📚 Modulus Functions | 绝对值函数
The modulus function, often called the absolute value function, is a fundamental concept in IB Mathematics. It measures the distance of a number from zero on the real number line, always returning a non‑negative value. Understanding modulus is essential for solving equations and inequalities, analysing graphs, and modelling real‑world situations where only magnitude matters, irrespective of sign.
模函数，又称绝对值函数，是 IB 数学中的基础概念。它度量一个数在实数轴上到零的距离，总是返回非负值。掌握模函数对求解方程与不等式、分析图像，以及为仅关心大小而不计符号的实际情境建模至关重要。
1. Definition and Piecewise Form | 定义与分段形式
The modulus of a real number x, denoted |x|, is defined piecewise: it equals x when x is greater than or equal to zero, and it equals −x when x is less than zero. This definition removes any negative sign and guarantees a non‑negative output.
实数 x 的模记为 |x|，以分段形式定义：当 x 大于或等于零时等于 x，当 x 小于零时等于 −x。这一定义消除了负号，保证输出为非负值。
|x| = x for x ≥ 0, and |x| = −x for x < 0.
当 x ≥ 0 时 |x| = x，当 x < 0 时 |x| = −x。
2. Basic Properties | 基本性质
Modulus obeys several algebraic rules that simplify manipulation. The absolute value of any real number is always non‑negative. The product of absolute values equals the absolute value of the product, and similarly for quotients, provided the denominator is non‑zero. The square root of x² is |x|, not simply x.
模运算满足若干代数规则，可简化运算。任意实数的绝对值总是非负。两数绝对值的乘积等于乘积的绝对值；类似地，商的绝对值等于绝对值的商，只要分母不为零。x² 的平方根是 |x|，而不仅仅是 x。
- |a| ≥ 0, and |a| = 0 ⟺ a = 0
- |ab| = |a| |b|
- |a/b| = |a| / |b| (b ≠ 0)
- |aⁿ| = |a|ⁿ
- √a² = |a|
- Triangle inequality: |a + b| ≤ |a| + |b|
对应的中文性质：|a| ≥ 0 且 |a| = 0 当且仅当 a = 0；|ab| = |a| |b|；|a/b| = |a|/|b| (b ≠ 0)；|aⁿ| = |a|ⁿ；√a² = |a|；三角不等式：|a + b| ≤ |a| + |b|。
3. Graph of y = |f(x)| | 函数 y = |f(x)| 的图像
To sketch y = |f(x)|, start with the graph of y = f(x). Any portion of the graph that lies below the x‑axis is reflected in the x‑axis. Parts on or above the x‑axis remain unchanged. This transformation ensures the output is never negative.
绘制 y = |f(x)| 时，先从 y = f(x) 的图像出发。图像中位于 x 轴下方的部分沿 x 轴向上翻折，在 x 轴上或上方的部分保持不变。这一变换确保输出值永不为负。
For example, y = |x − 3| is a V‑ (3,0). The line y = x − 3 for x ≥ 3 stays, while the part for x < 3 is reflected upwards.
例如，y = |x − 3| 是一条顶点在 (3,0) 的 V 形曲线。x ≥ 3 时直线 y = x − 3 保留，而 x < 3 的部分被向上翻折。
4. Graph of y = f(|x|) | 函数 y = f(|x|) 的图像
The graph of y = f(|x|) is obtained by taking the right‑hand side of y = f(x) (for x ≥ 0) and reflecting it in the y‑axis. The left‑d side of the original graph (x < 0) is discarded, and the result is symmetric about the y‑axis. This composition makes the function even.
y = f(|x|) 的图像可通过取 y = f(x) 在右侧 (x ≥ 0) 的部分并将其关于 y 轴反射得到。原图左侧 (x < 0) 被舍去，最终图像关于 y 轴对称。这一复合使函数成为偶函数。
For instance, y = (|x| − 2)² produces a parabola that has been “folded” at the y‑axis; the minimum remains at (2,0) for x ≥ 0 and a mirror image appears for negative x.
例如，y = (|x| − 2)² 生成一条在 y 轴处“对折”的抛物线；最小值在 x ≥ 0 时位于 (2,0)，而负 x 区域呈现镜像。
5. Solving Equations with a Single Modulus | 含单个绝对值方程的求解
To solve |expression| absolute value cannot be negative. If k = 0, then expression = 0 is the only possibility.
求解 |表达式| = k，其中 k 是正常数，可将方程改写为两个独立的线性方程：表达式 = k 或 表达式 = −k。若 k 为负，方程无解，因为绝对值不可能为负。若 k = 0，则只能有 表达式 = 0。
Solve |2x − 5| = 7 → 2x − 5 = 7 or 2x − 5 = −7 → x = 6 or x = −1.
