📚 Composite Functions: Concept and Evaluation | 复合函数的概念与求值方法
In IB Mathematics, the concept of composite functions is fundamental to understanding how functions interact and combine. A composite function is formed when the output of one function becomes the input of another, creating a new function that represents this sequential application. This idea appears across both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses, at both Standard Level (SL) and Higher Level (HL).
在 IB 数学中，复合函数是理解函数之间如何相互作用与组合的基础概念。复合函数是将一个函数的输出作为另一个函数的输入，从而形成一个新的函数来表示这种依次应用的关系。这一概念出现在 数学分析与方法（AA）和 数学应用与解释（AI）两门课程中，涵盖标准级别（SL）和高级别（HL）。
1. What Is a Composite Function? | 什么是复合函数？
A composite function is written as f ° g, read as "f composed with g" or "f after g". It is defined as (f ° g)(x) = f(g(x)). This means we first apply the function g to x, obtaining g(x), and then apply the function f to that result. The order of application is crucial: f ° g is generally not the same as g ° f.
复合函数写作 f ° g，读作 "f 复合 g" 或 "先 g 后 f"。其定义为 (f ° g)(x) = f(g(x))。也就是说，我们先将函数 g 作用于 x，得到 g(x)，然后再将函数 f 作用于该结果。应用顺序至关重要：f ° g 通常不等于 g ° f。
For example, if f(x) = 2x + 1 and g(x) = x2, then (f ° g)(x) = f(g(x)) = f(x2) = 2x2 + 1. In contrast, (g ° f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1. Clearly, these two composite functions are not the same.
例如，若 f(x) = 2x + 1 且 g(x) = x2，则 (f ° g)(x) = f(g(x)) = f(x2) = 2x2 + 1。相比之下，(g ° f)(x) = g(f(x)) = g(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1。显然，这两个复合函数并不相同。
2. Notation and Domain Considerations | 符号表示与定义域注意事项
In IB notation, the composition is often written as (f ° g)(x) or simply f(g(x)). Some textbooks and exam papers use the notation f ° g to mean "first apply g, then apply f". It is essential to read the definition carefully in your textbook and exam to avoid confusion, as some older curricula used the opposite convention.
在 IB 的记法中，复合函数常写作 (f ° g)(x) 或简写为 f(g(x))。有些教材和试卷用 f ° g 表示 "先应用 g，再应用 f"。务必仔细阅读教材和试卷中的定义，以避免混淆，因为某些较早的课程版本使用相反的约定。
The domain of a composite function f ° g is the set of all x in the domain of g such that g(x) lies in the domain of f. In other words, we must ensure that every intermediate output g(x) is a valid input for f.
复合函数 f ° g 的定义域是：所有属于 g 的定义域且使 g(x) 属于 f 的定义域的 x 的集合。换句话说，我们必须确保每一个中间结果 g(x) 都是 f 的合法输入。
For instance, if f(x) = √x and g(x) = x - 3, then (f ° g)(x) = √x - 3. The domain of g is all real numbers ℝ, but the domain of f is x ≥ 0. Therefore, we require x - 3 ≥ 0, so the domain of f ° g is x ≥ 3.
例如，若 f(x) = √x 且 g(x) = x - 3，则 (f ° g)(x) = √x - 3。g 的定义域为全体实数 ℝ，但 f 的定义域为 x ≥ 0。因此需满足 x - 3 ≥ 0，故 f ° g 的定义域为 x ≥ 3。
3. Evaluating Composite Functions Numerically | 数值求值复合函数
When asked to evaluate a composite function at a specific value, such as (f ° g)(2), the most reliable approach is to work step by step from the inside out. First compute g(2), then substitute that result into f.
当需要对复合函数在特定数值处求值时，例如 (f ° g)(2)，最可靠的方法是从内到外逐步计算。先求 g(2)，然后将结果代入 f。
Consider f(x) = 3x - 5 and g(x) = x + 4. To evaluate (f ° g)(2):
考虑 f(x) = 3x - 5 和 g(x) = x + 4。求 (f ° g)(2)：
- Step 1: Compute g(2) = 2 + 4 = 6.
- 第一步：计算 g(2) = 2 + 4 = 6。
- Step 2: Substitute into f: f(6) = 3(6) - 5 = 18 - 5 = 13.
