📚 Exercise 2F: Composite and Inverse Functions | 练习2F：复合函数与反函数
Welcome to the focused revision guide for Exercise 2F, which explores two foundational ideas in IB Mathematics: composite functions and inverse functions. Understanding how to build new functions from existing ones and how to reverse the mapping process is essential for success in both Analysis & Approaches and Applications & Interpretation. This article breaks down every key concept with clear explanations, practical examples, and exam-style hints so you can tackle Exercise 2F with confidence.
欢迎来到练习 2F 的专项复习指南，本练习探讨 IB 数学中两个基础概念：复合函数与反函数。理解如何从已有函数构建新函数，以及如何逆转映射过程，对于在分析与方法和应用与解释课程中取得成功至关重要。本文将通过清晰的解释、实用示例和考试风格提示，分解每一个关键概念，帮助你有信心地攻克练习 2F。
1. Function Refresher: Notation and Core Ideas | 函数回顾：符号与核心思想
² + 3. The set of all allowable input values is called the domain, and the set of all possible output values is called the range.
函数是一种规则，它为每个输入值 x 分配恰好一个输出值 f(x)。在 IB 中，我们通常写作 f : x → f(x)，并使用如 f(x) = x² + 3 这样的记号。所有允许的输入值的集合称为定义域，所有可能的输出值的集合称为值域。
When working with composite and inverse functions, you must always keep domain restrictions in mind. For example, the square root function √x – 2 requires x – hecking domains, as IB exam questions frequently test your ability to infer natural domains from expressions or context.
在处理复合函数和反函数时，必须时刻牢记定义域的限制。例如，平方根函数 √x – 2 要求 x – 2 ≥ 0，因此它的定义域是 x ≥ 2。绝不要跳过对定义域的检查，因为 IB 考试题目经常测试你从表达式或上下文推断自然定义域的能力。
Additionally, recall that a function can be represented verbally, algebraically, graphically, or numerically. Composite and inverse concepts rely heavily on algebraic manipulation, but you should also be able to interpret them from graphs and tables.
另外，回忆一下，函数可以用文字、代数、图形或数值方式表示。复合与反函数概念在很大程度上依赖于代数操作，但你也应该能够从图形和表格中解读它们。
2. Domain and Range Revisited | 再次审视定义域和值域
e set of inputs for which the function is defined. For rational functions like f(x) = 1/(x – 3), the denominator cannot be zero, so x ≠ 3. For even-root functions such as f(x) = √2x + 1, the radicand must be non-negative, giving x ≥ -1/2.
函数的定义域是使函数有意义的输入值集合。对于有理函数，如 f(x) = 1/(x – 3)，分母不能为零，因此 x ≠ 3。对于偶次根式函数，如 f(x) = √2x + 1，被开方数必须非负，得到 x ≥ -1/2。
y. For example, if f(x) = 2/(x + 1), rewriting as y = 2/(x + 1) and solving gives x = 2/y – 1, so y cannot be 0.
值域描述函数能够产生的输出值集合。为了求值域，通常先将 y 解出 x，然后找出对 y 的限制。例如，如果 f(x) = 2/(x + 1)，重写为 y = 2/(x + 1) 并求解得 x = 2/y – 1，因此 y 不能为 0。
A quick technique: sketch the graph of the function whenever possible. The domain becomes the set of x‑coordinates where the graph exists, and the range is the set of y‑coordinates covered by the graph. This visual approach helps avoid common algebraic mistakes.
一个快捷技巧：尽可能画出函数的大致图形。定义域就变为图形存在的 x 坐标集合，值域则是图形覆盖的 y 坐标集合。这种可视化方法有助于避免常见的代数错误。
3. Introducing Composite Functions | 引入复合函数
e function is applied to the output of another. We denote the composition of f and g as (f ∘ g)(x) = f(g(x)). Read it as “f of g of x.” The inner function g is applied first, and then f works on the result.
复合函数是通过将一个函数作用于另一个函数的输出而得到的。我们将 f 与 g 的复合记为 (f ∘ g)(x) = f(g(x))，读作 “f of g of x”。先应用内层函数 g，然后 f 再作用于其结果。
For example, if f(x) = x² and g(x) = 3x + 1, then (f ∘ u simply replace the input variable in f with the entire expression of g(x). Similarly, (g ∘ f)(x) = g(x²) = 3x² + 1, showing that order matters greatly.
