📚 Review Set 7B: Key Functions & Graphs for IB Mathematics | IB 数学函数与图像核心复习题组 7B
This article unpacks the essential concepts tested in a typical Review Set 7B for IB Mathematics, focusing on functions, transformations, equations, and mathematical modelling. Whether you are preparing for an end‑of‑topic test or building fluency for the final examination, the following sections break down the content into manageable skills, common pitfalls, and exam‑ready strategies.
这篇文章旨在解析 IB 数学中一份典型的 Review Set 7B 所涵盖的核心考点，重点关注函数、图像变换、方程求解与数学建模。无论你正在准备阶段测验还是为最终大考积累熟练度，以下各节将把内容拆解为可管理的小项，并穿插常见错误与应试策略。
1. Function Notation and Domain/Range | 函数记号及其定义域与值域
A function is a rule that assigns exactly one output to each input. Recognising f(x) notation and determining the set of permissible x-values (domain) and resulting y-values (range) is a fundamental starting point. Pay close attention to square roots, denominators, and logarithms, as they naturally restrict the domain.
函数是一种将每个输入对应到唯一输出的规则。熟练掌握 f(x) 记号，并能够判断容许的 x 值集合（定义域）和生成的 y 值集合（值域），是所有分析的根本。特别注意根号、分母和对数，它们天然地限制着定义域。
- Watch for: f(x) = √x−3 requires x ≥ 3; g(x) = 1/(x−2) requires x ≠ 2.
- 注意：f(x) = √x−3 的定义域为 x ≥ 3；g(x) = 1/(x−2) 要求 x ≠ 2。
- Range from a graph: Use the vertical extent; for quadratic functions, the vertex determines the minimum or maximum y-value.
- 由图像读值域：观察竖直范围；二次函数由顶点决定 y 的最小或最大值。
2. Composite Functions | 复合函数
A composite function (f ∘ to the output of another. Start with the inside function and work outwards. The domain of the composite must be restricted so that g(x) lies in the domain of f.
复合函数 (f ∘ g)(x) = f(g(x)) 将内部函数的输出作为外部函数的输入。运算时由内向外逐层处理。定义域需要同时确保 g(x) 落在 f 的定义域内。
- Example: If f(x)=√x and g(x)=x−4, then (f ∘ g)(x)=√x−4 → domain x ≥ 4.
- 例子：设 f(x)=√x，g(x)=x−4，则 (f ∘ g)(x)=√x−4，定义域 x ≥ 4。
- Check explicitly: (f ∘ g)(x) ≠ (g ∘ f)(x) in general.
- 明确检查：一般来说 (f ∘ g)(x) 不等于 (g ∘ f)(x)。
3. Inverse Functions | 反函数
for y. Graphically, the inverse is a reflection in the line y = x. A function must be one‑to‑ ntal line test) to have an inverse over its entire domain; otherwise, restrict the domain.
反函数 f⁻¹(x) 颠倒原函数的对应关系。求反函数时在 y = f(x) 中交换 x 与 y 并解出 y。从图像上看，反函数是原函数关于直线 y = x 的反射。函数必须是单射（通过水平线检验）才能在整个定义域上拥有反函数，否则需要先限定定义域。
- Key property: f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.
- 核心性质：f⁻¹(f(x)) = x 且 f(f⁻¹(x)) = x。
- If f(x) = (2x+1)/(x−3), swap and solve: x = (2y+1)/(y−3) → cross‑multiply to obtain y = (3x+1)/(x−2).
- 若 f(x) = (2x+1)/(x−3)，交换后解出：y = (3x+1)/(x−2)。
4. Graph Transformations: Translations | 图像变换：平移
es horizontally by c units (to the right if c > 0). These are often tested in combination.
平移在保持图形形状不变的条件下移动图像。y = f(x) + d 使图像竖直移动 d 个单位（d > 0 时上移）。y = f(x − c) 使图像水平移动 c 个单位（c > 0 时右移）。考试常把两者组合考查。
- Horizontal shift: y = f(x + 2) shifts left by 2.
- 水平平移：y = f(x + 2) 向左平移 2 个单位。
- Vertical shift: y = f(x) − 5 shifts down by 5.
