📚 Mastering IB Math Exercise 2K: Composite and Inverse Functions | 掌握IB数学练习2K：复合函数与反函数
Exercise 2K typically appears in IB Mathematics textbooks at the stage where students consolidate their understanding of composite and inverse functions. This exercise is a gateway to mastering function manipulation, a skill that runs through the entire IB syllabus, from algebra and calculus to modelling and probability. Whether you are following the AA or AI route, the ability to form f(g(x)), find f⁻¹(x) and work with restricted domains is essential. This article unpacks every concept you need to tackle Exercise 2K confidently, blending clear English explanations with parallel Chinese notes, step-by-step worked examples and exam-style tips.
练习2K通常出现在IB数学教材中函数部分的巩固阶段，要求学生熟练掌握复合函数与反函数。这道练习是掌握函数操作的敲门砖，相关技能贯穿整个IB大纲，从代数、微积分到建模和概率统计。无论你选择AA还是AI方向，构建f(g(x))、求解f⁻¹(x)以及处理定义域限制都是必备基本功。本文深度拆解应对练习2K所需的全部概念，采用中英对照讲解，并配有分步解题示范和考试技巧。
1. Understanding Functions | 理解函数
A function is a rule that assigns each input exactly one output. In IB Mathematics, we often write f : x ↦ f(x) to describe the mapping. The vertical line test determines whether a graph represents a function: if any vertical line cuts the graph more than once, the relation is not a function. Functions can be one-to-one (each y-value comes from exactly one x-value) or many-to-one, where different x-values can give the same y.
函数是一种为每个输入指定唯一输出的规则。在IB数学中，我们常用f : x ↦ f(x)表述这种映射。垂直线测试可用于判断图像是否为函数：如果任意一条垂直线与图像相交超过一次，则该关系不是函数。函数可以是一对一的（每个y值恰好由一个x值产生），也可以是多对一的，即不同的x值可以给出相同的y值。
2. Domain and Range Refresher | 定义域与值域复习
rking with Exercise 2K, you must often identify the maximal natural domain (e.g. x ≥ 0 for √x) and restrict domains to ensure an inverse exists. Notation used includes interval notation [a, b], set-builder notation {x | x > 2} and inequality forms.
函数的定义域是所有允许输入值的集合，值域是所有可能输出值的集合。处理练习2K时，你需要能识别最大自然定义域（例如√x的定义域为x ≥ 0），并且通过限制定义域来保证反函数存在。常用的表示法包括区间表示[a, b]、集合构造式{x | x > 2}以及不等式形式。
reate an invertible function, we restrict the domain to x ≥ 0, giving range y ≥ 0 as well.
对于简单的二次函数f(x) = x²，自然定义域为全体实数，但值域为y ≥ 0。要构造可逆函数，我们会将定义域限制为x ≥ 0，此时值域同样为y ≥ 0。
3. Function Notation and Mappings | 函数符号与映射
IB frequently uses mapping notation such as f : x ↦ given value or expression, e.g. f(4) = 2(4) + 3 = 11, or f(a + 1) = 2(a + 1) + 3 = 2a + 5.
IB考试中常用映射符号，例如f : x ↦ 2x + 3，或简写为f(x) = 2x + 3。理解这种符号对读懂练习2K题目至关重要。计算函数值时，我们用给定数值或表达式替换x，如f(4) = 11，或f(a + 1) = 2a + 5。
4. Composite Functions | 复合函数
The composite function (f ∘ g)(x) is defined as f(g(x)). You first apply g to x, then apply f to the result. The domain of f ∘ g consists of x-values that lie in the domain of g and for which g(x) lies in the domain of f. Exercise 2K often tests this by asking you to form composite functions and state their domains.
复合函数(f ∘ g)(x)的定义为f(g(x))：先将g作用于x，再将f作用于其结果。f ∘ g的定义域由那些既在g的定义域内，又使g(x)落在f定义域内的x值组成。练习2K常会要求构造复合函数并写出其定义域。
(f ∘ g)(x) = f(g(x))
For example, if f(x) = √x and g(x) = x − 4, then (f ∘ g)(x) = √x − 4 all real numbers, but the output g(x) must be ≥ 0 for f to accept it. Therefore x − 4 ≥ 0 ⇒ x ≥ 4.
