📚 The Modulus Function | 模函数
The modulus function, often called the absolute value function, is a cornerstone of IB Mathematics. It appears in equations, inequalities, graph sketching, and transformations. Mastery of |x| and its properties is essential for success in both Analysis & Approaches and Applications & Interpretation. This article provides a thorough, bilingual revision of every key aspect.
模函数，也称绝对值函数，是 IB 数学中的一块基石。它出现在方程、不等式、图像绘制及变换中。掌握 |x| 及其性质对于在分析与方法、应用与解释两门课程中取得成功至关重要。本文提供全面、双语的复习，涵盖每个关键要点。
1. Introduction to the Modulus Function | 模函数简介
The modulus of a real number x, denoted |x|, represents its distance from zero on the number line. It is always non‑negative. In IB, the modulus function is used to measure absolute differences, define error tolerances, and solve equations where both positive and negative possibilities must be considered.
实数 x 的模，记作 |x|，表示该数在数轴上到零的距离。它总是非负的。在 IB 中，模函数用于衡量绝对差值、定义误差容限以及求解需要同时考虑正负两种可能的方程。
2. Definition and Notation | 定义与符号
Formally, the modulus function is defined piecewise: |x| = x if x ≥ 0, and |x| ion. In handwritten work, the bars must be drawn clearly to avoid confusion with brackets.
正式地，模函数采用分段定义：当 x ≥ 0 时，|x| = x；当 x < 0 时，|x| = −x。符号由两条竖线组成。在计算器上，通过 ABS 功能调出。手写时竖线必须清晰，以免与括号混淆。
3. Basic Properties | 基本性质
Several properties are vital for manipulation: |x| ≥ 0 for all real x; |x| = 0 if and only if x = 0; |−x| = |x|; |xy| = |x||y|; |x/y| = |x|/|y| (y ≠ 0). The triangle inequality |x + y| ≤ |x| + |y| is also used in proof questions. Note that |x²| = x² because x² ≥ 0.
以下几个性质对运算至关重要：对所有实数 x，|x| ≥ 0；|x| = 0 当且仅当 x = 0；|−x| = |x|；|xy| = |x||y|；|x/y| = |x|/|y|（y ≠ 0）。三角不等式 |x + y| ≤ |x| + |y| 也用于证明题。注意 |x²| = x²，因为 x² ≥ 0。
4. Graphs of Modulus Functions | 模函数图像
The graph of y = |x| is a V‑shape with vertex at the origin, symmetric about the y‑axis. For y = |f(x)|, any part of the graph below the x‑axis is reflected upwards. For y = f(|x|), the graph for x < 0 is the reflection of the graph for x > 0 across the y‑axis. Sketching these quickly is a core skill.
y = |x| 的图像是以原点为顶点的 V 形，关于 y 轴对称。对于 y = |f(x)|，图像中位于 x 轴下方的部分被向上翻折。对于 y = f(|x|)，x < 0 部分的图像是 x > 0 部分图像关于 y 轴的镜像。快速绘制这些图像是核心技能。
| Transformation Type | Effect |
| y = |f(x)| | Erase below x‑axis, mirror above |
| y = f(|x|) | Discard left side for x<0, reflect right side |
| 变换类型 | 效果 |
| y = |f(x)| | 擦除 x 轴下方部分，向上翻转 |
| y = f(|x|) | 丢弃 x<0 的左半侧，镜像右半侧 |
5. Solving Equations with Modulus | 解含模的方程
To solve an equation such as |2x − 3| heck solutions in the original equation because extraneous roots can sometimes appear. For |ax + b| = |cx + d| ng both sides is a reliable method that avoids casework, leading to (ax + b)² = (cx + d)².
要解如 |2x − 3| = 5 的方程，需分成两种情形：2x − 3 = 5 或 2x − 3 = −5。务必代入原方程检验解，因为有时会出现增根。对于 |ax + b| = |cx + d|，两边平方是避免分类讨论的可靠方法，可得 (ax + b)² = (cx + d)²。
Example: Solve |x − 4| = 2x + 1. Case 1: x − 4 = 2x + 1 ⇒ x = −5. Check: |−9| = 9, RHS = −9 → reject. Case 2: x − 4 = −(2x + 1) ⇒ 3x = 3 ⇒ x = 1. Check: |−3| = 3, RHS = 3 → accept. Solution set: {1}.
示例：解 |x − 4| = 2x + 1。情况 1：x − 4 = 2x + 1 ⇒ x = −5。检验：|−9| = 9，右边 = −9 → 舍弃。情况 2：x − 4 = −(2x + 1) ⇒ 3x = 3 ⇒ x = 1。检验：|−3| = 3，右边 = 3 → 接受。解集：{1}。
6. Solving Inequalities with Modulus | 解含模的不等式
For |x| < a (a > 0), the solution is −a < x < a. For |x| > a, the result is x < −a or x > a. With linear expressions, rewrite as a double inequality for 'less than' problems, or two separate inequalities for 'greater than' problems. Graphical interpretation on a number line solidifies understanding.
对于 |x| < a（a > 0），解为 −a < x < a。对于 |x| > a，结果为 x < −a 或 x > a。处理线性表达式时，“小于”型问题改写为双重不等式，“大于”型问题拆成两个单独不等式。在数轴上进行几何解释可巩固理解。
More complex inequalities such as |2x − 5| and signs. Critical values are found by setting the inside expression equal to ± the right‑ eck that solutions satisfy any necessary conditions, like 3x ≥ 0, to avoid invalid ranges.
