📚 Exercise 2B: Functions, Domain and Range | 练习2B：函数、定义域与值域
This guide covers the essential skills tested in IB Mathematics Exercise 2B, including function evaluation, identifying domain and range, working with composite and inverse functions, and interpreting mapping diagrams. Each section pairs an English explanation with a Chinese translation to support bilingual learners, and all examples are fully worked through to mirror the style of exam-style questions.
本指南涵盖 IB 数学练习 2B 中考查的核心技能，包括求函数值、确定定义域与值域、处理复合函数与反函数，以及解读映射图。每个部分配有英文解释和中文翻译，以辅助双语学习者，所有例题均提供完整解答，与考试题型风格一致。
1. Understanding Function Notation | 理解函数符号
epresents the input and f(x) is the corresponding output. For example, f(x) = 2x + 3 means 'multiply the input by 2 and then add 3'.
函数是一种为每个输入值分配唯一输出值的规则。在 IB 数学中，我们将其写作 f(x) = 表达式，其中 x 表示输入值，f(x) 是对应的输出值。例如，f(x) = 2x + 3 表示“将输入值乘以 2 再加 3”。
When we write f(4), we replace every x with 4. Always use parentheses carefully; f(x + h) means the input is the whole quantity (x + h). This distinction is especially important when simplifying difference quotients or evaluating pieceswise functions.
当我们写 f(4) 时，我们用 4 替换每个 x。务必仔细使用括号；f(x + h) 表示输入是整个量 (x + h)。在简化差商或求分段函数值时，这一区分尤为重要。
- f(a) → substitute x = a | f(a) → 代入 x = a
- f(g(x)) → input is g(x) | f(g(x)) → 输入是 g(x)
- f(x + 1) = 2(x + 1) + 3 = 2x + 5 (correct expansion) | f(x + 1) = 2(x + 1) + 3 = 2x + 5（正确展开）
2. Evaluating Functions at a Point | 在某一点求函数值
e order of operations: brackets, exponents, multiplication/division, addition/subtraction.
要在某一点求函数值，直接将给定的数字或表达式代入函数关系式。对于 f(x) = x² − 4x + 1，f(3) = 3² − 4(3) + 1 = 9 − 12 + 1 = −2。务必遵循运算顺序：括号、指数、乘除、加减。
= a² + 2ah + h² − 4a − 4h + 1. Practice this to gain fluency with algebraic manipulation.
有些题目要求计算 f(a + h) 并化简，这在引入微分时很常见。对于同一个二次函数，f(a + h) = (a + h)² − 4(a + h) + 1 = a² + 2ah + h² − 4a − 4h + 1。练习此类运算可提升代数操作熟练度。
Example: If f(x) = 3x − 2, find f(−1). Solution: f(−1) = 3(−1) − 2 = −3 − 2 = −5.
例题：若 f(x) = 3x − 2，求 f(−1)。解：f(−1) = 3(−1) − 2 = −3 − 2 = −5。
3. Finding the Domain of a Function | 求函数的定义域
The domain of a function is the set of all allowable input values (x-values) for which the function is defined. For functions given by a formula, start by assuming the domain is all real numbers ℝ, then exclude any values that cause division by zero or the square root of a negative number.
函数的定义域是函数有定义的所有允许输入值（x 值）的集合。对于由公式给出的函数，首先假设定义域为全体实数 ℝ，然后排除任何导致除以零或负数开平方的值。
| Function type / 函数类型 | Domain restriction / 定义域限制 |
|---|---|
| Polynomial / 多项式 (e.g., x³ − 5x) | All real numbers ℝ / 全体实数 |
| Rational / 有理式 (e.g., 1/(x − 2)) | Denominator ≠ 0; x ≠ 2 / 分母 ≠ 0 |
| Square root / 平方根 (e.g., √x + 4) | Radicand ≥ 0; x ≥ −4 / 被开方数 ≥ 0 |
| Logarithmic / 对数 (e.g., ln(x − 1)) | Argument > 0; x > 1 / 真数 > 0 |
nvolves multiple restrictions, combine them using inequalities or set notation. For f(x) = √x + 2/(x − 3), we need x + 2 ≥ 0 → x ≥ −2, and x − 3 ≠ 0 → x ≠ 3. The domain is [−2, 3) ∪ (3, ∞).
