Vertical transformations y = af(x) can be seen as h(x) = a·f(x) = (multiplication by a) applied after f.

类似地，y = f(kx) 表示水平伸缩，伸缩因子为 1/k。此时 g(x) = kx。纵向变换 y = af(x) 可视为在 f 之后应用乘以 a 的运算。

Composite transformations: The order of transformations matters. For example, to obtain y = f(2x + 1), you can either:

复合变换：变换的顺序很重要。例如，要得到 y = f(2x + 1)，你可以选择：

First translate by −1 horizontally, then stretch by 1/2 horizontally; or
先水平平移 −1，再水平伸缩 1/2；或
First stretch by 1/2, then translate by −1/2 horizontally.
先水平伸缩 1/2，再水平平移 −1/2。

This reflects the non-commutativity of composition: the transformation sequence matters. Understanding this as a composition of the linear function 2x + 1 with f clarifies the correct order.

这反映了复合运算的非交换性：变换的先后顺序非常重要。将 2x + 1 这个线性函数与 f 的复合关系理解清楚，有助于确定正确的变换顺序。

9. Inverse Functions and Composition | 反函数与复合

A key property linking inverse functions and composition is that if f is a one-to-one function with inverse f⁻¹, then:

反函数与复合之间的一个关键性质是：若 f 是一一对应函数，且其反函数为 f⁻¹，则：

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

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

This is often used in exam questions to verify whether two functions are inverses of each other. Simply compose them in both orders and check that both composites equal x.

考试中常用这一点来验证两个函数是否互为反函数。只需按两个顺序分别进行复合，并检查两个复合函数是否都等于 x。

Example: Show that f(x) = 3x + 2 and g(x) = (x − 2)/3 are inverses.

例：证明 f(x) = 3x + 2 与 g(x) = (x − 2)/3 互为反函数。

First, (f ∘ g)(x) = f((x − 2)/3) = 3·((x − 2)/3) + 2 = x − 2 + 2 = x. Second, (g ∘ f)(x) = g(3x + 2) = ((3x + 2) − 2)/3 = 3x/3 = x. Since both composites equal x, f and g are indeed inverse functions.

首先，(f ∘ g)(x) = f((x − 2)/3) = 3·((x − 2)/3) + 2 = x − 2 + 2 = x。其次，(g ∘ f)(x) = g(3x + 2) = ((3x + 2) − 2)/3 = 3x/3 = x。由于两个复合函数均等于 x，因此 f 与 g 确实互为反函数。

The graph of an inverse function is the reflection of the original function’s graph in the line y = x. This geometric interpretation is also related to composition: composing a function with its inverse returns the identity function, whose graph is the line y = x.

反函数的图像是原函数图像关于直线 y = x 的对称图形。这一几何解释也与复合有关：函数与其反函数复合后得到恒等函数，其图像就是直线 y = x。

10. Common Exam Pitfalls and Tips | 常见考试易错点与技巧

Here are some frequent mistakes students make with composite functions, along with advice on how to avoid them.

以下是一些学生在复合函数问题上常犯的错误，以及相应的避免建议。

Pitfall 1: Swapping the order of composition. Always identify which function is applied first by looking at which one appears on the right in (f ∘ g)(x).
易错点 1：颠倒复合顺序。在 (f ∘ g)(x) 中，始终通过观察哪个函数在右侧来确定先应用哪个函数。
Pitfall 2: Forgetting to restrict the domain of the composite. Always check for values that make the inner function or the outer function undefined.
易错点 2：忘记限制复合函数的定义域。始终检查使内层函数或外层函数无定义的数值。
Pitfall 3: Assuming that (f + g)(x) is the same as (f ∘ g)(x). The former is addition, the latter is composition.
易错点 3：误以为 (f + g)(x) 与 (f ∘ g)(x) 相同。前者是加法运算，后者是复合运算。
Pitfall 4: Misinterpreting f²(x). This notation usually means (f(x))², not f(f(x)). In some contexts, however, f²(x) might mean f(f(x)) (iteration). Read the question carefully.
易错点 4：误解 f²(x) 的含义。此记法通常表示 (f(x))²，而非 f(f(x))。但在某些上下文中，f²(x) 也可能表示 f(f(x))（迭代）。请仔细阅读题目。
Tip: When sketching composite graphs, always annotate key points such as intersections with the axes and turning points.
技巧：绘制复合函数图像时，务必标注关键点，如与坐标轴的交点和转折点。