例如，如果 f(x) = x²，g(x) = 3x + 1，则 (f ∘ g)(x) = f(3x + 1) = (3x + 1)²。你只需把 f 中的输入变量替换为整个 g(x) 的表达式。类似地，(g ∘ f)(x) = g(x²) = 3x² + 1，这表明顺序至关重要。
Always use parentheses and substitute carefully. IB questions may hide tricky expressions where you must simplify the result, like expanding (3x + 1)² to 9x² + 6x + 1. Practice both forming and simplifying composite functions until the process feels automatic.
始终使用括号并仔细代入。IB 题目可能会隐藏一些棘手的表达式，你需要对结果进行化简，例如将 (3x + 1)² 展开为 9x² + 6x + 1。反复练习构造和化简复合函数，直到过程感觉自然而然。
4. Finding the Domain of a Composite Function | 求复合函数的定义域
The domain of (f ∘ g)(x) is not simply the domain of g. It consists of all x in the domain of g such that g(x) is in the domain of f. In other words, you need both g to be defined at x and f to be defined at g(x).
(f ∘ g)(x) 的定义域不仅仅是 g 的定义域。它由所有这样的 x 组成：x 属于 g 的定义域，且 g(x) 属于 f 的定义域。换言之，要求 g 在 x 处有定义，同时 f 在 g(x) 处有定义。
hen find the values x that make g(x) satisfy the restrictions of f. For example, if f(x) = √x (domain x ≥ 0) and g(x) = x – 2 (domain all real numbers), then (f ∘ g)(x) = √x – 2. The composite requires x – 2 ≥ 0, so final domain is x ≥ 2.
一种结构化方法：首先记下 g(x) 的限制条件，然后找出使 g(x) 满足 f 限制条件的 x 值。例如，如果 f(x) = √x（定义域 x ≥ 0），g(x) = x – 2（定义域全体实数），则 (f ∘ g)(x) = √x – 2。该复合函数要求 x – 2 ≥ 0，因此最终定义域为 x ≥ 2。
When both functions have restrictions, combine them. Suppose f(x) = 1/x (x ≠ 0) and g(x) = √x – 1 (x ≥ 1). For (f ∘ g)(x) = 1/√x – 1, g(x) is defined for x ≥ 1, but we also need g(x) ≠ 0, which means √x – 1 ≠ 0, so x ≠ 1. Hence the domain is x strictly greater than 1.
当两个函数都有限制时，将它们合并。假设 f(x) = 1/x (x ≠ 0)，g(x) = √x – 1 (x ≥ 1)。对于 (f ∘ g)(x) = 1/√x – 1，g(x) 在 x ≥ 1 时有定义，但我们还需要 g(x) ≠ 0，这意味着 √x – 1 ≠ 0，所以 x ≠ 1。因此定义域为 x 严格大于 1。
is expressed in set notation or interval notation in IB exams. For instance, the domain {x ∈ ℝ | x > 1} or in interval notation (1, ∞).
记住，IB 考试中定义域用集合符号或区间符号表示。例如，定义域 {x ∈ ℝ | x > 1} 或在区间符号 (1, ∞)。
5. Decomposing Composite Functions | 分解复合函数
(x)). This skill is especially useful in differentiation using the chain rule and in modelling real-world situations.
分解是逆过程：给定一个函数 h(x)，找出两个函数 f 和 g，使得 h(x) = f(g(x))。这一技能在使用链式法则进行微分以及建模实际情境时特别有用。
A typical strategy is to look for an “inner” u⁴ (the outer part). Always verify by composing them back: f(g(x)) = (g(x))⁴ = (2x + 5)⁴.
一个典型策略是寻找一个重复出现或进行了幂运算的“内层”表达式。例如，如果 h(x) = (2x + 5)⁴，你可以令 g(x) = 2x + 5（内层部分），f(u) = u⁴（外层部分）。始终通过复合回去来验证：f(g(x)) = (g(x))⁴ = (2x + 5)⁴。
For h(x) = sin(3x² – 1), an obvious decomposition is g(x) = 3x² – IB often asks you to decompose a function as part of a larger calculus or algebra problem.
