📚 Exercise 2L: Mastering Function Transformations and Equation Solving | 练习2L：掌握函数变换与方程求解
Exercise 2L in the IB Mathematics: Analysis and Approaches course is a cumulative set of review questions designed to test and deepen your understanding of functions, their transformations, and the solving of associated equations. This exercise brings together concepts such as domain and range, composite and inverse functions, graph shifts, stretches, reflections, and the algebraic solution of exponential, logarithmic, and trigonometric equations. Working through these problems systematically will not only prepare you for end‑of‑chapter assessments but also build the fluency required for the internal assessment and final examinations.
练习2L 是 IB 数学：分析与方法课程中的一组综合性复习题，旨在检验和加深你对函数、函数变换及相关方程求解的理解。该练习汇集了定义域与值域、复合函数与反函数、图像平移、伸缩与反射，以及指数方程、对数方程和三角方程的代数求解等概念。系统地完成这些题目不仅能为章节末测试做好准备，还能培养内部评估和最终考试所需的熟练度。
1. Function Notation and Domain | 函数符号与定义域
A function f assigns exactly one output to each input from its domain. In IB problems, you are often given f(x) = √x – 2 or g(x) = 1/(x² – 4) and asked to state the largest possible domain. For f(x), the expression under the square root must be non‑negative, so x – 2 ≥ 0 ⇒ x ≥ 2. The domain can be written as [2, ∞). For g(x), the denominator cannot be zero, so x² – 4 ≠ 0 ⇒ x ≠ ±2, giving domain ℝ \ {–2, 2}.
函数 f 为定义域中的每个输入恰好分配一个输出。在 IB 题目中，你常会遇到 f(x) = √x – 2 或 g(x) = 1/(x² – 4)，要求写出最大可能定义域。对于 f(x)，根号内的表达式必须非负，因此 x – 2 ≥ 0 ⇒ x ≥ 2，定义域可写作 [2, ∞)。对于 g(x)，分母不能为零，故 x² – 4 ≠ 0 ⇒ x ≠ ±2，定义域为 ℝ \ {–2, 2}。
When a function involves both a square root and a denominator, you must satisfy all restrictions simultaneously. For instance, h(x) = √x+1 / (x–3) requires x+1 ≥ 0 and x ≠ 3, so the domain is [–1, 3) ∪ (3, ∞). Always express intervals using correct notation and be mindful of included or excluded endpoints.
当函数同时包含根号和分母时，必须同时满足所有限制条件。例如，h(x) = √x+1 / (x–3) 要求 x+1 ≥ 0 且 x ≠ 3，因此定义域为 [–1, 3) ∪ (3, ∞)。请始终使用正确的区间记号，并注意端点的包含与排除。
2. Composite Functions and Their Domains | 复合函数及其定义域
A composite function (f ∘ g)(x) = f(g(x)) exists only when the range of g falls within the domain of f. To find the domain of f(g(x)), first determine the domain of the inner function g, then reject any x for which g(x) is not in the domain of f. For example, if f(x) = ln x and g(x) = x – 3, then f(g(x)) = ln(x – 3). The natural log requires x – 3 > 0, so the composite domain is x > 3.
复合函数 (f ∘ g)(x) = f(g(x)) 仅在 g 的值域包含在 f 的定义域内时才存在。要求 f(g(x)) 的定义域，先确定内层函数 g 的定义域，然后剔除使得 g(x) 不在 f 定义域内的 x。例如，若 f(x) = ln x 而 g(x) = x – 3，则 f(g(x)) = ln(x – 3)。自然对数要求 x – 3 > 0，故复合定义域为 x > 3。
When working with more complex composites, it helps to sketch the flow of inputs. If g(x) = √x and f(x) = 1/(x–2), then f(g(x)) = 1/(√x – 2). Here, √x is defined for x ≥ 0, and we additionally need √x – 2 ≠ 0, i.e. x ≠ 4. The composite domain is thus [0, 4) ∪ (4, ∞). Always verify both constraints.
处理更复杂的复合函数时，画出输入流有助于理解。若 g(x) = √x，f(x) = 1/(x–2)，则 f(g(x)) = 1/(√x – 2)。这里 √x 在 x ≥ 0 时有定义，此外还需 √x – 2 ≠ 0，即 x ≠ 4。因此复合定义域为 [0, 4) ∪ (4, ∞)。务必同时验证两个约束条件。
3. Inverse Functions and Graphical Symmetry | 反函数与图形对称性
The inverse function f⁻¹ reverses the effect of f. To find an inverse, write y = f(x), swap x and y, and solve for y. For a function to have an inverse that is itself a function, f must be one‑to‑one on its domain – a condition often checked by the horizontal line test. The graphs of f and f⁻¹ are reflections of each other across the line y = x.