求解 |2x − 5| = 7 → 2x − 5 = 7 或 2x − 5 = −7 → x = 6 或 x = −1。
Always check the solutions in the original equation, especially when the expression inside contains variables that may introduce extraneous roots, though with pure linear expressions inside a single modulus no extraneous roots appear.
务必将解代入原方程检验，特别是内部含变量可能引入增根时，但对单个绝对值内为纯线性表达式的情形，不会产生增根。
6. Equations Involving Two Moduli | 含两个绝对值的方程
When an equation has the form |f(x)| = |g(x)|, squaring both sides is a convenient method because (|A| ly, one can consider the critical points where f(x) = 0 or g(x) = 0 and examine intervals.
当方程为 |f(x)| = |g(x)| 时，两边平方是一种便捷方法，因为 (|A|)² = A²。平方消去绝对值符号，得到一个二次或线性方程。亦可考虑 f(x) = 0 或 g(x) = 0 的临界点并分段讨论。
|2x − 1| = |x + 4| ² + 8x + 16 → 3x² − 12x − 15 = 0 → x² − 4x − 5 = 0 → (x − 5)(x + 1) = 0 → x = 5 or x = −1.
|2x − 1| = |x + 4| 4x + 1 = x² + 8x + 16 → 3x² − 12x − 15 = 0 → x² − 4x − 5 = 0 → (x − 5)(x + 1) = 0 → x = 5 或 x = −1。
Both solutions satisfy the original equation because squaring does not introduce extraneous solutions for absolute values on both sides; however, verification is good practice.
两者都满足原方程，因为两边取绝对值后平方不会引入增根；但验证仍是良好习惯。
7. Solving Modulus Inequalities: |f(x)| < a and > a | 解 |f(x)| < a 与 > a 的不等式
For a > 0, the inequality |f(x)| rt of the function lies between horizontal lines y = −a and y = a. The converse inequality |f(x)| > a splits into two separate regions: f(x) < −a or f(x) > a.
对 a > 0，不等式 |f(x)| < a 表示 f(x) 到零的距离小于 a。这可以改写为复合不等式 −a < f(x) < a。在图像上，满足条件的函数部分位于两条水平线 y = −a 与 y = a 之间。反向不等式 |f(x)| > a 分裂为两个分离的区域：f(x) < −a 或 f(x) > a。
|3x + 2| < 5 → −5 < 3x + 2 < 5 → −7 < 3x < 3 → −7/3 < x < 1.
|3x + 2| < 5 → −5 < 3x + 2 < 5 → −7 < 3x < 3 → −7/3 < x < 1。
When a ≤ 0, |f(x)| < a has no solution, while |f(x)| > a is true for all real numbers except possibly those making f(x) undefined.
当 a ≤ 0 时，|f(x)| < a 无解，而 |f(x)| > a 对所有使 f(x) 有定义的实数均成立。
8. Inequalities with a Modulus on One Side and a Linear Expression on the Other | 一边为模另一边为线性表达式的不等式
For |f(x)| ≥ g(x) or |f(x)| with the original inequality. Alternatively, one can use graphical methods by plotting y = |f(x)| and y = g(x) and identifying regions where the modulus graph is above or below the line.
对于 |f(x)| ≥ g(x) 或 |f(x)| ≤ g(x) 这类不等式，求解策略是分两种情形讨论：f(x) ≥ 0 时与 f(x) < 0 时，并与原不等式取交集。也可采用图像法，绘制 y = |f(x)| 和 y = g(x)，找出模函数图像位于直线上方或下方的区域。
Take |x − 2| → −4x ≤ −1 → x ≥ 0.25. Intersection with x < 2 yields 0.25 ≤ x < 2. The union is x ≥ 0.25.
以 |x − 2| ≤ 3x + 1 为例。当 x ≥ 2 时，x − 2 ≤ 3x + 1 → −2x ≤ 3 → x ≥ −1.5，与 x ≥ 2 取交集得 x ≥ 2。当 x < 2 时，−(x − 2) ≤ 3x + 1 → −x + 2 ≤ 3x + 1 → −4x ≤ −1 → x ≥ 0.25，与 x < 2 取交集得 0.25 ≤ x < 2。取并集得 x ≥ 0.25。
9. Sum of Absolute Values Inequalities | 绝对值之和的不等式
Inequalities involving sums such as |x − a| + |x − b| ere the expressions inside the moduli change sign. Divide the number line into intervals, remove the absolute value signs using appropriate signs, and solve the resulting linear inequality on each interval. Finally, combine the intervals that satisfy the inequality.