对于 h(x) = sin(3x² – 1)，一个显然的分解是 g(x) = 3x² – 1，f(u) = sin u。当存在多个正确答案时，选择使后续分析更简单的那个。IB 经常要求你在一个较大的微积分或代数问题中分解函数。
6. Defining Inverse Functions | 定义反函数
An inverse function undoes the action of the original function. If f maps a to b, then the inverse function f⁻¹ maps b back to a. For the inverse to exist, f must be one‑to‑one (injective) on its domain, meaning each y‑value comes from exactly one x‑value.
反函数撤销原函数的作用。如果 f 将 a 映射到 b，那么反函数 f⁻¹ 将 b 映射回 a。为了使反函数存在，f 在其定义域上必须是一一映射（单射），这意味着每个 y 值恰好来自一个 x 值。
Graphically, a function is one‑to‑ ts the graph at most once. For example, f(x) = x³ passes the test, so it has an inverse on ℝ not on its natural domain, so we restrict the domain to x ≥ 0 to define a usable inverse.
图形上，如果函数通过水平线检验，它就是一一函数：任何水平线与图形最多相交一次。例如，f(x) = x³ 通过了检验，因此在 ℝ 上有反函数。但 f(x) = x² 在其自然定义域上没有通过，因此我们将定义域限制为 x ≥ 0 来定义一个可用的反函数。
The notation f⁻¹(x) does not mean 1/f(x). It is the inverse mapping. Avoid this common exam blunder. Also, the domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹.
符号 f⁻¹(x) 并不表示 1/f(x)。它表示逆映射。避免这个考试中常见的错误。此外，f 的定义域成为 f⁻¹ 的值域，f 的值域成为 f⁻¹ 的定义域。
7. Finding Inverse Functions Step by Step | 逐步求反函数
y, then solve for y. This new y is f⁻¹(x). Always check that the original function is one‑to‑one over the given domain.
用代数方法求反函数的标准方法是：写出 y = f(x)，交换 x 和 y，然后解出 y。这个新的 y 就是 f⁻¹(x)。务必要检查原函数在给定定义域上是否是一一映射。
Example: find the inverse of f(x) = (2x – 1)/(x + 3), defined for x ≠ -3. Write y = (2x – 1)/(x + 3). Swap: x = (2y – 1)/(y + 3). Multiply both sides by (y + 3): x(y + 3) = 2y – 1 → xy + 3x = 2y – 1. Collect y‑terms on one side: xy – 2y = -3x – 1 → y(x – 2) = -3x – 1. Finally, y = (-3x – 1)/(x – 2), so f⁻¹(x) = (-3x – 1)/(x – 2), with domain x ≠ 2.
示例：求 f(x) = (2x – 1)/(x + 3) 的反函数，该函数在 x ≠ -3 上有定义。写出 y = (2x – 1)/(x + 3)。交换：x = (2y – 1)/(y + 3)。两边乘以 (y + 3)：x(y + 3) = 2y – 1 → xy + 3x = 2y – 1。将含 y 的项移到一边：xy – 2y = -3x – 1 → y(x – 2) = -3x – 1。最后，y = (-3x – 1)/(x – 2)，因此 f⁻¹(x) = (-3x – 1)/(x – 2)，定义域为 x ≠ 2。
For simpler functions such as f(x) = 3x – 7, swapping gives x = 3y – adical, and exponential functions to become fluent in the algebraic manipulation required.
对于更简单的函数，如 f(x) = 3x – 7，交换得 x = 3y – 7，解出 y：y = (x + 7)/3，因此 f⁻¹(x) = (x + 7)/3。通过练习有理函数、根式函数和指数函数，熟练掌握所需的代数操作。
8. Domain and Range of Inverse Functions | 反函数的定义域与值域
had a restricted domain, the range often becomes easier to determine. For instance, f(x) = √x – 4 with domain x ≥ 4 has range y ≥ 0. Hence f⁻¹(x) has domain x ≥ 0.
一旦得到 f⁻¹，你必须说明它的定义域，也就是原函数 f 的值域。如果原函数有受限的定义域，值域往往更容易确定。例如，f(x) = √x – 4 定义域为 x ≥ 4，值域为 y ≥ 0。因此 f⁻¹(x) 的定义域为 x ≥ 0。
omains. These composition identities give a powerful check. For the previous rational example, substituting f⁻¹(x) into f should simplify to x, confirming the inverse is correct.