反函数 f⁻¹ 会逆转 f 的作用。求反函数时，写出 y = f(x)，交换 x 和 y，再解出 y。要使反函数本身也是函数，f 在其定义域上必须是一一映射——这一条件通常通过水平线检验来判断。f 与 f⁻¹ 的图像关于直线 y = x 对称。
When solving y = x² for x, we obtain x = ±√y, which is not a single‑valued function unless we restrict the domain. For f(x) = x² with domain x ≥ 0, the inverse is f⁻¹(x) = √x. The domain of f⁻¹ is the range of f. Always state both the rule and the domain of the inverse function, as required by IB mark schemes.
当解 y = x² 求 x 时，会得到 x = ±√y，除非限制定义域，否则这并非单值函数。对于定义域为 x ≥ 0 的 f(x) = x²，其反函数为 f⁻¹(x) = √x。f⁻¹ 的定义域是 f 的值域。按照 IB 评分标准，务必同时写出反函数的表达式和定义域。
4. Transformations of Functions: Translations | 函数变换：平移
Translations shift the graph of a function without altering its shape. A vertical translation of k units is given by f(x) + k: adding k moves the graph up, subtracting moves it down. A horizontal translation of h units is given by f(x – h): replacing x with x – h shifts the graph right by h, while f(x + h) shifts it left. The order of operations matters when multiple transformations are applied.
平移会移动函数图像而不改变其形状。垂直平移 k 个单位表示为 f(x) + k：加上 k 图像上移，减去则下移。水平平移 h 个单位表示为 f(x – h)：将 x 替换为 x – h 会使图像右移 h，而 f(x + h) 则使其左移。当施加多个变换时，运算顺序至关重要。
For example, starting from y = x³, the graph of y = (x – 2)³ + 3 is the original graph shifted 2 units right and 3 units up. Key points such as the origin (0,0) move to (2,3). When describing transformations in an exam, use clear language like ' translate by the vector (2, 3) ' or ' horizontal translation right by 2 followed by vertical translation up by 3 '.
例如，从 y = x³ 出发，y = (x – 2)³ + 3 的图像是原图像向右平移 2 个单位、向上平移 3 个单位所得。关键点如原点 (0,0) 移动至 (2,3)。在考试中描述变换时，请使用清晰的语言，如“按向量 (2, 3) 平移”或“先向右水平平移 2 个单位，再向上垂直平移 3 个单位”。
5. Transformations: Stretches and Reflections | 变换：伸缩与反射
Vertical stretches and compressions are represented by a f(x). If |a| > 1, the graph stretches away from the x‑axis; if 0 < |a| < 1, it compresses towards the x‑axis. A negative a also reflects the graph in the x‑axis. Horizontal stretches are given by f(bx): when |b| > 1, the graph compresses horizontally, and when 0 < |b| < 1, it stretches horizontally.
垂直伸缩由 a f(x) 表示。若 |a| > 1，图像沿 y 轴方向远离 x 轴拉伸；若 0 < |a| < 1，则向 x 轴压缩。负的 a 还会使图像关于 x 轴反射。水平伸缩由 f(bx) 给出：当 |b| > 1 时图像水平压缩，当 0 < |b| < 1 时水平拉伸。
Consider the transformation from f(x) = sin x to g(x) = 3 sin(2x). The coefficient 3 gives a vertical stretch by factor 3, while the 2 inside the argument gives a horizontal compression by factor ½. Reflections occur when coefficients are negative: f(–x) reflects in the y‑axis, and –f(x) reflects in the x‑axis. Combining these transformations requires careful ordering: horizontal shifts are affected by horizontal stretches.
考虑从 f(x) = sin x 到 g(x) = 3 sin(2x) 的变换。系数 3 给出了垂直拉伸，倍数为 3，而幅角内的 2 给出了水平压缩，倍数为 ½。当系数为负时产生反射：f(–x) 关于 y 轴反射，–f(x) 关于 x 轴反射。组合这些变换时需要谨慎排序：水平平移会受到水平伸缩的影响。
6. Solving Equations with Exponentials and Logarithms | 解指数与对数方程
Exponential equations such as 2ˣ⁺¹ = 8 can be solved by expressing both sides with the same base. Since 8 = 2³, we have x+1 = 3 ⇒ x = 2. When bases differ, use logarithms. For 3ˣ = 5, take ln of both sides: ln(3ˣ) = ln 5 ⇒ x ln 3 = ln 5 ⇒ x = ln 5 / ln 3. Always check that the solution lies within the domain of any logarithmic expressions involved.
形如 2ˣ⁺¹ = 8 的指数方程可以通过将两边表示为同底数来求解。因为 8 = 2³，故 x+1 = 3 ⇒ x = 2。当底数不同时，需使用对数。对于 3ˣ = 5，两边取自然对数：ln(3ˣ) = ln 5 ⇒ x ln 3 = ln 5 ⇒ x = ln 5 / ln 3。务必检验解是否在所含对数表达式的定义域内。
Logarithmic equations often require the laws logₐ(xy) = logₐ x + logₐ y and logₐ(xⁿ) = n logₐ x. To solve log₂(x+1) + log₂(x–1) = 3, combine: log₂[(x+1)(x–1)] = 3 ⇒ (x+1)(x–1) = 2³ = 8 ⇒ x² – 1 = 8 ⇒ x² = 9 ⇒ x = ±3. However, x = –3 is extraneous because it makes the original log arguments negative. The only valid solution is x = 3.