更复杂的不等式，如 |2x − 5| ≤ 3x，需考虑定义域和符号来求解。通过设内部表达式等于 ± 右侧表达式找出临界值。始终检查解是否满足必要条件，例如 3x ≥ 0，以避免无效范围。
7. Modulus and Transformations | 模与图像变换
IB often asks students to sketch y = |f(x)| and y = f(|x|) starting from a given function shape. For y = |f(x)|, the portions of the original graph that are negative are reflected in the x‑axis. For y = f(|x| and the left side is replaced by the mirror image of the right side. Combinations like y = |f(|x|)| require applying transformations stepwise.
IB 常要求从给定函数形状出发，绘制 y = |f(x)| 和 y = f(|x|) 的图像。对于 y = |f(x)|，原图中为负的部分关于 x 轴翻折。对于 y = f(|x|)，x ≥ 0 部分的图像保持不变，左侧代之以右侧的镜像。如 y = |f(|x|)| 这样的组合，需要逐级施加变换。
8. Piecewise Representation | 分段表示
unctions without the absolute bars often simplifies differentiation and integration. Write |x − a| as a piecewise function: x − a for x ≥ a, and −(x − a) for x < a. For |3x + 6| x + 6 when x ≥ −2 and −3x − 6 when x < −2. This approach is key for extended questions.
将模函数去掉绝对值符号表示成分段形式，通常能简化微分和积分。将 |x − a| 写成分段函数：当 x ≥ a 时为 x − a，当 x < a 时为 −(x − a)。对于 |3x + 6|，分段点为 x = −2，因此当 x ≥ −2 时为 3x + 6，当 x < −2 时为 −3x − 6。这种方法是解答综合题的关键。
9. Modulus in Quadratic Equations | 二次方程中的模
Quadratic expressions inside modulus, like |x² − 4x| gative. First solve x² − 4x = 0 → x = 0, 4. The sign changes at these points. The equation |x² − 4x| each solved within the relevant domain. Graphically, it is equivalent to intersecting y = |x² − 4x| with y = 3.
模内含有二次表达式，如 |x² − 4x|，需要找出二次式为正和为负的区间。先解 x² − 4x = 0 → x = 0, 4。符号在这些点处改变。方程 |x² − 4x| = 3 产生四种情形：x² − 4x = 3 或 x² − 4x = −3，每种在相应定义域内求解。从图形上看，等价于求 y = |x² − 4x| 与 y = 3 的交点。
10. Modulus of Linear Expressions and Regions | 线性表达式的模与区域
Modulus inequalities define regions in the Cartesian plane. For example, |y| ≤ |x| describes the region between the lines y = x and y = −x. The area formed by |x| + |y| preting these geometrically tests the ability to connect algebraic and graphical thinking.
模不等式可在笛卡尔平面上定义区域。例如，|y| ≤ |x| 描述的是直线 y = x 与 y = −x 之间的区域。由 |x| + |y| ≤ 4 构成的区域是顶点在 (±4, 0) 和 (0, ±4) 的菱形（旋转 45° 的正方形）。对此类区域进行几何解释，考查连接代数思维与图形思维的能力。
11. Common Mistakes and Tips | 常见错误与提示
Students often forget that √x² = |x|, not simply x. When solving |ax + b| side modulus includes fractions or roots. IB marking schemes reward clear case breakdowns.
学生常忘记 √x² = |x| 而非仅为 x。解 |ax + b| = cx + d 时，忽略强制 cx + d ≥ 0 会引入增根。解不等式前先画草图可极大减少符号错误。当模内表达式包含分式或根式时，始终要考虑定义域。IB 评分方案奖励清晰的分类讨论。
12. Exam-Style Example | 考试样题
IB Paper 1 (non‑calculator) example: Solve |2x + 1| = |x − 4| atisfy the original because squaring is reversible in this context. Solution set: {−5, 1}.
IB 试卷一（不可使用计算器）例题：解 |2x + 1| = |x − 4|。两边平方： → 4x² + 4x + 1 = x² − 8x + 16 → 3x² + 12x − 15 = 0 → x² + 4x − 5 = 0 → (x + 5)(x − 1) = 0。于是 x = −5 或 x = 1。由于在此处平方可逆，两者皆满足原方程。解集：{−5, 1}。
Another common task: Sketch y = |x² − 9|. The inside x² − 9 is a parabola with roots at ±3. For |x| ≥ 3, the graph is y = x² − 9; for |x| < 3, reflect the negative part, so y = 9 − x². The graph is a W‑shaped curve touching the x‑axis at (−3,0) and (3,0).
另一常见任务：绘制 y = |x² − 9| 的图像。内层 x² − 9 是根为 ±3 的抛物线。当 |x| ≥ 3 时，图像为 y = x² − 9；当 |x| < 3 时，将负的部分翻折，于是 y = 9 − x²。图像是 W 形曲线，在 (−3,0) 和 (3,0) 处接触 x 轴。
Exam tip: When inequalities involve a modulus on one side and a variable expression on the other, always set up cases with the condition that the right‑hand side is non‑negative, then combine results with a number line for the final answer.
考试提示：当不等式一侧含模，另一侧含变量表达式时，务必在分类时设定右侧非负的条件，然后结合数轴给出最终答案。
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