📚 Modulus Functions | 模函数
The modulus function, often called the absolute value function, is a fundamental concept in mathematics that measures the distance of a number from zero on the real number line. It is denoted by |x| and always returns a non-negative value, regardless of the sign of the input. This function appears extensively in algebra, coordinate geometry, inequalities, and calculus, and a solid understanding of its properties and associated solving techniques is essential for success in the IB Mathematics curriculum.
模函数，通常也称为绝对值函数，是数学中的一个基本概念，它衡量实数轴上某一点到零的距离。记作 |x|，无论输入是正是负，输出始终是非负值。该函数在代数、坐标几何、不等式及微积分中广泛出现，透彻理解它的性质以及相关的求解技巧，对于在 IB 数学课程中取得好成绩至关重要。
1. Definition and Notation | 定义与记法
For any real number x, the modulus of x is defined piecewise as follows:
对于任意实数 x，模的定义采用分段形式如下：
|x| = x if x ≥ 0
|x| = −x if x < 0
In words, if x is non-negative, the modulus simply returns x itself; if x is negative, it returns the opposite sign, effectively multiplying by −1 to produce a positive number. An alternative geometric interpretation is that |x| represents the distance from the point x to the origin.
通俗地说，若 x 非负，模直接返回 x；若 x 为负，则返回其相反数，相当于乘以 −1 得到一个正数。另一种几何解释是，|x| 表示点 x 到原点的距离。
Key algebraic properties of the modulus function include:
模函数的关键代数性质有：
- Non-negativity: |x| ≥ 0 for all real x.
- 非负性：对一切实数 x，|x| ≥ 0。
- |x| = 0 if and only if x = 0.
- |x| = 0 当且仅当 x = 0。
- Symmetry: |−x| = |x|.
- 对称性：|−x| = |x|。
- Multiplicativity: |xy| = |x||y|.
- 乘法性质：|xy| = |x||y|。
- Triangle inequality: |x + y| ≤ |x| + |y|.
- 三角不等式：|x + y| ≤ |x| + |y|。
2. Graph of y = |x| | y = |x| 的图像
The basic modulus function y = |x| y = −x. The vertex is at the origin (0,0), where the slope changes abruptly from −1 to 1.
基本模函数 y = |x| 的图像呈 V 形。当 x ≥ 0 时，图像与直线 y = x 重合；当 x < 0 时，与直线 y = −x 重合。顶点位于原点 (0,0)，该点处斜率从 −1 突变到 1。
Key features of the graph y = |x|:
图像 y = |x| 的主要特征：
- Domain: all real numbers (−∞, ∞).
- 定义域：全体实数 (−∞, ∞)。
- Range: y ≥ 0, i.e. [0, ∞).
- 值域：y ≥ 0，即 [0, ∞)。
- Axis of symmetry: the y-axis (x = 0).
- 对称轴：y 轴 (x = 0)。
- Increasing on [0, ∞) and decreasing on (−∞, 0].
- 在 [0, ∞) 上单调递增，在 (−∞, 0] 上单调递减。
The sharp corner at the vertex is not differentiable, which is a crucial point when studying modulus functions in calculus.
顶点处的尖角不可导，这是在微积分中研究模函数时需要特别留意的一点。
3. Transformations: y = |f(x)| and y = f(|x|) | 变换：y = |f(x)| 和 y = f(|x|)
Two common transformations involving the modulus sign are y = |f(x)| and y = f(|x|). They behave quite differently and often cause confusion.
涉及模符号的两种常见变换是 y = |f(x)| 和 y = f(|x|)。它们表现截然不同，容易混淆。
For y = |f(x)| ns above the x-axis remain unchanged. This transformation keeps all y-values non-negative.
对于 y = |f(x)|，原函数 y = f(x) 图像中所有位于 x 轴下方的部分都会关于 x 轴向上翻折，变为正；x 轴上方的部分保持不变。这一变换令所有 y 值非负。
For y = f(|x| even, i.e. symmetric about the y-axis. The input is first made positive before applying f.
对于 y = f(|x|)，x ≥ 0 的图像被保留，x < 0 的部分则通过将右侧 (x ≥ 0) 图像关于 y 轴对称翻折得到。所得图像为偶函数，即关于 y 轴对称。在施加 f 之前，输入值先被取正。
Example: If f(x) = x − 2, then y = |x − 2| has its V-shaped vertex at (2,0). Meanwhile, y = |x| − 2 keeps the V shape vertex at (0,−2), and then y = ||x| − 2| would reflect the part below the x-axis upwards.
