📚 IB Math Review Set 5C: Functions, Graphs and Equations | IB数学复习集5C：函数、图像与方程
Review Set 5C brings together the most important ideas from the functions and equations topics in the IB Mathematics course. It challenges students to apply definitions of functions, domain and range, work with composite and inverse functions, tackle graph transformations, and solve exponential, logarithmic and trigonometric equations. This article works through these key areas, offering clear explanations, common exam-style examples and practical tips to help you build confidence and accuracy.
复习集5C汇集了IB数学课程中函数与方程部分的核心概念。它要求学生灵活运用函数定义、定义域与值域、复合函数与反函数、图像变换，并能求解指数、对数与三角方程。本文将对上述重点内容逐一展开，结合典型考题讲解，帮助你建立清晰的解题思路，提升应试信心与准确率。
1. Understanding Functions: Domain and Range | 理解函数：定义域与值域
A function is a relation where each input x corresponds to exactly one output y ) has domain x ≥ 3, and its range is y ≥ 0 because the square root function yields non-negative results. When the function is rational, like g(x) = 1/(x - 2), the domain is all real numbers except x = 2, and the range is all real numbers except 0, since the reciprocal can never equal zero.
函数是指每个x输入值都恰好对应一个y输出值的关系。定义域是所有可能输入值的集合，值域是所有可能输出值的集合。在IB题目中，通常需要通过排除使分母为零或使平方根内表达式为负的值来确定定义域。例如，f(x) = √(x - 3) 的定义域是 x ≥ 3，值域为 y ≥ 0，因为平方根函数结果非负。对于有理函数如 g(x) = 1/(x - 2)，定义域是除 x = 2 外的所有实数，值域是除 0 外的所有实数，因为倒数永远不等于零。
2. Composite and Inverse Functions | 复合函数与反函数
on: f⁻¹(x) satisfies f(f⁻¹(x)) = x, provided f is one-to-one. To find an inverse, swap x and y in the equation y = f(x) and solve for y. For example, if f(x) = 2x + 3, setting x = 2y + 3 gives the inverse f⁻¹(x) = (x - 3)/2. The graph of an inverse is the reflection of the original function across the line y = x.
复合函数将两个函数组合起来，把一个函数的输出作为另一个函数的输入。记号 f(g(x)) 表示先施加 g，再施加 f。复合函数 f(g(x)) 的定义域包含所有满足 g(x) 属于 f 定义域的 x。反函数则逆转函数的作用：f⁻¹(x) 满足 f(f⁻¹(x)) = x，前提是 f 是一一对应的。求反函数时，将 y = f(x) 中的 x 与 y 互换，再解出 y。例如，若 f(x) = 2x + 3，令 x = 2y + 3 可得反函数 f⁻¹(x) = (x - 3)/2。反函数的图像是原函数关于直线 y = x 的反射。
3. Graph Transformations: Shifts, Stretches and Reflections | 图像变换：平移、伸缩与反射
) changes under modifications is essential for sketching quickly and accurately. The following table summarises the main transformations you will encounter in Review Set 5C.
理解 y = f(x) 的图像在参数变化时如何变动，是快速准确作图的根本。下表总结了复习集5C中会碰到的主要变换类型。
| Transformation | Equation | Effect |
|---|---|---|
| Vertical shift up by d | y = f(x) + d | graph moves up d units |
| Horizontal shift left by c | y = f(x + c) | graph moves left c units |
| Vertical stretch by factor a | y = a f(x), a > 1 | y-values multiplied by a |
| Horizontal stretch by factor 1/b | y = f(bx), b > 1 | horizontal compression |
| Reflection in x-axis | y = -f(x) | graph flipped upside down |
| Reflection in y-axis | y = f(-x) | graph mirrored left-right |
) shows a shift to the right by π/3, a horizontal compression by factor 1/3, a vertical stretch by 2, reflection in the x-axis, and a vertical shift up by 1.