- 第二步：代入 f：f(6) = 3(6) - 5 = 18 - 5 = 13。
Thus (f ° g)(2) = 13. This "inside-out" method is the most intuitive and least error-prone, especially when functions are given as tables or graphs.
因此 (f ° g)(2) = 13。这种 "由内向外" 的方法最为直观且不易出错，特别是当函数以表格或图像形式给出时。
4. Evaluating Composite Functions from a Table | 利用表格求复合函数值
IB exams frequently present functions using tables of values. In such cases, you cannot compute an algebraic rule, but you can still evaluate composites by carefully reading the table.
IB 考试中经常用数值表格给出函数。在这种情况下，无法求出代数表达式，但仍可仔细查阅表格来计算复合函数值。
Consider the following table:
考虑以下表格：
| x | 0 | 1 | 2 | 3 |
| f(x) | 2 | 0 | 3 | 1 |
| g(x) | 1 | 3 | 0 | 2 |
To find (f ° g)(1): first compute g(1) = 3 from the table. Then find f(3) = 1. Therefore (f ° g)(1) = 1.
为求 (f ° g)(1)：先从表中查得 g(1) = 3，再查找 f(3) = 1。因此 (f ° g)(1) = 1。
Similarly, to find (g ° f)(0): first find f(0) = 2, then g(2) = 0. Thus (g ° f)(0) = 0. Practice reading tables in both directions, as exams often include such questions to test your understanding of the order of composition.
类似地，求 (g ° f)(0)：先找 f(0) = 2，则 g(2) = 0。因此 (g ° f)(0) = 0。务必练习双向查表，因为考试中常有此类问题来考察你对复合顺序的理解。
5. Composite Functions Using Graphs | 利用图像求复合函数值
When functions are given graphically, you can evaluate composites by locating points on the graphs. For (f ° g)(a), first find g(a) on the graph of y = g(x), then use that y-value as the input to the graph of f.
当函数以图像给出时，可以通过在图上定位点来计算复合函数。对于 (f ° g)(a)，先在 y = g(x) 的图像上找到 g(a)，再将这个 y 值作为 f 图像的输入值。
This technique is especially useful in the calculator paper, where you may be asked to "use the graphs to find" a composite value. Always trace the path: x → g(x) → f(g(x)), and clearly label your steps in the working.
这种技巧在使用计算器的试卷中特别有用，题目可能要求 "利用图像求" 复合函数值。务必追踪路径：x → g(x) → f(g(x))，并清晰地在解题过程中标注每一步。
6. Algebraic Expansion of Composite Functions | 复合函数的代数展开
Often, an IB question will ask you to find an algebraic expression for (f ° g)(x) and then simplify it. This requires careful substitution and expansion.
IB 考题经常会要求你求 (f ° g)(x) 的代数表达式并化简。这需要仔细代入和展开。
For example, let f(x) = x2 + 1 and g(x) = 2x - 3. Then:
例如，设 f(x) = x2 + 1，g(x) = 2x - 3。则：
(f ° g)(x) = f(2x - 3) = (2x - 3)2 + 1 = 4x2 - 12x + 9 + 1 = 4x2 - 12x + 10
Notice how the binomial expansion (2x - 3)2 = 4x2 - 12x + 9 was used. You should be comfortable expanding squares and simplifying expressions.
注意这里使用了二项式展开 (2x - 3)2 = 4x2 - 12x + 9。你应该熟练掌握平方展开和表达式的化简。
Always write intermediate steps in your exam working to avoid losing marks. Even if the final answer is correct, showing the substitution f(g(x)) explicitly demonstrates your understanding.
考试中务必写出中间步骤以免失分。即使最终答案正确，明确写出 f(g(x)) 这一步也能展现你的理解。
7. Decomposing a Composite Function | 复合函数的分解
A reverse type of question asks you to express a given function h(x) as a composition of two or more simpler functions, i.e., find f and g such that h(x) = f(g(x)). This is called decomposition and is a common IB question type.
一种反向题型要求你将给定的函数 h(x) 表示为两个或更多简单函数的复合，即找到 f 和 g，使得 h(x) = f(g(x))。这称为分解，是 IB 常见题型。
For example, given h(x) = (3x + 2)5, we can choose the "inner function" g(x) = 3x + 2 and the "outer function" f(x) = x5. Then f(g(x)) = (3x + 2)5 = h(x).