例如，若f(x) = √x，g(x) = x − 4，则(f ∘ g)(x) = √x − 4。此处g的定义域为全体实数，但g(x)必须≥0才能作为f的输入，因此x − 4 ≥ 0，即x ≥ 4。
5. Inverse Functions | 反函数
problems frequently ask you to find f⁻¹(x) algebraically and to verify that f(f⁻¹(x)) = x.
反函数f⁻¹可以逆转f的作用。如果f(a) = b，则f⁻¹(b) = a。f⁻¹作为函数存在的前提条件是f在其定义域上必须是一对一的。f⁻¹的定义域是f的值域，f⁻¹的值域是f的定义域。练习2K的题目经常要求用代数方法求解f⁻¹(x)，并验证f(f⁻¹(x)) = x。
f⁻¹(f(x)) = x and f(f⁻¹(x)) = x
6. Finding Inverses Algebraically | 求反函数的代数步骤
for y. The result is y = f⁻¹(x). This is the core skill tested in many Exercise 2K items.
求函数y = f(x)的反函数时，先交换x和y，再解出y，所得即为y = f⁻¹(x)。这是练习2K中多次考察的核心技能。
Example: f(x) = 2x + 3. Write y = 2x + 3. Swap: x = 2y + 3. Solve: x − 3 = 2y ⇒ y = (x − 3)/2. So f⁻¹(x) = (x − 3)/2.
例如：f(x) = 2x + 3。设y = 2x + 3，交换得x = 2y + 3，解出y = (x − 3)/2，因此f⁻¹(x) = (x − 3)/2。
Always check: f⁻¹(f(x)) = ( (2x+3) − 3 )/2 = 2x/2 = x. Works perfectly.
务必检验：f⁻¹(f(x)) = ( (2x+3) − 3 )/2 = x，完全吻合。
7. Domain and Range of Inverse Functions | 反函数的定义域和值域
read off the domain of f⁻¹ directly from the range of f. In Exercise 2K you may be asked: "Find f⁻¹ and state its domain." For f(x) = √x − 2 x ≥ 2 and range y ≥ 0, the inverse is f⁻¹(x) = x² + 2, with domain x ≥ 0 and range y ≥ 2.
由于定义域与值域的角色互换，通常可以直接从f的值域读出f⁻¹的定义域。练习2K中可能出现这样的提问：“求f⁻¹并写出其定义域”。对于f(x) = √x − 2，定义域x ≥ 2，值域y ≥ 0，其反函数为f⁻¹(x) = x² + 2，定义域为x ≥ 0，值域为y ≥ 2。
8. Graphical Interpretation of Inverses | 反函数的图形解释
etry helps visualise why a function must be one-to-one to have an inverse that is also a function. If a horizontal line cuts the original graph more than once, the inverse will fail the vertical line test unless the domain is restricted. Many IB questions, including those in Exercise 2K, ask you to sketch both graphs and label the line of symmetry.
y = f⁻¹(x)的图像是y = f(x)关于直线y = x的镜像。这种对称性有助于直观理解为何函数必须一对一才能使其反函数同样是函数。如果原图像被任意水平线切割超过一次，那么反函数将无法通过垂直线测试，除非限制定义域。许多IB考题（包括练习2K中的题目）会要求画出原函数与反函数的图像并标出对称线。
9. Transformations of Functions (Link to Exercise 2K) | 函数变换（与练习2K关联）
Some versions of Exercise 2K integrate transformations such as translations, reflections and stretches. Understanding the effect of adding constants inside or outside the function is essential. The table below summarises the most common transformations seen in IB exams.
部分教材的练习2K会融入平移、反射、伸缩等变换内容。掌握在函数内部或外部添加常数所带来的影响至关重要。下表总结了IB考试中最常见的变换类型。
| Transformation | Effect on y = f(x) |
| Vertical translation k units up | y = f(x) + k |
| Horizontal translation h units right | y = f(x − h) |
| Reflection across x-axis | y = −f(x) |
| Reflection across y-axis | y = f(−x) |
| Vertical stretch by factor a | y = a f(x) |
| Horizontal stretch by factor 1/b | y = f(bx) |
typically leading to g⁻¹(x) = f⁻¹(x − 2). Practice from Exercise 2K connects these ideas.
复合函数和反函数会与这些变换相互作用。例如，如果g(x) = f(x) + 2，那么求g⁻¹(x)就需要逆转平移，通常会得出g⁻¹(x) = f⁻¹(x − 2)。练习2K的题目会将这两类概念联系起来。
10. Solving Equations with Composite and Inverse Functions | 解含复合与反函数的方程
given f⁻¹(2x + 1) = 4, apply f to both sides to get 2x + 1 = f(4) = 11, leading to x = 5.