示例：若 f(x) = x − 2，则 y = |x − 2| 的 V 形顶点在 (2,0)。而 y = |x| − 2 保持 V 形但顶点在 (0,−2)，继而 y = ||x| − 2| 会将 x 轴下方的部分向上翻折。
4. Solving Modulus Equations | 解模方程
Solving equations involving a single modulus like |ax + b| that the expression inside can equal either c or −c. Thus, we split into two linear equations:
求解含单个模的方程如 |ax + b| = c，其中 c ≥ 0，其原理为内部表达式可以等于 c 或 −c。于是，我们拆成两个线性方程：
ax + b = c or ax + b = −c
Each solution must be checked in the original equation, especially if derived after squaring, though for linear cases checks rarely reveal extraneous solutions. If c < 0, the equation has no real solution because an absolute value cannot be negative.
每一个解都需要代入原方程验证，特别是在平方后求解的情况下，但对于线性情形，很少出现增根。若 c < 0，方程无实数解，因为绝对值不可能为负。
Example: Solve |2x − 1| = 5.
示例：解 |2x − 1| = 5。
- 2x − 1 = 5 → 2x = 6 → x = 3.
- 2x − 1 = −5 → 2x = −4 → x = −2.
Both x = 3 and x = −2 satisfy the original equation.
x = 3 和 x = −2 都满足原方程。
5. Solving Modulus Inequalities | 解模不等式
Inequalities of the form |ax + b| < c and |ax + b| > c (with c > 0) are handled by turning them into compound inequalities without modulus.
形如 |ax + b| < c 和 |ax + b| > c（其中 c > 0）的不等式，可通过转化为不含模的复合不等式来求解。
| Case 1: |ax + b| < c | Equivalent to −c < ax + b < c. |
| 情形一：|ax + b| < c | 等价于 −c < ax + b < c。 |
| Case 2: |ax + b| > c | Equivalent to ax + b < −c or ax + b > c. |
| 情形二：|ax + b| > c | 等价于 ax + b < −c 或 ax + b > c。 |
When c = 0, the inequalities |ax + b| < 0 have no solution (since absolute value is non-negative), while |ax + b| > 0 holds for all x except the root where ax + b = 0.
当 c = 0 时，不等式 |ax + b| < 0 无解（因绝对值非负），而 |ax + b| > 0 对所有 x 成立，除了使 ax + b = 0 的根。
Example: Solve |3x + 2| ≤ 7. This yields −7 ≤ 3x + 2 ≤ 7 → −9 ≤ 3x ≤ 5 → −3 ≤ x ≤ 5/3.
示例：解 |3x + 2| ≤ 7。这得出 −7 ≤ 3x + 2 ≤ 7 → −9 ≤ 3x ≤ 5 → −3 ≤ x ≤ 5/3。
rict inequalities, the same logical steps apply with careful attention to ≤ and ≥ symbols.
对于严格或非严格不等式，同样的逻辑步骤适用，但需要留意 ≤ 和 ≥ 符号。
6. Equations with Two Modulus Expressions | 含有两个绝对值表达式的方程
Equations like |ax + b| = |cx + d| lence without introducing extraneous roots from the modulus itself: (ax + b)² = (cx + d)².
形如 |ax + b| = |cx + d| 的方程，可通过两边平方来求解，因为两边均非负，平方不会因为模本身引入增根：(ax + b)² = (cx + d)²。
Alternatively, we can use the principle that |u| = |v| quations: ax + b = cx + d and ax + b = −(cx + d). Both approaches lead to the same result.
此外，也可利用原理：|u| = |v| 当且仅当 u = v 或 u = −v。这给出两个线性方程：ax + b = cx + d 及 ax + b = −(cx + d)。两种方法会得到相同的结果。
Example: |2x − 1| = |x + 3| 4x + 1 = x² + 6x + 9 → 3x² − 10x − 8 = 0 → (3x + 2)(x − 4) = 0, giving x = −2/3 or x = 4.