例如，给 h(x) = (3x + 2)5，可选择 "内层函数" g(x) = 3x + 2，"外层函数" f(x) = x5。则 f(g(x)) = (3x + 2)5 = h(x)。
Another example: h(x) = 1⁄x2 + 1. Choose g(x) = x2 + 1 and f(x) = 1x. Then f(g(x)) = 1⁄x2 + 1.
再如：h(x) = 1⁄x2 + 1。选 g(x) = x2 + 1 和 f(x) = 1x。则 f(g(x)) = 1⁄x2 + 1。
There may be more than one correct answer to a decomposition question. Choose the simplest decomposition that involves elementary functions, and always verify your answer by recomposing.
分解题的答案可能不唯一。选择涉及基本函数的最简分解，并且一定要通过重新复合来验证你的答案。
8. Composite Functions with Inverses | 复合函数与反函数
A special relationship exists between a function and its inverse: f-1(f(x)) = x, and f(f-1(x)) = x, for all x in the respective domains. This property is fundamentally a composite function relationship.
函数与其反函数之间存在特殊关系：f-1(f(x)) = x，且 f(f-1(x)) = x，对所有属于相应定义域的 x 均成立。该性质本质上是一种复合函数关系。
IB questions may ask you to "show that g is the inverse of f" by computing both f(g(x)) and g(f(x)) and showing that they both simplify to x. This is a rigorous proof of the inverse relationship.
IB 题目可能要求你 "证明 g 是 f 的反函数"，即计算 f(g(x)) 和 g(f(x))，并证明两者都化简为 x。这是证明反函数关系的严谨方法。
For example, let f(x) = x + 12 and g(x) = 2x - 1. To verify they are inverses:
例如，设 f(x) = x + 12 和 g(x) = 2x - 1。验证它们互为反函数：
f(g(x)) = f(2x - 1) = (2x - 1) + 12 = 2x2 = x
g(f(x)) = g(x + 12) = 2(x + 12) - 1 = (x + 1) - 1 = x
Since both composites yield x, the functions are indeed inverses of each other.
由于两个复合结果都等于 x，这两个函数确实互为反函数。
9. Domain and Range of Composite Functions | 复合函数的定义域与值域
Determining the domain of a composite function can be tricky. The rule is: for (f ° g)(x) = f(g(x)), the domain consists of all x in the domain of g such that g(x) is in the domain of f.
确定复合函数的定义域可能较为棘手。规则是：对于 (f ° g)(x) = f(g(x))，其定义域包含所有属于 g 的定义域且使 g(x) 属于 f 的定义域的 x。
Consider f(x) = 1x and g(x) = x - 2. The domain of f is x ≠ 0. Since (f ° g)(x) = 1x - 2, we must have x - 2 ≠ 0, hence x ≠ 2. The domain of g is ℝ, so the domain of the composite is {x ∈ ℝ ∣ x ≠ 2}.
考虑 f(x) = 1x 和 g(x) = x - 2。f 的定义域为 x ≠ 0。由于 (f ° g)(x) = 1x - 2，必须有 x - 2 ≠ 0，即 x ≠ 2。g 的定义域为 ℝ，因此复合函数的定义域为 {x ∈ ℝ ∣ x ≠ 2}。
The range of a composite function is the set of all possible outputs f(g(x)). This is often easier to determine after sketching the graph of the composite function or by considering the range of the inner function and how the outer function transforms it.
复合函数的值域是所有可能的输出值 f(g(x)) 的集合。这通常在画出复合函数图像或先考虑内层函数的值域再分析外层函数如何变换后更容易确定。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One of the most frequent mistakes students make is reversing the order of composition. Remember that f ° g means f(g(x)), NOT g(f(x)). Always check the order specified in the question.
学生最常见的错误之一是弄反复合的顺序。记住：f ° g 表示 f(g(x))，而不是 g(f(x))。始终检查题目中指定的顺序。
Another common error is forgetting to consider the domain of the inner function and the restrictions imposed by the outer function. For example, if f(x) = √x and g(x) = x - 5, then (f ° g)(x) = √x - 5, and the domain is x ≥ 5, not all real numbers.