始终验证 f(f⁻¹(x)) = x 和 f⁻¹(f(x)) = x 对于在合适定义域内的 x 成立。这些复合恒等式提供了强有力的检验。对于前面的有理例子，将 f⁻¹(x) 代入 f 应简化为 x，从而确认反函数正确。
e y = x. This geometric relationship helps you quickly spot errors in algebraic solutions.
在同一坐标系中画出 f 和 f⁻¹ 作为视觉检验：它们的图形关于直线 y = x 对称。这种几何关系帮你快速发现代数解法中的错误。
9. Interpreting Composition with Inverses | 理解逆复合
ns and inverse functions interact beautifully. For any invertible function f, we have (f⁻¹ ∘ f)(x) = x and (f ∘ main. This property is often used in equations and in simplifying complicated expressions.
复合函数和反函数有着优美的相互作用。对于任何可逆函数 f，有 (f⁻¹ ∘ f)(x) = x 和 (f ∘ f⁻¹)(x) = x，条件是 x 属于相应的定义域。这一性质常用于方程以及简化复杂表达式。
Example: If f(x) = ln(x) and g(x) = eˣ, then (f ∘ g)(x) = ln(eˣ) = x, and (g ∘ f)(x) = eln x ach other, so their composition in either order yields the identity function on the appropriate domain.
示例：如果 f(x) = ln(x)，g(x) = eˣ，则 (f ∘ g)(x) = ln(eˣ) = x，而 (g ∘ f)(x) = eln x = x。这两个函数互为反函数，因此按任意顺序复合都会在合适的定义域上得到恒等函数。
olving equations like f(g(x)) = a, you may apply inverses strategically. For example, if e2x + 1 = 5, taking the natural log of both sides “undoes” is simply applying the inverse function ln to both sides because ln is the inverse of exp.
在解如 f(g(x)) = a 这样的方程时，你可以策略性地应用反函数。例如，若 e2x + 1 = 5，对两边取自然对数就“撤销”了指数函数：2x + 1 = ln 5。这其实就是对两边应用反函数 ln，因为 ln 是指数函数的反函数。
Always check for extraneous solutions caused by domain narrowing. Even when using inverses, ensure your final answer lies in the original domain of the nested functions.
务必检查由于定义域缩小导致的增根。即使使用反函数，也要确保最终答案位于嵌套函数的原有定义域内。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
One typical mistake is forgetting to swap x and y when calculating inverses, leaving the answer in terms of y instead of x. Another is failing to write the inverse function with its restricted domain. IB examiners expect both the algebraic expression and the domain statement.
一个典型错误是在计算反函数时忘记交换 x 和 y，导致答案仍然用 y 表示而不是 x。另一个错误是忘记写出反函数及其受限定义域。IB 考官期望同时给出代数表达式和定义域声明。
l. Mixing these can cost valuable marks, especially in calculus questions involving inverse trigonometric functions.
学生也经常混淆符号 f⁻¹(x) 与 [f(x)]⁻¹。记住：f⁻¹(x) 是反函数，而 [f(x)]⁻¹ = 1/f(x) 是倒数。混淆这些符号可能导致严重失分，尤其是在涉及反三角函数的微积分题目中。
When composing functions, always work inside‑out, and use parentheses generously. For (f ∘ g ∘ h)(x), start with h(x), then apply g, then f. Misplacing the order will give an entirely wrong result. If you are unsure, test with a simple number to catch slip‑ups.
进行函数复合时，始终从内向外操作，并大方地使用括号。对于 (f ∘ g ∘ h)(x)，先从 h(x) 开始，然后应用 g，最后应用 f。搞错顺序会得到完全错误的结果。如果不确定，用一个简单数字测试来捕捉失误。
Finally, practice with past IB questions on composite and inverse functions. They often combine these concepts with transformations, quadratics, exponentials, and rational functions. Building confidence here will pay off across multiple topics in the syllabus.