当函数包含多个限制时，用不等式或集合符号将其合并。对于 f(x) = √x + 2/(x − 3)，我们要求 x + 2 ≥ 0 → x ≥ −2，且 x − 3 ≠ 0 → x ≠ 3。定义域为 [−2, 3) ∪ (3, ∞)。
4. Finding the Range from a Graph | 从图像求值域
= x², the graph is a parabola opening upward with vertex at (0,0), so the range is y ≥ 0.
值域是函数可以产生的所有输出值（y 值）的集合。定义域通常通过代数方法求得，而值域往往通过绘制图像并观察纵向范围更容易确定。对于 f(x) = x²，图像是开口向上、顶点在 (0,0) 的抛物线，因此值域为 y ≥ 0。
= 2x + 1, as x runs through all real numbers, y also takes all real numbers, giving range ℝ. However, restricted domains change the range. If the domain of f(x) = 2x + 1 is {x | 0 ≤ x ≤ 3}, then the range becomes {y | 1 ≤ y ≤ 7}.
对于如 f(x) = 2x + 1 的线性函数，当 x 取遍所有实数时，y 也取遍所有实数，值域为 ℝ。然而，限制定义域会改变值域。若 f(x) = 2x + 1 的定义域为 {x | 0 ≤ x ≤ 3}，则值域变为 {y | 1 ≤ y ≤ 7}。
Key idea: Always examine the shape and turning point of the graph when finding the range of a quadratic, exponential, or trigonometric function.
关键思想：求二次函数、指数函数或三角函数的值域时，务必考察图像的形状和转折点。
5. Composite Functions: f(g(x)) | 复合函数：f(g(x))
combines two functions by using the output of one as the input of another. The notation (f ∘ g)(x) = f(g(x)) means 'apply g first, then apply f to the result'. Order matters: f(g(x)) is usually different from g(f(x)).
复合函数通过将一个函数的输出作为另一个函数的输入，将两个函数组合起来。符号 (f ∘ g)(x) = f(g(x)) 表示“先应用 g，然后对结果应用 f”。顺序很重要：f(g(x)) 通常不同于 g(f(x))。
f the composite consists of all x in the domain of g such that g(x) is in the domain of f.
构造复合函数时，将外函数中的每个 x 替换为整个内函数表达式。对于 f(x) = 2x + 1 和 g(x) = x²，f(g(x)) = 2(x²) + 1 = 2x² + 1。复合函数的定义域由 g 的定义域中所有满足 g(x) 在 f 的定义域内的 x 组成。
- Example 1: f(x) = √x, g(x) = x − 4 → f(g(x)) = √x − 4, domain x ≥ 4. | 例 1：f(x) = √x, g(x) = x − 4 → f(g(x)) = √x − 4，定义域 x ≥ 4。
- Example 2: If f(x) = 1/x and g(x) = x + 2, then g(f(x)) = 1/x + 2, domain x ≠ 0. | 例 2：若 f(x) = 1/x 且 g(x) = x + 2，则 g(f(x)) = 1/x + 2，定义域 x ≠ 0。
6. Inverse Functions: Determining f⁻¹(x) | 反函数：求 f⁻¹(x)
ses the effect of the original function. For a function to have an inverse, it must be one‑to‑ t). The process: write y = f(x), swap x and y, then solve for y. The result is y = f⁻¹(x).
反函数，记作 f⁻¹(x)，会逆转原函数的效果。函数要有反函数，必须是一一函数（通过水平线检验）。具体步骤：写出 y = f(x)，交换 x 和 y，然后解出 y。结果即为 y = f⁻¹(x)。
Example: Find the inverse of f(x) = 2x − 3.
lution: y = 2x − 3 → swap: x = 2y − 3 → 2y = x + 3 → y = (x + 3)/2, so f⁻¹(x) = (x + 3)/2.
例题：求 f(x) = 2x − 3 的反函数。
解：y = 2x − 3 → 交换：x = 2y − 3 → 2y = x + 3 → y = (x + 3)/2，故 f⁻¹(x) = (x + 3)/2。
x. In the example, f((x+3)/2) = 2((x+3)/2) − 3 = x + 3 − 3 = x. This confirms correctness.