另一个常见错误是忘记考虑内层函数的定义域以及外层函数对输入的限制。例如，若 f(x) = √x 且 g(x) = x - 5，则 (f ° g)(x) = √x - 5，定义域为 x ≥ 5，而不是全体实数。
Exam tips:
考试小贴士：
- Always write the composite in expanded form before simplifying, e.g., f(g(x)) = f(2x + 1) = ...
- 在化简前始终先写出复合的展开形式，例如 f(g(x)) = f(2x + 1) = ...
- When using a GDC, verify your algebraic answer by computing the composite at a specific value of x.
- 使用图形计算器（GDC）时，可以通过在特定 x 值处计算复合函数来验证你的代数答案。
- Read the question carefully: it may ask for (f ° g)(2), or it may ask for the expression of (f ° g)(x). These require different approaches.
- 仔细审题：题目可能要求 (f ° g)(2)，也可能要求 (f ° g)(x) 的表达式。两者需要不同的处理方法。
- For decomposition, always check your answer by recomposing the functions.
- 对于分解题，务必通过重新复合来检验你的答案。
11. Worked Example: Full IB-Style Problem | 典型例题：完整 IB 风格题型
Let us solve a complete IB-style problem step by step.
让我们一步步解决一道完整的 IB 风格题目。
Problem: Let f(x) = 2x - 1 for x ∈ ℝ, x ≠ 1, and g(x) = x2 + 1 for x ∈ ℝ.
题目：设 f(x) = 2x - 1，其中 x ∈ ℝ, x ≠ 1，且 g(x) = x2 + 1，其中 x ∈ ℝ。
(a) Find (f ° g)(x).
（a）求 (f ° g)(x)。
Solution:
解答：
(f ° g)(x) = f(g(x)) = f(x2 + 1) = 2⁄(x2 + 1) - 1 = 2⁄x2
(b) State the domain of f ° g.
（b）写出 f ° g 的定义域。
Since the denominator x2 ≠ 0, we must have x ≠ 0. The domain of g is all real numbers, so the domain of f ° g is {x ∈ ℝ ∣ x ≠ 0}.
由于分母 x2 ≠ 0，所以必须 x ≠ 0。g 的定义域为全体实数，因此 f ° g 的定义域为 {x ∈ ℝ ∣ x ≠ 0}。
(c) Find (g ° f)(x), and state its domain.
（c）求 (g ° f)(x)，并写出其定义域。
(g ° f)(x) = g(f(x)) = g(2x - 1) = (2x - 1)2 + 1 = 4⁄(x - 1)2 + 1
The domain of f is x ≠ 1. For the composite to be defined, x - 1 ≠ 0, so x ≠ 1. No other restrictions arise, so the domain is {x ∈ ℝ ∣ x ≠ 1}.
f 的定义域为 x ≠ 1。要使得复合函数有定义，需 x - 1 ≠ 0，即 x ≠ 1。无其他限制，故定义域为 {x ∈ ℝ ∣ x ≠ 1}。
(d) Solve (f ° g)(x) = 8.
（d）解方程 (f ° g)(x) = 8。
2⁄x2 = 8 ⇒ x2 = 14 ⇒ x = ± 12
Both solutions are in the domain x ≠ 0, so x = 12 or x = -12.
两个解都在定义域 x ≠ 0 内，故 x = 12 或 x = -12。
12. Summary and Key Takeaways | 总结与核心要点
Composite functions are a central topic in IB Mathematics. The key ideas to remember are: the order of composition matters; (f ° g)(x) = f(g(x)); the domain of the composite is restricted by both the inner and outer functions; and decomposition requires identifying an inner and outer function.
复合函数是 IB 数学的核心内容。需要记住的关键要点有：复合的顺序很重要；(f ° g)(x) = f(g(x))；复合函数的定义域受内层和外层函数的共同限制；分解则需要识别内层和外层函数。
Mastering composite functions not only helps you score marks in direct questions but also builds a solid foundation for topics such as transformations, inverse functions, and calculus, all of which rely on the ability to combine and manipulate functions.
掌握复合函数不仅能帮助你在直接相关题目中得分，还为函数变换、反函数和微积分等主题奠定了坚实基础，这些主题都依赖于组合和操作函数的能力。
Practice with a variety of representations—algebraic, tabular, and graphical—to become fully prepared for any format in your IB exams.
多练习各种表示形式——代数、表格和图形——以便对 IB 考试中的任何形式做好充分准备。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图