- 竖直平移：y = f(x) − 5 向下平移 5 个单位。
y = f(x − h) + k moves the graph h units horizontally and k units vertically.
y = f(x − h) + k 将图像水平平移 h 个单位，竖直平移 k 个单位。
5. Graph Transformations: Stretches and Reflections | 图像变换：伸缩与反射
ge the steepness and reflections flip the graph. y = a f(x) stretches vertically by factor |a|; if a is negative, it also reflects in the x‑axis. y = f(bx) stretches horizontally by factor 1/|b|; if b is negative, reflect in the y‑axis.
伸缩改变图像陡峭程度，反射则翻转图像。y = a f(x) 将图像纵向伸缩 |a| 倍；若 a 为负，还关于 x 轴反射。y = f(bx) 将图像横向伸缩 1/|b| 倍；若 b 为负，关于 y 轴反射。
| Transformation | Notation | Effect |
|---|---|---|
| Vertical stretch | y = 2 f(x) | Stretch ×2 in y‑direction |
| Horizontal compression | y = f(3x) | Compress ×1/3 in x‑direction |
| Reflection in x‑axis | y = −f(x) | Flip upside‑down |
| Reflection in y‑axis | y = f(−x) | Flip left‑right |
6. Quadratic Functions and the Discriminant | 二次函数与判别式
one repeated real root; Δ < 0 gives no real roots. In IB problems, linking discriminant to the number of intersections between a quadratic and a line is common.
一般二次式 f(x) = ax² + bx + c 可写成顶点式 a(x − h)² + k。判别式 Δ = b² − 4ac 揭示根的性质：Δ > 0 有两个不等实根；Δ = 0 有一个重实根；Δ < 0 无实根。在 IB 题目中，常需要把判别式用于分析二次曲线与直线的交点个数。
- Vertex x‑coordinate: h = −b/(2a); substitute into f to find k.
- 顶点横坐标：h = −b/(2a)；代入函数求纵坐标 k。
- Given y = 2x² − 3x + 1, discriminant = (−3)² − 4·2·1 = 9 − 8 = 1 > 0 → two roots.
- 对于 y = 2x² − 3x + 1，判别式 = (−3)² − 4·2·1 = 1 > 0，有两个不相等的实根。
7. Exponential and Logarithmic Equations | 指数方程与对数方程
Equations involving aˣ require logarithms for their solution. The definition aˣ = b ⇔ ogₐ(x/y) = logₐ x − logₐ y, logₐ xⁿ = n logₐ x. Always check that arguments stay positive.
含有指数 aˣ 的方程需要借助对数求解。定义 aˣ = b ⇔ x = logₐ b 是求解的桥梁。牢记对数运算法则：logₐ(xy) = logₐ x + logₐ y，logₐ(x/y) = logₐ x − logₐ y，logₐ xⁿ = n logₐ x。注意始终保证真数为正。
Solve 2ˣ⁺¹ = 5 → x + 1 = log₂ 5 → x = log₂ 5 − 1.
解 2ˣ⁺¹ = 5 → x + 1 = log₂ 5 → x = log₂ 5 − 1。
- Alternatively, use natural logs: (x+1) ln 2 = ln 5.
- 也可用自然对数：(x+1) ln 2 = ln 5。
- For log₃(x+4) = 2, rewrite as 3² = x+4 → x = 5.
- 对于 log₃(x+4) = 2，改写为 3² = x+4 → x = 5。
8. Solving Trigonometric Equations | 求解三角方程
r tangent value. Apply identities such as sin²θ + cos²θ = 1 when a quadratic form appears.
IB 题目通常要求在给定区间（如 0 ≤ x ≤ 2π）内求出所有解。利用单位圆或 CAST 图示捕捉所有具备特定正弦、余弦或正切值的角。当出现二次形式时，灵活使用 sin²θ + cos²θ = 1 等恒等式化简。
- Example: Solve 2 sin x − 1 = 0 for 0 ≤ x ≤ 2π → sin x = 0.5 → x = π/6, 5π/6.