组合多个变换时，次序很关键：通常先处理水平平移，再处理伸缩和反射，最后处理垂直平移。以绘制 y = -2 sin(3x - π) + 1 为例，先将自变量因式分解：3x - π = 3(x - π/3)，这表明图像向右平移 π/3，水平压缩至 1/3，垂直拉伸至 2 倍，关于 x 轴反射，最后向上平移 1 个单位。
4. Exponential and Logarithmic Functions | 指数函数与对数函数
Exponential functions of the form f(x) = a·b&# tant base b and appear in growth and decay models. The natural exponential function y = e&# he natural logarithm y = ln x, are especially important. The fundamental relationship is eln x = x for x > 0, and ln(eψ) = x for all real x. Graphs of y = eψ and y = ln x are reflections across y = x. When solving equations like e2x e in exponential form: x - 1 = 2³ = 8, giving x = 9, then check the original domain x > 1.
形如 f(x) = a·bψ 的指数函数有固定底数 b，出现在增长与衰减模型中。自然指数函数 y = eψ 和它的反函数自然对数 y = ln x 尤为重要。基本关系是：当 x > 0 时 eln x = x，且对所有实数 x 有 ln(eψ) = x。y = eψ 和 y = ln x 的图像关于直线 y = x 对称。求解方程 e2x = 5 时，两边取自然对数得 2x = ln 5，所以 x = (ln 5)/2。对数方程如 log₂(x - 1) = 3 改成指数形式：x - 1 = 2³ = 8，得 x = 9，再检验原定义域 x > 1 即可。
5. Trigonometric Functions and Their Properties | 三角函数及其性质
zontal shift of the sine curve: cos x = sin(x + π/2). The tangent graph has period π and vertical asymptotes at x = π/2 + kπ. The transformation y = A sin(Bx - C) + D modifies amplitude to |A|, period to (2π)/|B| Remember to use the CAST diagram or unit circle to find all solutions in a given interval.
三个基本三角函数 sin、cos 和 tan 都是周期函数。在IB考试中需要掌握它们的图像、周期、振幅和重要恒等式。y = sin x 的图像振幅为1，周期为2π，与 x 轴相交于 π 的整数倍处。y = cos x 是正弦曲线的水平平移：cos x = sin(x + π/2)。正切函数的周期为 π，在 x = π/2 + kπ 处有竖直渐近线。对于变换 y = A sin(Bx - C) + D，振幅变为 |A|，周期为 (2π)/|B|，相位平移为 C/B，垂直平移为 D。求解基础三角方程如 sin θ = 1/2，在 [0, 2π) 内得到主值 θ = π/6 和 5π/6。记住使用 CAST 图或单位圆来找出给定区间内的所有解。
6. Solving Equations Graphically and Algebraically | 图像法与代数法解方程
In Review Set 5C you often see equations that link different function types, such as |x - 2| = 3 or e&# t both sides and look for intersection points. When mixing exponential and linear functions, exact solutions may require logarithms, but graphical methods help confirm the number of solutions and estimate their positions. The equation 2&# ly one solution because the increasing exponential curve crosses the decreasing line once.
在复习集5C中，常会遇到连接不同函数类型的方程，例如 |x - 2| = 3 或 eψ = x + 1。这些题目可通过代数法或图像分析法求解。处理绝对值方程时，拆分为两种情况：x - 2 = 3 得 x = 5；x - 2 = -3 得 x = -1。图像法则需要画出两边函数，观察交点。当指数函数与一次函数混合时，精确解可能需借助对数，但图像法有助于确认解的个数并估算大致位置。例如方程 2ψ = -x + 5 有一个解，因为递增的指数曲线与递减直线恰好相交一次。
7. Applications in Real-World Modelling | 实际建模应用
modelling real-life situations. Exponential models describe population growth P(t) = P₀ ekt and radioactive decay N(t) = N₀ e-λt t - c) + d. Given data points or graph features, you determine the amplitude a, period, phase shift and vertical shift. Once the model is set, you can solve for specific times or values, always interpreting results in the real-world context and checking that they make sense (e.g. time must be positive, height cannot be negative).