始终通过验证 f(f⁻¹(x)) = x 和 f⁻¹(f(x)) = x 来检查反函数。本例中，f((x+3)/2) = 2((x+3)/2) − 3 = x + 3 − 3 = x，确认无误。
7. Domain and Range of Inverse Functions | 反函数的定义域与值域
The domain of f⁻¹ is exactly the range of f, and the range of f⁻¹ is the domain of f. This relationship provides a quick way to state the domain and range of an inverse without full analysis. If f has domain D and range R, then f⁻¹ has domain R and range D.
f⁻¹ 的定义域恰好是 f 的值域，f⁻¹ 的值域是 f 的定义域。这一关系提供了一种无需完整分析即可说明反函数定义域和值域的快捷方法。若 f 的定义域为 D，值域为 R，则 f⁻¹ 的定义域为 R，值域为 D。
For f(x) = x² with domain restricted to x ≥ 0, the range is y ≥ 0. The inverse is f⁻¹(x) = √x atching the swapped domain and range. This is why restricting the domain is crucial for obtaining a valid inverse of a quadratic.
对于定义域限制为 x ≥ 0 的 f(x) = x²，其值域为 y ≥ 0。反函数为 f⁻¹(x) = √x，定义域为 x ≥ 0，值域为 y ≥ 0，与交换后的定义域和值域一致。这就是为何限制定义域对于获得二次函数的有效反函数至关重要。
| Function / 函数 | f domain / f 定义域 | f range / f 值域 | f⁻¹ domain / 反函数定义域 |
|---|---|---|---|
| f(x) = eˣ | ℝ | y > 0 | x > 0 |
| f(x) = ln x | x > 0 | ℝ | ℝ |
| f(x) = √x − 1 | x ≥ 1 | y ≥ 0 | x ≥ 0 |
8. Using Mapping Diagrams | 使用映射图
Mapping diagrams illustrate how each element of the domain is paired with an element of the range. A function is a mapping where every input has exactly one arrow leaving it. If an input has no arrow or multiple arrows, the relation is not a function.
映射图展示了定义域中的每个元素如何与值域中的一个元素配对。函数是一种映射，其中每个输入值恰好有一条箭头射出。如果输入值没有箭头或有多条箭头，则该关系不是函数。
In Exercise 2B, you may be given a mapping diagram and asked whether it represents a function, a one‑to‑one function, or an onto function. Draw arrows carefully and count: if two arrows point to the same output, it is many‑to‑one (still a function). If an output has no incoming arrow, the function is not onto.
在练习 2B 中，你可能会遇到给定的映射图，并被问及它是否表示一个函数、一一函数或满射函数。仔细绘制箭头并计数：如果两个箭头指向同一个输出，则为多对一（仍是函数）。如果某个输出没有箭头射入，则该函数不是满射。
- One‑to‑one (injective): Different inputs map to different outputs. | 一一映射（单射）：不同的输入映射到不同的输出。
- Many‑to‑one: Two or more inputs map to the same output. | 多对一：两个或多个输入映射到同一个输出。
- Onto (surjective): Every element of the codomain has at least one incoming arrow. | 满射：陪域中的每个元素都至少有一条射入箭头。
9. One‑to‑One and Many‑to‑One Functions | 一一函数与多一函数
A function is one‑to‑one if each y‑value corresponds to exactly one unique x‑ l line test: if any horizontal line cuts the graph more than once, the function is not one‑to‑ example, f(x) = x³ passes the test, but f(x) = sin x does not without domain restriction.
如果每个 y 值对应唯一一个 x 值，则该函数为一一函数。图形上可通过水平线检验来判断：若任意一条水平线与图像相交多于一次，则该函数不是一一函数。例如，f(x) = x³ 通过检验，而 f(x) = sin x 在不限制定义域时不通过。
Many‑to‑one functions, like quadratics, map two different x‑values to the same y‑ but can have an inverse if the domain is restricted to a branch where the function is one‑to‑one (e.g., x ≥ 0 for f(x) = x²).
多一函数（如二次函数）将两个不同的 x 值映射到同一个 y 值。它们在整个定义域上没有反函数，但如果将定义域限制在函数为一一映射的分支上（例如对 f(x) = x² 限制 x ≥ 0），则可以拥有反函数。
Remember: Only one‑to‑one functions have inverse functions. | 记住：只有一一函数才有反函数。
10. Common Mistakes in Exercise 2B | 练习2B中的常见错误
One frequent error is forgetting to check the domain when forming composite functions. For f(g(x)), you must ensure x is in the domain of g and that g(x) is in the domain of f. Neglecting this can lead to an incorrect domain for the composite.
