📚 Review Set B: Functions and Calculus Practice | IB数学复习套题B：函数与微积分练习
This review set is designed for IB Mathematics: Analysis and Approaches students who want to strengthen their understanding of functions and introductory calculus. The set covers core concepts such as domain, range, composite and inverse functions, limits, differentiation rules, tangents, stationery points, integration and area under a curve. Each section presents a brief explanation followed by a worked example with step‑by‑step reasoning. Working through these problems will help you identify gaps in your knowledge and prepare effectively for topic tests and the final examination.
本复习套题专为 IB 数学：分析与方法课程的学生设计，旨在巩固函数与基础微积分的理解。内容涵盖定义域、值域、复合函数、反函数、极限、求导法则、切线、驻点、积分以及曲线下的面积等核心概念。每个小节先给出简要说明，再通过详细的例题分析展示解题步骤。通过练习这些题目，你可以发现知识盲点，为专题测试和最终考试做好充分准备。
1. Domain and Range | 定义域与值域
The domain of a function is the set of all possible input values (x‑values) for which the function is defined, while the range is the set of all possible output values (y‑values). For square‑root functions, the expression inside the root must be non‑negative; for rational functions, the denominator must not equal zero.
函数的定义域是使函数有意义的全体输入值（x 值）的集合，值域则是全体可能的输出值（y 值）的集合。对于平方根函数，根号内的表达式必须非负；对于有理函数，分母不能为零。
Example: Find the domain and range of f(x) = √4 − x².
例题：求 f(x) = √4 − x² 的定义域和值域。
Solution: We require 4 − x² ≥ 0 ⇒ x² ≤ 4 ⇒ when x = ±2, giving f = 0; the maximum occurs at x = 0, giving f = 2. The range is [0, 2].
解答：需满足 4 − x² ≥ 0 ⇒ x² ≤ 4 ⇒ −2 ≤ x ≤ 2，因此定义域为 [−2, 2]。函数的最小值在 x = ±2 处取得，f = 0；最大值在 x = 0 处取得，f = 2。值域为 [0, 2]。
2. Composite Functions | 复合函数
A composite function (g ◦ f)(x) = g(f(x)) applies f first, then g. The domain of g ◦ f consists of x‑values in the domain of f for which f(x) lies in the domain of g. Always check the ‘inner’ function’s domain first.
复合函数 (g ◦ f)(x) = g(f(x)) 先作用 f，再作用 g。g ◦ f 的定义域由 f 定义域中使得 f(x) 落入 g 定义域的那些 x 组成。务必先检查“内层”函数的定义域。
Example: Let f(x) = ln(x − 1) and g(x) = √x. Find (g ◦ f)(x) and its domain.
例题：设 f(x) = ln(x − 1)，g(x) = √x。求 (g ◦ f)(x) 及其定义域。
Solution: (g ◦ f)(x) = √(ln(x − 1)). f(x) requires x − 1 > 0 ⇒ x > 1. Then f(x) must be in the domain of g, so ln(x − 1) ≥ 0 ⇒ x − 1 ≥ 1 ⇒ x ≥ 2. Domain: [2, ∞).
解答：(g ◦ f)(x) = √(ln(x − 1))。f(x) 要求 x − 1 > 0 ⇒ x > 1。接着 f(x) 须在 g 的定义域中，故 ln(x − 1) ≥ 0 ⇒ x − 1 ≥ 1 ⇒ x ≥ 2。定义域为 [2, +∞)。
3. Inverse Functions | 反函数
e function f⁻¹ reverses the effect of f, so that f⁻¹(f(x)) = x. It exists only if f is one‑to‑ he line y = x. To find the inverse, swap x and y in the equation y = f(x) and solve for y.
反函数 f⁻¹ 能逆转 f 的作用，满足 f⁻¹(f(x)) = x。只有当 f 在其定义域上是一一映射时，反函数才存在。从图像上看，反函数可以通过将 f 的图像关于直线 y = x 反射得到。求反函数时，把方程 y = f(x) 中的 x 与 y 互换，再解出 y。
Example: Find f⁻¹(x) for f(x) = 3e²ˣ + 1, x ∈ ℝ.