示例：|2x − 1| = |x + 3|。平方：2x − 1)² = (x + 3)² → 4x² − 4x + 1 = x² + 6x + 9 → 3x² − 10x − 8 = 0 → (3x + 2)(x − 4) = 0，得 x = −2/3 或 x = 4。
Always check solutions for more complex cases where squaring might introduce extraneous solutions due to domain restrictions, but with pure modulus equations this method is safe.
对于更复杂的情形，平方可能因定义域限制产生增根，应始终验证解，但对纯模方程，此方法安全。
7. Squaring Method for Inequalities | 不等式平方法
For inequalities involving two modulus expressions, e.g. |ax + b| > |cx + d|, we can square both sides because both sides are non-negative, and the inequality direction is preserved. Thus, we obtain (ax + b)² > (cx + d)², which simplifies to a quadratic inequality that can be solved via factorisation or sign analysis.
对于含有两个模表达的不等式，如 |ax + b| > |cx + d|，可两边平方，因为两边均非负，且不等号方向保持不变。从而得到 (ax + b)² > (cx + d)²，它可简化为一个二次不等式，通过因式分解或符号分析求解。
Care must be taken when applying the squaring method to single-modulus inequalities. For instance, |x| < 2 is equivalent to x² < 4, which is true, but it is usually simpler to use the compound inequality approach. The squaring technique is most useful when comparing two absolute values.
将平方法用于单模不等式时必须小心。例如 |x| < 2 等价于 x² < 4，虽成立，但通常使用复合不等式方法更简单。平方技巧在比较两个绝对值时最为有用。
Example: Solve |x − 2| > |2x + 1| 8x − 3 → 3x² + 8x − 3 < 0. Factorising: (3x − 1)(x + 3) < 0. The solution is −3 < x < 1/3.
示例：解 |x − 2| > |2x + 1|。平方得 (x − 2)² > (2x + 1)² → x² − 4x + 4 > 4x² + 4x + 1 → 0 > 3x² + 8x − 3 → 3x² + 8x − 3 < 0。因式分解：(3x − 1)(x + 3) < 0。解为 −3 < x < 1/3。
8. Sketching Modulus Graphs | 绘制模函数图像
Sketching graphs of functions involving modulus signs is best done systematically. A recommended approach is:
绘制含模函数图像最好系统地进行。推荐步骤如下：
- Step 1: Start with the basic non-modulus function. Sketch its graph lightly or mentally.
- 第一步：从基本的不含模函数入手，画出其草图。
- Step 2: Identify the intervals where the inner expression changes sign (critical points).
- 第二步：找出内部表达变号的区间（临界点）。
- Step 3: Apply the modulus transformation as needed: for y = |f(x)|, reflect the negative parts upward; for y = f(|x|), reflect the right-hand side to the left.
- 第三步：根据需要施加模变换：对于 y = |f(x)|，将负值部分向上翻折；对于 y = f(|x|)，将右侧图形镜像到左侧。
- Step 4: Mark key points such as intercepts and vertices, and ensure the graph matches the expected symmetry and range.
- 第四步：标出截距、顶点等关键点，确保图像符合预期的对称性和值域。
lus signs are nested, work from the innermost function outward. For example, to sketch y = ||x| − 2|, first draw y = |x|, then shift it down by 2 to get y = |x| − 2, and finally reflect the part below the x-axis upwards.
当有嵌套模符号时，应从最内层函数向外处理。例如，要绘制 y = ||x| − 2|，先画出 y = |x|，然后下移 2 个单位得 y = |x| − 2，最后将 x 轴下方部分向上翻折。
9. Applications of Modulus | 模函数的应用
The modulus function models many real-world situations involving absolute deviation or distance. In error analysis, the inequality |measured value − true value| < tolerance describes acceptable measurement errors. In geometry, the distance between two points on a number line is given by |a − b|.
模函数可用来模拟许多涉及绝对偏差或距离的现实场景。在误差分析中，不等式 |测量值 − 真实值| < 容差描述可接受的测量误差。在几何中，数轴上两点间的距离由 |a − b| 给出。
lity at corners often requires checking left-hand and right-hand derivatives. The function |x| is differentiable everywhere except at x = 0, where a cusp occurs.
在微积分中，模函数出现在分段定义的函数里，它们在尖点处的可导性通常需要通过左右导数检验。函数 |x| 除 x = 0 处出现尖点外处处可导。
Modulus is also essential in defining the concept of neighbourhoods and limits in mathematical analysis, e.g. |x − a| < δ.