一个常见错误是在构造复合函数时忘记检查定义域。对于 f(g(x))，必须确保 x 在 g 的定义域内，且 g(x) 在 f 的定义域内。忽略这一点会导致复合函数的定义域出错。
Another common mistake is mishandling the inverse process. Students often solve for y before swapping x and y, or they forget to restrict the domain when necessary. Always write 'swap x and y' as a distinct step, and check your result by composition.
另一个常见错误是处理反函数步骤不当。学生常常在交换 x 和 y 之前就解出 y，或者在必要时忘记限制定义域。始终将“交换 x 和 y”作为一个独立步骤写出，并通过复合运算检查结果。
Confusing domain and range with the notation of inverse functions is also common. f⁻¹(x) ≠ 1/f(x). The superscript −1 denotes the inverse function, not the reciprocal. Practice reading the notation correctly to avoid errors in meaning.
混淆定义域和值域与反函数符号也很常见。f⁻¹(x) ≠ 1/f(x)。上标 −1 表示反函数，而非倒数。练习正确解读符号，以避免含义上的错误。
11. Practice Problems with Solutions | 附解答的练习题
Test your understanding with these short problems that mirror Exercise 2B style.
用以下模仿练习 2B 风格的简答题测试你的理解。
- Given f(x) = 4 − x², find f(−√3). | 已知 f(x) = 4 − x²，求 f(−√3)。
Solution: f(−√3) = 4 − (−√3)² = 4 − 3 = 1. | 解：f(−√3) = 4 − 3 = 1。 - State the domain of h(x) = √2x − 8. | 写出 h(x) = √2x − 8 的定义域。
2x − 8 ≥ 0 → 2x ≥ 8 → x ≥ 4. Domain: {x | x ≥ 4}. | 定义域：{x | x ≥ 4}。 - If f(x) = 1/(x + 1) and g(x) = 2x, find (f ∘ g)(x) and its domain. | 若 f(x) = 1/(x + 1) 且 g(x) = 2x，求 (f ∘ g)(x) 及其定义域。
(f ∘ g)(x) = 1/(2x + 1). Domain: 2x + 1 ≠ 0 → x ≠ −½. | (f ∘ g)(x) = 1/(2x + 1)，定义域：x ≠ −½。 - Find the inverse of f(x) = (x − 2)/3. | 求 f(x) = (x − 2)/3 的反函数。
y = (x − 2)/3 → swap: x = (y − 2)/3 → 3x = y − 2 → y = 3x + 2. So f⁻¹(x) = 3x + 2. | f⁻¹(x) = 3x + 2。 - h inputs {1,2,3} and outputs {a,b} where 1→a, 2→a, 3→b is a function and whether it is one‑to‑one. | 判断输入 {1,2,3}、输出 {a,b} 且 1→a, 2→a, 3→b 的映射图是否为函数，以及是否是一一映射。
It is a function (each input has exactly one arrow). It is not one‑to‑one because 1 and 2 both map to a. | 它是一个函数（每个输入恰好有一条箭头）。它不是一一映射，因为 1 和 2 都映射到 a。
12. Summary and Key Takeaways | 总结与关键要点
Exercise 2B builds the foundation for all advanced function topics in the IB course. Mastery of function notation, domain and range detection, and composite and inverse manipulation will support your understanding of transformations, calculus, and equation solving later on. Keep a structured approach: always define the domain first, check restrictions, and use graphs as a verification tool.
练习 2B 为 IB 课程中所有高级函数主题奠定基础。掌握函数符号、定义域与值域的确定以及复合与反函数的操作，将有助于后续理解图像变换、微积分和方程求解。保持有条理的方法：始终先确定定义域，检查限制条件，并将图像用作验证工具。
Remember that the domain and range swap for inverse functions, composite domain requires careful intersection, and only one‑to‑one functions possess inverses without domain restriction. Regular practice with a variety of function types—polynomial, rational, radical, exponential—will make these skills automatic.
记住，反函数的定义域与值域会互换，复合函数的定义域需要仔细求交集，且只有一一函数在不限制定义域的情况下才拥有反函数。通过多种函数类型（多项式、有理式、根式、指数）的定期练习，这些技能将变得自动化。
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