最后，用过去的 IB 题目练习复合函数和反函数。这些题目经常将这些概念与变换、二次函数、指数函数和有理函数结合在一起。在此建立信心将使整个教学大纲中的多个主题受益。
11. Worked Exercise-Style Problem Set | 练习题样题集
Let’s consolidate with a few Exercise 2F‑style problems. Try to solve them before reading the solutions.
让我们通过几道练习 2F 风格的题目进行巩固。在阅读解答之前，请先尝试自己求解。
Problem 1: Given f(x) = 2x – 5 and g(x) = x² + 1, find (f ∘ g)(x) and (g ∘ f)(x). Then evaluate (f ∘ g)(–2).
题目 1： 已知 f(x) = 2x – 5，g(x) = x² + 1，求 (f ∘ g)(x) 和 (g ∘ f)(x)。然后计算 (f ∘ g)(–2)。
Solution: (f ∘ g)(x) = f(x² + 1) = 2(x² + 1) – 5 = 2x² – 3. (g ∘ f)(x) = g(2x – 5) = (2x – 5)² + 1 = 4x² – 20x + 26. Evaluating: (f ∘ g)(–2) = 2(–2)² – 3 = 8 – 3 = 5.
解答： (f ∘ g)(x) = f(x² + 1) = 2(x² + 1) – 5 = 2x² – 3。(g ∘ f)(x) = g(2x – 5) = (2x – 5)² + 1 = 4x² – 20x + 26。求值：(f ∘ g)(–2) = 2(–2)² – 3 = 8 – 3 = 5。
Problem 2: Determine the inverse of f(x) = √3x – 6 with domain x ≥ 2, and state its domain and range.
题目 2： 求 f(x) = √3x – 6（定义域 x ≥ 2）的反函数，并写出其定义域和值域。
Solution: Write y = √3x – 6. Swap: x = √3y – 6. Square both sides: x² = 3y – verse has domain x ≥ 0 and range y ≥ 2. So f⁻¹(x) = (x² + 6)/3, domain x ≥ 0, range y ≥ 2.
解答： 写出 y = √3x – 6。交换：x = √3y – 6。两边平方：x² = 3y – 6 → 3y = x² + 6 → y = (x² + 6)/3 = f⁻¹(x)。原函数 f 的值域为 y ≥ 0，因此反函数的定义域为 x ≥ 0，值域为 y ≥ 2。所以 f⁻¹(x) = (x² + 6)/3，定义域 x ≥ 0，值域 y ≥ 2。
Problem 3: For f(x) = 1/(x + 2) and g(x) = 3x – 1, find the domain of (f ∘ g)(x).
题目 3： 对于 f(x) = 1/(x + 2) 和 g(x) = 3x – 1，求 (f ∘ g)(x) 的定义域。
Solution: (f ∘ g)(x) = f(3x – 1) = 1/((3x – all real numbers. However, the denominator 3x + 1 cannot be zero, so x ≠ -1/3. Domain: {x ∈ ℝ | x ≠ -1/3}.
解答： (f ∘ g)(x) = f(3x – 1) = 1/((3x – 1) + 2) = 1/(3x + 1)。g(x) 在所有实数上有定义。但分母 3x + 1 不能为零，所以 x ≠ -1/3。定义域：{x ∈ ℝ | x ≠ -1/3}。
12. Summary and Final Checklist | 总结与最终清单
braically, and stating their domains. A quick self-check: can you explain why f(x) = x² on ℝ does not have an inverse, but on x ≤ 0 it does? Can you write the domain of (f ∘ g)(x) when f(x) = ln x and g(x) = 2x – 5?
至此，你应该对核心步骤感到自如：构造并化简复合函数，确定复合定义域，分解函数，用代数方法求反函数，以及说明它们的定义域。快速自检：你能解释为什么 f(x) = x² 在 ℝ 上没有反函数，而在 x ≤ 0 上有？你能写出当 f(x) = ln x 且 g(x) = 2x – 5 时 (f ∘ g)(x) 的定义域吗？
Keep this revision sheet handy while working through Exercise 2F. Identify each topic within the exercise questions, apply the techniques step by step, and you’ll find that composite and inverse functions become a genuine strength in your IB Mathematics toolkit.
在做练习 2F 时，将这份复习资料放在手边。在练习题中识别每个主题，逐步应用这些技巧，你会发现复合函数与反函数将成为你 IB 数学工具箱中真正的强项。
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