对数方程常需用到法则 logₐ(xy) = logₐ x + logₐ y 和 logₐ(xⁿ) = n logₐ x。要求解 log₂(x+1) + log₂(x–1) = 3，先合并：log₂[(x+1)(x–1)] = 3 ⇒ (x+1)(x–1) = 2³ = 8 ⇒ x² – 1 = 8 ⇒ x² = 9 ⇒ x = ±3。然而，x = –3 是增根，因为它会使原对数中的真数变为负数。唯一有效的解是 x = 3。
7. Solving Trigonometric Equations in a Given Interval | 在给定区间内解三角方程
Trigonometric equations such as 2 sin x – 1 = 0 for 0 ≤ x ≤ 2π require knowledge of the unit circle. Rearranging gives sin x = ½. The principal solution is x = π/6. Since sine is positive in the first and second quadrants, the second solution is x = π – π/6 = 5π/6. Always state all solutions within the specified interval.
求解三角方程，如 2 sin x – 1 = 0 在区间 0 ≤ x ≤ 2π 内，需要了解单位圆。移项得 sin x = ½。主解为 x = π/6。由于正弦在第一和第二象限为正，第二个解为 x = π – π/6 = 5π/6。务必给出指定区间内的所有解。
When solving equations like cos 2x = 0.5 for 0 ≤ x ≤ 2π, first solve for the double angle. Let θ = 2x, then cos θ = 0.5 ⇒ θ = π/3, 5π/3, 7π/3, … (general solution). Divide by 2 to recover x: x = π/6, 5π/6, 7π/6, 11π/6. Using graphical methods or CAST diagrams can help visualise the periodic nature and avoid missing solutions.
当求解如 cos 2x = 0.5 在 0 ≤ x ≤ 2π 内的方程时，先对倍角求解。令 θ = 2x，则 cos θ = 0.5 ⇒ θ = π/3, 5π/3, 7π/3, …（通解）。除以 2 得到 x：x = π/6, 5π/6, 7π/6, 11π/6。借助图像法或 CAST 图有助于直观理解周期性，避免遗漏解。
8. Using the Discriminant for Intersections | 利用判别式处理交点问题
The discriminant Δ = b² – 4ac of a quadratic ax² + bx + c = 0 determines the number of real roots. In the context of functions, setting f(x) = g(x) often yields a quadratic. The discriminant tells you whether the graphs intersect in two distinct points (Δ > 0), touch at one point (Δ = 0), or do not intersect (Δ < 0). This is a powerful tool for finding tangency conditions or parameter ranges.
二次方程 ax² + bx + c = 0 的判别式 Δ = b² – 4ac 决定了实根的数量。在函数情境中，令 f(x) = g(x) 通常会得到一个二次方程。判别式可告知图像是否交于两个不同点（Δ > 0）、相切于一点（Δ = 0）或不交（Δ < 0）。这是求相切条件或参数范围的强有力工具。
For example, to find the value of k for which y = x + k is tangent to y = x² – 2x, set x + k = x² – 2x ⇒ x² – 3x – k = 0. Tangency requires Δ = 0: (–3)² – 4(1)(–k) = 0 ⇒ 9 + 4k = 0 ⇒ k = –9/4. This method avoids lengthy calculus and is frequently tested in IB papers.
例如，要使得 y = x + k 与 y = x² – 2x 相切，令 x + k = x² – 2x ⇒ x² – 3x – k = 0。相切需要 Δ = 0：(–3)² – 4(1)(–k) = 0 ⇒ 9 + 4k = 0 ⇒ k = –9/4。这一方法避免了冗长的微积分运算，在 IB 试卷中经常考查。
9. Applying Functions to Modelling Problems | 函数在建模问题中的应用
Functions are used to model real‑world scenarios such as population growth, radioactive decay, or revenue optimisation. An exponential model P(t) = P₀ eᵏᵗ might describe bacteria growth, while a quadratic model could represent the trajectory of a projectile. IB questions often require you to interpret parameters, solve for time, or find maximum/minimum values using the function’s vertex or derivative.
函数被用于对现实世界的情景建模，如种群增长、放射性衰变或收益优化。指数模型 P(t) = P₀ eᵏᵗ 可描述细菌的繁殖，而二次模型可表示抛射体的轨迹。IB 题目常要求你解释参数、求解时间，或利用函数的顶点或导数求最大值或最小值。
For instance, a company’s profit P(x) in thousands of dollars is modelled by P(x) = –2x² + 40x – 150, where x is the number of items sold (in hundreds). The maximum profit occurs at the vertex. Comple
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