在数学分析中，模对于定义邻域和极限等概念至关重要，比如 |x − a| < δ。
10. Common Mistakes and Tips | 常见错误与提示
Students frequently make the following errors when handling modulus problems:
学生在处理模函数问题时，常犯以下错误：
- Forgetting to consider both cases when solving |ax + b| = c; always split into ax + b = c AND ax + b = −c.
- 忘记考虑两种情况：解 |ax + b| = c 时，应始终拆为 ax + b = c 和 ax + b = −c。
- Misinterpreting |−x|: It is equal to |x|, not −x.
- 误解 |−x|：它等于 |x|，而非 −x。
- Applying squaring without checking validity domain when expressions involve variables under a square root or denominator.
- 未检查定义域有效性就平方，特别是表达式涉及根号或分母时。
- Confusing y = |f(x)| with y = f(|x|): The former affects the output (reflecting in x-axis), the latter affects the input (reflecting in y-axis).
- 混淆 y = |f(x)| 与 y = f(|x|)：前者影响输出（关于 x 轴翻折），后者影响输入（关于 y 轴翻折）。
- Assuming |x + y| = |x| + |y|: This is only generally true when x and y have the same sign; otherwise the triangle inequality is strict.
- 假设 |x + y| = |x| + |y|：只有当 x 和 y 同号时才普遍成立；否则三角不等式为严格不等。
Tip: When solving modulus inequalities, it is often helpful to visualize the problem with a rough graph to confirm the algebraic solution.
提示：解模不等式时，画一个粗略的草图来验证代数解往往很有帮助。
11. Worked Practice Example | 完整练习示例
Let's solve the equation |2x − 5| = |x + 1| and represent the solutions on a number line.
我们求解方程 |2x − 5| = |x + 1| 并在数轴上表示解。
Method: x + 24 = 0. Divide by 3: no, factorising gives (3x − 4)(x − 6) = 0. Thus x = 4/3 or x = 6.
方法：两边平方：(2x − 5)² = (x + 1)² → 4x² − 20x + 25 = x² + 2x + 1 → 3x² − 22x + 24 = 0。因式分解得 (3x − 4)(x − 6) = 0。因此 x = 4/3 或 x = 6。
Check: For x = 4/3, LHS = |8/3 − 5| = |−7/3| = 7/3; RHS = |4/3 + 1| = 7/3. For x = 6, LHS = |12 − 5| = 7; RHS = |6 + 1| = 7. Both are valid.
检验：x = 4/3 时，左边 = |8/3 − 5| = |−7/3| = 7/3；右边 = |4/3 + 1| = 7/3。x = 6 时，左边 = |12 − 5| = 7；右边 = |6 + 1| = 7。两者均成立。
Sketching the inequality |x − 3| < 2|x| 3 > 0 → (x + 3)(x − 1) > 0, solution: x < −3 or x > 1. Always test intervals: x = −4 gives |−7| < 2|−4| → 7 < 8 true. So final solution: (−∞, −3) ∪ (1, ∞).
绘制不等式 |x − 3| < 2|x| 亦可平方求解： 6x + 9 < 4x² → 0 < 3x² + 6x − 9 → 3x² + 6x − 9 > 0 → x² + 2x − 3 > 0 → (x + 3)(x − 1) > 0，解为 x < −3 或 x > 1。务必测试区间：x = −4 时 |−7| < 2|−4| → 7 < 8 成立。最终解：(−∞, −3) ∪ (1, ∞)。
12. Summary | 总结
The modulus function, defined as |x| cornerstone of algebraic manipulation. Its graph is V-shaped, and transformations like y = |f(x)| and y = f(|x| s or squaring, always respecting the non-negative nature of absolute values. With practice, these techniques become intuitive, allowing you to tackle a wide range of IB Mathematics problems confidently.
模函数定义为 |x| 等于到零的距离，是代数运算的基石。它的图像呈 V 形，变换如 y = |f(x)| 和 y = f(|x|) 分别通过对输出或输入进行翻折。求解含模的方程与不等式依赖于分情况讨论或两边平方，并始终尊重绝对值的非负性。通过练习，这些技巧会变得直观，使你能够自信地应对各类 IB 数学问题。
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