📚 Exercise 3C | 练习 3C
This article is designed to help IB Mathematics students master the core ideas behind Exercise 3C, which typically focuses on composite functions. We will explore notation, evaluation, algebraic manipulation, domain and range, decomposition, graphical interpretation, common pitfalls, and exam-style problems. Each concept is presented with clear explanations followed by a matching Chinese translation, ensuring you build both conceptual understanding and bilingual fluency.
本文专为 IB 数学学生打造，旨在彻底掌握练习 3C 的核心内容——复合函数。我们将逐步讲解符号表示、求值方法、代数构建、定义域与值域、函数分解、图像意义、常见错误以及 IB 真题风格的题目。每个要点均先提供英文阐述，紧接着给出对应的中文解释，帮助你既夯实概念又提升双语表达能力。
1. Understanding Composite Functions | 理解复合函数
A composite function is formed when the output of one function becomes the input of another. If we have two functions, f and g, the composite function f(g(x)) means we first apply g to x, then apply f to the result. The order matters: f(g(x)) is generally different from g(f(x)). This operation chains two processes into a single rule. For instance, if g adds 3 to a number and f doubles the result, then f(g(x)) takes x, adds 3, then doubles that sum, giving 2(x+3).
复合函数是指将一个函数的输出作为另一个函数的输入所形成的新函数。对于两个函数 f 和 g，复合函数 f(g(x)) 表示先对 x 应用 g，再将结果代入 f。运算顺序至关重要：f(g(x)) 通常与 g(f(x)) 不同。这一操作将两个过程串联为一个统一的规则。例如，若 g 对输入加 3，f 对结果乘以 2，则 f(g(x)) 先加 3，再将和翻倍，得到 2(x+3)。
2. Notation and Evaluation | 符号与求值
There are two common ways to denote a composite function: f∘g(x) or f(g(x)). Both are read as 'f of g of x'. The small circle ∘ evaluate a composite at a specific value, work from the inside out. For example, if f(x)=√x and g(x)=x²+1, to find f(g(2)) we first compute g(2)=5, then f(5)=√5. Always handle the innermost function first, then move outward.
表示复合函数有两种常见方式：f∘g(x) 或 f(g(x))，均读作“f of g of x”。小圆圈符号 ∘ 表示函数依次实施。要对某一特定值求复合函数的值，应从内向外计算。例如，设 f(x)=√x，g(x)=x²+1，要求 f(g(2))，先算 g(2)=5，再算 f(5)=√5。务必先处理最内层函数，再逐层向外。
3. Algebra of Composite Functions | 复合函数的代数运算
gn errors. If f(x)=1/(x−1) and g(x)=x², then f(g(x))=1/(x²−1). Always factor and simplify where possible — this helps later when identifying domain restrictions.
要获得 f(g(x)) 的简化表达式，只需将 f 中的自变量全部替换为 g(x) 的整体表达式。例如，设 f(x)=2x−3，g(x)=x/4，则 f(g(x))=2(x/4)−3=(x/2)−3。当处理较复杂的函数时，务必使用括号以避免符号错误。若 f(x)=1/(x−1)，g(x)=x²，则 f(g(x))=1/(x²−1)。只要可能，应进行因式分解与化简，这将有助于后续识别定义域限制。
4. Domain Considerations | 定义域的考量
is in the domain of f. This two-step restriction often catches students out. Suppose f(x)=√x (domain x≥0) and g(x)=x−4 (all real numbers). For f(g(x))=√x−4 lways test both conditions: first ensure the inner function exists, then check that its output can be accepted by the outer function.
复合函数 f(g(x)) 的定义域由所有满足以下条件的 x 构成：x 在 g 的定义域内，且 g(x) 落在 f 的定义域内。这种双层限制常令学生失误。假设 f(x)=√x（定义域 x≥0），g(x)=x−4（全体实数）。要使 f(g(x))=√x−4 有效，必须满足 x−4≥0→x≥4，因此复合函数的定义域为 [4,∞)。务必检验两个条件：先确保内层函数有意义，再验证其输出能被外层函数接受。
5. Range of Composite Functions | 复合函数的值域
Finding the range of a composite function requires tracing how the inner function' th the range of g for the allowed domain, then apply f to that set. For example, if f(x)=eˣ ² (domain all reals), then the range of g is (−∞,0]. Applying f, we obtain output values eʸ where y≤0, giving the range (0,1]. A sketch or number‑line analysis often helps visualise the mapping.