练习2K中常有类似f(g(x)) = k或f⁻¹(x) = g(x)的方程。推荐策略是先化简复合函数，或在等式两边同时施加反函数。例如，已知f(x) = 3x − 1且f⁻¹(2x + 1) = 4，两边同时作用f，得到2x + 1 = f(4) = 11，进而解得x = 5。
Always verify that solutions lie in the appropriate domains, especially when square roots or denominators are present.
务必验证解是否落在正确的定义域内，尤其是当函数中出现了根号或分母时。
11. Common Pitfalls and Exam Strategies | 常见陷阱与考试策略
Many students confuse (f ∘ the domain before finding an inverse. When solving f(f⁻¹(x)) = x, avoid circular reasoning— ons. If the question asks for f⁻¹ and its domain, both must be correct to earn full marks.
许多学生将(f ∘ g)(x)误认为是f(x)·g(x)，要记住复合不是乘积。另一常见错误是忘记在求反函数前限制定义域。解f(f⁻¹(x)) = x时避免循环论证——它应作为验证手段。考试中要逐步展示所有推导步骤，清晰注明定义域限制。如果题目要求求f⁻¹及其定义域，两者均正确才能得满分。
For Exercise 2K timing, allocate about 1–1.5 minutes per mark. If a question seems too long, check whether an algebraic simplification or a graph sketch can shorten the process.
对于练习2K的时间分配，建议每分值预留1–1.5分钟。若题目过长，可考虑通过代数简化或画图来加快速度。
12. Practice Questions (Style of Exercise 2K) | 模拟练习（练习2K风格）
Try these typical Exercise 2K-style problems to test your understanding. Solutions are outlined below each problem.
尝试以下典型练习2K风格的题目来检验理解程度，每题下方附有简要解答思路。
Q1: f(x) = 4 − 3x, g(x) = 2x + 5. Find (f ∘ g)(x) and (g ∘ f)(x). State their domains.
Q1：f(x) = 4 − 3x, g(x) = 2x + 5，求(f ∘ g)(x)和(g ∘ f)(x)，并指出定义域。
Solution: (f ∘ g)(x) = f(g(x)) = 4 − 3(2x + 5) = 4 − 6x − 15 = −6x − 11. (g ∘ 5 = 8 − 6x + 5 = 13 − 6x. Both are linear functions, so the domain is all real numbers, ℝ.
解答：(f ∘ g)(x) = −6x − 11，(g ∘ f)(x) = 13 − 6x，两者均为线性函数，定义域为全体实数ℝ。
Q2: Given f(x) = (x + 1)/(x − 2), x ≠ 2. Find f⁻¹(x) and its domain.
Q2：已知f(x) = (x + 1)/(x − 2), x ≠ 2，求f⁻¹(x)及其定义域。
Solution: Let y = (x+1)/(x−2). Swap x and y: x = (y+1)/(y−2). Cross-multiply: x(y−2) = y+1 ⇒ xy − 2x = y+1 ⇒ xy − y = 2x+1 ⇒ y(x−1) = 2x+1 ⇒ y = (2x+1)/(x−1). So f⁻¹(x) = (2x+1)/(x−1), domain x ≠ 1.
解答：f⁻¹(x) = (2x+1)/(x−1)，定义域x ≠ 1。
Q3: Given f(x) = √x + 3 and g(x) = x² − 1, find the domain of f(g(x)) and evaluate (f ∘ g)(2).
Q3：已知f(x) = √x + 3, g(x) = x² − 1，求f(g(x))的定义域并计算(f ∘ g)(2)。
Solution: f(g(x)) = √(g(x) + 3) = √x² − 1 + 3 = √x² + 2 s always ≥ 2 for all real x, the radicand is never negative. Domain is all real numbers, ℝ. (f ∘ g)(2) = √4 + 2 = √6.
解答：f(g(x)) = √x² + 2，由于x² + 2对所有实数x都≥2，定义域为全体实数ℝ。(f ∘ g)(2) = √6。
Q4: The graph of f passes through (3, 5). What point must lie on the graph of f⁻¹?
Q4：f的图像经过点(3, 5)，f⁻¹的图像必过哪个点？
Answer: (5, 3), because the inverse swaps coordinates.
答案：(5, 3)，因为反函数会交换坐标。
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