函数是模拟现实情境的有力工具。指数模型可描述人口增长 P(t) = P₀ ekt 和放射性衰变 N(t) = N₀ e-λt。三角函数可用于模拟周期性现象，如潮汐、日温度变化或摩天轮高度。在这类问题中，识别正确的函数形式至关重要：潮汐可用 H(t) = a cos(bt - c) + d 建模。根据数据点或图像特征，可确定振幅 a、周期、相位平移和垂直平移。建立模型后，即可求解特定时刻或数值，但必须将结果放回现实语境中解释，并检查是否合理（例如时间须为正，高度不可为负）。
8. Common Pitfalls and Exam Tips | 常见错误与考试技巧
s; when solving log₂(x + 1) - log₂(x - 3) = 1, you must ensure x + 1 > 0 and x - 3 > 0, meaning x > 3. In graph transformations, students sometimes confuse the direction of horizontal shifts: f(x + 2) shifts left by 2, not right. A useful strategy is to apply transformations step by step, marking key points on the graph, and verifying with a simple coordinate check. Also, when solving trigonometric equations, always write the general solution first or carefully list all solutions within the specified interval to avoid losing marks from missing values.
函数题中的错误往往源于忽略定义域限制。例如求反函数时，务必依据原函数的值域标出反函数的定义域。另一个常见错误是忘记对数只对正自变量有定义；在解 log₂(x + 1) - log₂(x - 3) = 1 时，必须保证 x + 1 > 0 且 x - 3 > 0，即 x > 3。图像变换中，学生有时会混淆水平平移方向：f(x + 2) 表示向左平移2个单位，而非向右。一条实用策略是逐步施加变换，在图上标出关键点，并用简单的坐标来验证。此外，解三角方程时，应先写出通解，或仔细列出指定区间内的所有解，避免因遗漏值而丢分。
9. Practice Problems from Review Set 5C | 复习集5C典型习题分析
Let us examine three representative exercises.
让我们分析三道代表性习题。
Problem 1: Given f(x) = √(2x - 4) and g(x) = x² - 1, find the domain of (f ∘ g)(x).
Solution (EN): ² - 6). The expression under the square root must be non-negative: 2x² - 6 ≥ 0 ⇒ x² ≥ 3, so x ≤ -√3 or x ≥ √3. Domain: (-∞, -√3] ∪ [√3, ∞).
中文解答： 先求复合函数 (f ∘ g)(x) = √(2x² - 6)。根号内须非负，即 2x² - 6 ≥ 0，得 x² ≥ 3，故定义域为 (-∞, -√3] ∪ [√3, ∞)。
Problem 2: Solve for x: log₃(x - 1) + log₃(x + 2) = 2.
Solution (EN): Approx. solutions: (-1 + 3√5)/2 ≈ 2.85 (valid), the negative root is ≈ -3.85 (rejected). Final answer: x = (-1 + 3√5)/2.
中文解答： 合并对数得 log₃[(x - 1)(x + 2)] = 2。改写为指数形式：(x - 1)(x + 2) = 9。展开得 x² + x - 11 = 0。求根公式：x = [-1 ± √45]/2 = [-1 ± 3√5]/2。检验定义域 x > 1，正根 (-1 + 3√5)/2 ≈ 2.85 合要求，负根舍弃。解为 x = (-1 + 3√5)/2。
Problem 3: f y = 3 cos(2x - π) - 1 is obtained from y = cos x by which sequence of transformations?
Solution (EN): ht by π/2, compress horizontally by factor 1/2, stretch vertically by factor 3, reflect? is positive. The full sequence: horizontal shift right π/2, horizontal compression (factor 1/2), vertical stretch (3), vertical shift down 1.
中文解答： 将表达式改写为 2(x - π/2)。从 y = cos x 开始，向右平移 π/2，水平压缩至 1/2，垂直拉伸至 3 倍，由于系数为正，无反射，最后向下平移 1。变换顺序：水平右移 π/2 → 水平压缩 → 垂直拉伸 → 垂直下移 1。
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