求复合函数的值域需要追踪内层函数的输出集如何被外层函数变换。首先在允许定义域下找出 g 的值域，再用 f 作用在该集合上。例如，设 f(x)=eˣ，g(x)=−x²（定义域为全体实数），则 g 的值域为 (−∞,0]。将 f 应用其上，得到 eʸ 且 y≤0，值域为 (0,1]。绘制草图或数轴分析常有助于直观观察映射过程。
6. Decomposing Functions | 分解函数
=3x+1 and f(x)=sin x. When dealing with roots, logs, or powers, look for an evident inside–e structure. Practice with examples like h(x)=ln(5x²−4) leads to g(x)=5x²−4 and f(x)=ln x.
分解是复合的逆过程：给出一个复杂的函数 h(x)，我们尝试将其表示为两个简单函数的复合 f(g(x))。这一技巧对运用链式法则求导以及解方程至关重要。对于 h(x)=sin(3x+1)，自然可选择 g(x)=3x+1，f(x)=sin x。在处理根式、对数或幂函数时，要寻找明显的“内层–外层”结构。通过练习如 h(x)=ln(5x²−4) 可得到 g(x)=5x²−4，f(x)=ln x。
7. Composite Functions with More Than Two Functions | 三个及以上函数的复合
ression f(g(h(x))) means apply h first, then g, then f. The associative property holds: (f∘g)∘h = f∘(g∘h). For example, with f(x)=√x s even more restrictive because we must satisfy conditions for h, then g, then f. In this case, we need x+2≠0 and 1/(x+2)>0, so x>−2 and x≠−2, giving x>−2.
复合可以拓展到三个或更多函数。f(g(h(x))) 表示先实施 h，再实施 g，最后实施 f。结合律成立：(f∘g)∘h = f∘(g∘h)。例如，设 f(x)=√x，g(x)=1/x，h(x)=x+2，可得 f(g(h(x)))=√(1/(x+2))。此时定义域限制更强，因为必须依次满足 h、g、f 的条件。本题中，需有 x+2≠0 且 1/(x+2)>0，因此 x>−2。
8. Graphical Interpretation | 图形解释
A composite function can be viewed as a sequence of transformations along the x‑axis and y‑ , and f transforms u to y=f(u), then the graph of y=f(g(x)) maps x through an intermediate 'hidden' riable. For linear cases, the effect is a new linear function with combined slope. For non‑ chain helps in sketching complicated graphs by breaking them into simple transformations.
复合函数可看作自变量沿 x 轴与 y 轴的一系列变换。若 g 将 x 映射为 u=g(x)，f 再将 u 映射为 y=f(u)，则 y=f(g(x)) 的图像便将 x 通过一个中间“隐藏”变量进行映射。对于线性情形，效果是一个新的线性函数，斜率为两者的组合。对于非线性情形，形状随最先应用的函数而变。理解这条变换链条有助于将复杂图形拆解为简单变换，从而轻松绘制草图。
9. Common Errors and How to Avoid | 常见错误及避免方法
The most frequent mistake is applying functions in the wrong order — confusing f(g(x)) with g(f(x)). A reliable habit is to always write the expression inside out: start with x, then enclose it in the inside function's rule, and finally enclose that result in the outer function. Another error is forgetting to check the domain of the inner function first, or naively simplifying a rational composite before identifying restrictions. Also, misusing brackets in algebra leads to sign errors. Always use parentheses when substituting.
最常见的错误是顺序颠倒——混淆 f(g(x)) 和 g(f(x))。一个可靠的习惯是始终从内向外书写表达式：以 x 开始，将其套入内层函数的规则中，最后再将结果套入外层函数。另一个错误是忘记首先检验内层函数的定义域，或者在未识别限制条件时就盲目化简有理函数复合式。此外，代数操作中误用括号会导致符号错误。代入时务必使用圆括号。
10. IB‑Style Practice Problems | IB 风格练习题
Now try these typical Exercise 3C problems adapted for IB Math Analysis & Approaches. For each, express the composite in simplest form and state its domain.