涉及如 |x − a| + |x − b| ≤ c 的和式不等式常见于 IB 高水平考试。最可靠的方法是确定临界值 x = a 与 x = b，即模内表达式变号的位置。将数轴划分为若干区间，按照各区间适当符号去掉绝对值，然后在每个区间上求解线性不等式。最后合并满足不等式的区间。
For |x + 1| + |x − 3| ≤ 7 → −2x ≤ 5 → x ≥ −2.5; intersection gives −2.5 ≤ x < −1. On −1 ≤ x < 3, first term non‑ e: x + 1 − x + 3 ≤ 7 → 4 ≤ 7, always true, so entire interval is valid. On x ≥ 3, both non‑ 7 → 2x − 2 ≤ 7 → 2x ≤ 9 → x ≤ 4.5; intersection gives 3 ≤ x ≤ 4.5. Union: −2.5 ≤ x ≤ 4.5.
对于 |x + 1| + |x − 3| ≤ 7，临界点为 −1 和 3。当 x < −1 时，两项皆为负：−x − 1 − x + 3 ≤ 7 → −2x + 2 ≤ 7 → −2x ≤ 5 → x ≥ −2.5；与区间取交得 −2.5 ≤ x < −1。在 −1 ≤ x < 3 时，第一项非负，第二项为负：x + 1 − x + 3 ≤ 7 → 4 ≤ 7 恒成立，因此整个区间成立。当 x ≥ 3 时，两项皆非负：x + 1 + x − 3 ≤ 7 → 2x − 2 ≤ 7 → 2x ≤ 9 → x ≤ 4.5，与区间取交得 3 ≤ x ≤ 4.5。取并集得 −2.5 ≤ x ≤ 4.5。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Students frequently forget that the square root of a squared variable is the absolute value, writing √x² = x instead of |x|. In inequalities, squaring both sides is only straightforward when both sides are non‑ us equations, always check the domain of any fractions or square roots inside the modulus.
学生常忘记平方后变量的平方根是绝对值，误写 √x² = x 而非 |x|。对于不等式，仅当两边非负时两边平方才直接可行；对于混合表达式，如果忽略符号，代数平方可能改变不等号方向。此外，求解模方程时，务必检查绝对值内部的分式或平方根的定义域。
A recommended approach in exams is to combine a quick sketch with analytical work. A rough graph can confirm the number of solutions and help avoid sign mistakes. When handling multiple moduli, clearly mark critical points on a number line and test intervals systematically.
考试中推荐结合快速草图与解析方法。一幅粗略图像可确认解的个数并有助于避免符号错误。处理多个模时，在数轴上清晰地标出临界点，并系统测试各个区间。
| Mistake / 错误 | Correct approach / 正确做法 |
|---|---|
| Assuming |x| = x for all x | Use piecewise definition; |x| = −x when x < 0 |
| Forgetting to reflect the negative part in y = |f(x)| | Always flip the below‑axis portion upwards |
| Squaring both sides without checking conditions | Ensure both sides have the same sign pattern or square only when safe |
| Solving |f(x)| < a by writing f(x) < a and f(x) < −a | Use compound inequality: −a < f(x) < a |
避免这些错误能显著提高解决模函数问题的准确度。
11. Modulus as a Distance and Modelling | 模作为距离及其建模
Geometrically, |x − a| represents the distance between x and a on the real line. This interpretation simplifies statements such as |x − 3| < 2, which describes all points x that are less than 2 units away from 3. Real‑life problems involving tolerance, error bounds, or absolute deviations are naturally expressed with moduli.
从几何角度看，|x − a| 表示实数轴上 x 与 a 之间的距离。这一解释有助于理解如 |x − 3| < 2 的表述，它描述所有到 3 的距离小于 2 个单位的点 x。涉及公差、误差界限或绝对偏差的实际问题很适合用模表示。
For example, a manufacturer requires that a rod's length L satisfies |L − 50| ls a tolerance of ±0.1 mm around the nominal length 50 mm. Solving yields 49.9 ≤ L ≤ 50.1.
例如，制造商要求杆的长度 L 满足 |L − 50| ≤ 0.1 mm。这模拟了围绕标称长度 50 mm 的 ±0.1 mm 公差。求解得出 49.9 ≤ L ≤ 50.1。
12. Transformations and Combinations with Modulus | 含模的组合变换
More involved functions such as y = a|b(x − h)| + k can be broken down by applying standard transformations to the basic V‑shape of y = |x| n, reflection, translation, and vertical stretch are handled in the usual order. Note that |b(x − h)| = |b| |x − h|, so the graph's slope is multiplied by |b|.
更复杂的函数如 y = a|b(x − h)| + k 可通过对基本 V 形 y = |x| 应用标准变换来分解处理。水平伸缩、反射、平移和垂直伸缩按常规顺序处理。注意 |b(x − h)| = |b| |x − h|，因此图像的斜率乘以 |b|。
Combining modulus with other functions tests deeper understanding. For instance, y = ||x| − 2| applies the modulus twice, creating a “W” shape. To plot it, first sketch the outer modulus's operation on the inner absolute value function.
模函数与其他函数结合可考察更深层次的理解。例如，y = ||x| − 2| 使用了两次模，形成 “W” 形。绘图时，先画出内层绝对值函数，再考虑外层模的作用。
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