- 例子：在 0 ≤ x ≤ 2π 上解 2 sin x − 1 = 0 → sin x = 0.5 → x = π/6, 5π/6。
- For 2 cos² x − cos x − 1 = 0, factor as (2 cos x + 1)(cos x − 1) = 0 and solve each.
- 对于 2 cos² x − cos x − 1 = 0，因式分解为 (2 cos x + 1)(cos x − 1) = 0，再分别求解。
9. Modelling with Functions | 函数建模
Modelling questions give a real‑world context and ask you to identify variables, write a function, and optimise or interpret values. Common scenarios include profit maximisation, projectile motion, and exponential growth/decay. Defining variables clearly and giving answers in context is essential for earning full marks.
建模题提供一个现实情境，要求你界定变量、写出函数，并进行优化或解释函数值。常见场景涵盖利润最大化、物体抛射和指数增长/衰减。清晰地定义变量并在上下文中给出最终答案，是获得满分的关键。
- Structure: Let x be …, then the quantity Q is given by Q(x) = … . Find maximum Q by locating vertex or using derivative.
- 结构：设 x 为 ……，则数量 Q 由 Q(x) = …… 给出。通过顶点或求导找到 Q 的最大值。
- cterial growth problem N(t) = N₀ eᵏᵗ, sketch graph and find time to double using ln 2 / k.
- 在细菌增长问题 N(t) = N₀ eᵏᵗ 中，画图并用 ln 2 / k 计算倍增时间。
10. Combining Transformations and Mapping | 组合变换与映射
y) under the transformation y = a f(b(x − c)) + d leads to the image ((x/b) + c, a y + d).
当多个变换作用在母函数上时，按照标准顺序处理：先进行水平方向的变换（括号内，同时处理平移和伸缩/反射），再进行竖直变换。在变换 y = a f(b(x − c)) + d 下，点 (x, y) 的映射结果为 ((x/b) + c, a y + d)。
- Order: (1) Horizontal stretch/reflection, (2) horizontal translation, (3) vertical stretch/reflection, (4) vertical translation.
- 顺序：(1) 水平伸缩/反射，(2) 水平平移，(3) 竖直伸缩/反射，(4) 竖直平移。
- bracket: y = f(2x − 6) must be written as f(2(x − 3)) to see the correct horizontal shift.
- 注意括号：y = f(2x − 6) 必须写成 f(2(x − 3)) 才能正确判断水平平移量。
11. Common Mistakes and How to Avoid Them | 常见错误与规避方法
Misreading domain restrictions and forgetting to check solutions against the original domain is a frequent slip. Students also confuse the order of transformations, mishandle negative signs in reflections, and apply logarithmic laws incorrectly. A quick sanity check with a test point can catch many errors.
误读定义域限制、忘记将解代回原方程检查是常见的疏忽。学生也容易混淆变换顺序、颠倒反射的负号方向，以及错误套用对数法则。用一个简单的测试点快速验算往往能发现很多错误。
- Inverse error: f⁻¹(x) does not mean 1/f(x).
- 反函数误区：f⁻¹(x) 不表示 1/f(x)。
- Log error: log(a + b) is not log a + log b.
- 对数误区：log(a + b) 不等于 log a + log b。
- Always state “for x > 0” when simplifying ln x² = 2 ln x.
- 化简 ln x² = 2 ln x 时务必附加条件 x > 0。
12. Approaching Review Set 7B Strategically | 策略性应对复习题组 7B
Treat each question as a mini investigation. Read carefully, sketch diagrams where appropriate, and label key points. Use the markscheme logic: one mark is often reserved for the correct domain or for stating the answer in the required form. After finishing, revisit transformation problems and verify with a calculator for selected checks.
把每道题当作小型探究题来处理。仔细读题，适时绘制草图，标注关键点。揣摩评分逻辑：经常有一分专门留给正确的定义域或答案的规范形式。完成后重新检查变换题，并挑选部分题目用计算器验证。
- Keep a formula sheet for quick reference to transformation rules and logarithm laws.
- 准备一张公式速查表，帮助回忆变换规则和对数法则。
- Time yourself: a typical review set should be completed under 45–60 minutes.
- 计时练习：一份典型的复习题组应能在 45–60 分钟内完成。
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