现在试解以下改编自 IB 数学分析与方法的典型练习 3C 题目。对每一小题，写出复合函数的最简形式并注明定义域。
| Functions | Composite to find |
|---|---|
| f(x)=4x−1, g(x)=1/x | f(g(x)) |
| f(x)=√x+5, g(x)=x²−9 | g(f(x)) |
| f(x)=eˣ, g(x)=ln x | f(g(x)) and state its domain |
| f(x)=1/(x−2), g(x)=3x+1 | f(g(x)) and find values of x for which it is undefined |
Solutions:
解答：
For f(x)=4x−1, g(x)=1/x: f(g(x))=4(1/x)−1 = 4/x − 1. Domain: x ≠ 0.
对于 f(x)=4x−1, g(x)=1/x: f(g(x))=4(1/x)−1 = 4/x − 1。定义域：x ≠ 0。
For f(x)=√x+5, g(x)=x²−9: g(f(x)) = (√x+5)² − 9 = x+5 − 9 = x − 4. But note: f(x) requires x+5 ≥ 0 → x ≥ −5. So domain is [−5, ∞).
对于 f(x)=√x+5, g(x)=x²−9: g(f(x)) = (√x+5)² − 9 = x+5 − 9 = x − 4。但注意 f(x) 要求 x+5 ≥ 0 → x ≥ −5，因此定义域为 [−5, ∞)。
For f(x)=eˣ, g(x)=ln x: f(g(x)) = eln x sitive numbers. So domain is (0, ∞), and the composite simplifies to y=x on that interval.
对于 f(x)=eˣ, g(x)=ln x: f(g(x)) = eln x = x，但必须限制 x>0，因 ln x 仅对正数有定义。故定义域为 (0, ∞)，在该区间上复合函数简化为 y=x。
−1=0 → x=1/3. Also g(x) is defined for all reals, so the domain is all reals except x=1/3.
对于 f(x)=1/(x−2), g(x)=3x+1: f(g(x)) = 1/[(3x+1)−2] = 1/(3x−1)。分母为零时无定义：3x−1=0 → x=1/3。同时 g(x) 对所有实数有定义，因此定义域为除 x=1/3 外的全体实数。
11. Linking Composite Functions to Inverse Functions | 复合函数与反函数的联系
k: f(f⁻¹(x))=2((x−3)/2)+3=x−3+3=x. Verify both directions to ensure true inverses. This compositional perspective reinforces the idea that inverses 'undo' each other.
一个函数与其反函数复合会得到恒等函数：f(f⁻¹(x)) = x 且 f⁻¹(f(x)) = x，前提是 x 处于合适的定义域内。这一性质在解方程时至关重要。在 IB 中，你必须能用复合的方式验证反函数。若 f(x)=2x+3，则 f⁻¹(x)=(x−3)/2。检验：f(f⁻¹(x))=2((x−3)/2)+3=x−3+3=x。务必双向验证以确保真正互逆。这种复合视角强化了反函数“相互抵消”的性质。
12. Summary and Revision Tips | 总结与复习建议
Mastering Exercise 3C requires fluency in both the mechanics and the conceptual screening of domains. Practice by varying the order of functions, sketching graphs of composites, and spotting hidden restrictions. When revising, write down a small library of standard functions and their composites, noting domain changes. Use the 'inside‑out' approach for evaluation, and always ask: 'What must be true for the chain to be valid?' Consistent practice with IB‑style questions builds the confidence needed for exam success.
掌握练习 3C 既需要熟练的代数操作，也需要对定义域进行概念性筛查。通过交换函数顺序、绘制复合函数图像、发现隐藏的限制条件来加强训练。复习时，可制作一个小型函数库，列出常见复合式及定义域的变化。运用“从内而外”的求值策略，并始终自问：“要使整个链条有效，必须满足什么条件？”坚持练习 IB 风格的题目，将为你赢取考试成功所需的信心。
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