11. Worked Exam-Style Problem | 考试风格例题精讲

Problem: The functions f and g are defined by f(x) = x² + 1 for x ∈ ℝ, and g(x) = √(x − 2) for x ≥ 2. (a) Find the composite function (f ∘ g)(x) and state its domain. (b) Find the composite function (g ∘ f)(x) and state its domain. (c) State the range of (f ∘ g)(x).

题目：函数 f 与 g 定义为 f(x) = x² + 1（x ∈ ℝ），g(x) = √(x − 2)（x ≥ 2）。(a) 求复合函数 (f ∘ g)(x) 并写出其定义域。(b) 求复合函数 (g ∘ f)(x) 并写出其定义域。(c) 写出 (f ∘ g)(x) 的值域。

Solution (a): (f ∘ g)(x) = f(g(x)) = (√(x − 2))² + 1 = x − 2 + 1 = x − 1. The inner function g requires x ≥ 2. The outer function f is defined for all real numbers, so the domain of the composite is x ≥ 2.

解答 (a)：(f ∘ g)(x) = f(g(x)) = (√(x − 2))² + 1 = x − 2 + 1 = x − 1。内层函数 g 要求 x ≥ 2。外层函数 f 对全体实数均有定义，因此复合函数的定义域为 x ≥ 2。

Solution (b): (g ∘ f)(x) = g(f(x)) = √((x² + 1) − 2) = √(x² − 1). For the square root to be defined, we require x² − 1 ≥ 0, i.e., x ≤ −1 or x ≥ 1. So the domain is (−∞, −1] ∪ [1, ∞).

解答 (b)：(g ∘ f)(x) = g(f(x)) = √((x² + 1) − 2) = √(x² − 1)。要使根号有意义，需要 x² − 1 ≥ 0，即 x ≤ −1 或 x ≥ 1。因此定义域为 (−∞, −1] ∪ [1, ∞)。

Solution (c): Since (f ∘ g)(x) = x − 1 for x ≥ 2, as x increases from 2 to ∞, the output increases from 1 to ∞. Hence the range is [1, ∞).

解答 (c)：由于 (f ∘ g)(x) = x − 1 且 x ≥ 2，当 x 从 2 增大到 ∞ 时，输出值从 1 增大到 ∞。因此值域为 [1, ∞)。

12. Summary | 总结

Composite functions are a core topic in A-Level Mathematics. The key ideas to remember are: the function on the right is applied first; domain restrictions come from both the inner and outer functions; the range is found by applying the outer function to the inner function’s range; and composition is not commutative. With consistent practice and attention to domain issues, you can confidently tackle any composite function question.

复合函数是 A-Level 数学的核心内容。需要记住的关键点包括：右侧的函数先被应用；定义域的限制来自内外两层函数；值域需通过将外层函数作用于内层函数的值域来求得；以及复合运算不满足交换律。通过持续练习并重视定义域问题，你定能从容应对任何复合函数题目。

Graphically, understanding the mapping chain from x to g(x) to f(g(x)) gives you a powerful tool for sketching functions that arise in both pure and applied mathematics. Master this concept, and you build a strong foundation for calculus, particularly the chain rule and related rates.

从图像角度而言，理解从 x 到 g(x) 再到 f(g(x)) 的映射链，将为你提供在纯数学与应用数学中绘制函数图像的强大工具。掌握这一概念，你就为微积分学习奠定了坚实基础，尤其是在链式法则和相关变化率方面。

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A-Level Mathematics: Operations and Graphs of Composite Functions | A-Level 数学：复合函数的运算与图像