例题：求 f(x) = 3e²ˣ + 1，x ∈ ℝ 的反函数 f⁻¹(x)。
Solution: Write y = 3e²ˣ + 1. Swap x and y: x = 3e²ʸ + 1 ⇒ x − 1 = 3e²ʸ ⇒ e²ʸ = (x − 1)/3. Take ln: 2y = ln((x − 1)/3) ⇒ y = ½ ln((x − 1)/3). Thus f⁻¹(x) = ½ ln((x − 1)/3). Domain of f⁻¹ is x > 1.
解答：设 y = 3e²ˣ + 1。互换 x 和 y：x = 3e²ʸ + 1 ⇒ x − 1 = 3e²ʸ ⇒ e²ʸ = (x − 1)/3。取自然对数：2y = ln((x − 1)/3) ⇒ y = ½ ln((x − 1)/3)。因此 f⁻¹(x) = ½ ln((x − 1)/3)。f⁻¹ 的定义域为 x > 1。
4. Limits and Continuity | 极限与连续性
A function f is continuous at a point a if limx→a imit as x → ±∞ is 0; if degrees are equal, the limit is the ratio of leading coefficients.
函数 f 在点 a 处连续，如果 limx→a f(x) = f(a)。无穷远处的极限用于描述函数的末端走势。对于有理函数，若分子次数低于分母，x → ±∞ 时极限为 0；若次数相等，则极限为最高次项系数的比值。
Example: Evaluate limx→∞ (3x² + 2x)/(5x² − 4).
例题：计算 limx→∞ (3x² + 2x)/(5x² − 4)。
Solution: ominator by x²: (3 + 2/x)/(5 − 4/x²). As x → ∞, 2/x → 0 and 4/x² → 0, so the limit is 3/5.
解答：分子、分母同除以 x²：(3 + 2/x)/(5 − 4/x²)。当 x → ∞ 时，2/x → 0，4/x² → 0，故极限为 3/5。
5. Differentiation from First Principles | 从第一原理求导
The derivative of a function f at x is defined as f'(x) = limh→0 [f(x+h) − f(x)] / h, provided the limit exists. This definition yields the gradient of the tangent at any point and is the foundation of differential calculus.
函数 f 在 x 处的导数定义为 f'(x) = limh→0 [f(x+h) − f(x)] / h，前提是极限存在。这一定义给出了任意点处切线的斜率，是微分学的基石。
Example: Use first principles to differentiate f(x) = x² − 4x.
例题：利用第一原理求 f(x) = x² − 4x 的导数。
Solution: h) − f(x) = 2xh + h² − 4h = h(2x + h − 4). Divide by h: 2x + h − 4. Take limit as h → 0: f'(x) = 2x − 4.
解答：f(x+h) = (x+h)² − 4(x+h) = x² + 2xh + h² − 4x − 4h。于是 f(x+h) − f(x) = 2xh + h² − 4h = h(2x + h − 4)。除以 h 得 2x + h − 4。取 h → 0 的极限：f'(x) = 2x − 4。
6. Chain, Product and Quotient Rules | 链式法则、乘积法则与商法则
al rules for differentiating more complicated functions: the chain rule d/dx [f(g(x))] = f'(g(x))·g'(x); the product rule d/dx [u·v] = u'v + uv'; and the quotient rule d/dx [u/v] = (u'v − uv') / v². Choosing the appropriate rule often simplifies the derivative significantly.
求复杂函数导数的三条核心法则：链式法则 d/dx [f(g(x))] = f'(g(x))·g'(x)；乘积法则 d/dx [u·v] = u'v + uv'；商法则 d/dx [u/v] = (u'v − uv') / v²。选择合适的法则往往能极大简化求导过程。
Example: Differentiate y = ln(3x² + 1) and y = x²·eˣ.
例题：求 y = ln(3x² + 1) 和 y = x²·eˣ 的导数。
Solution: or y = ln(3x² + 1), by the chain rule: dy/dx = (1/(3x²+1))·(6x) = 6x/(3x²+1). For y = x²·eˣ, use the product rule: u = x², v = eˣ, u' = 2x, v' = eˣ, so dy/dx = 2x·eˣ + x²·eˣ = eˣ·x(x + 2).
解答：对 y = ln(3x² + 1)，用链式法则：dy/dx = (1/(3x²+1))·(6x) = 6x/(3x²+1)。对 y = x²·eˣ，用乘积法则：u = x², v = eˣ, u' = 2x, v' = eˣ，故 dy/dx = 2x·eˣ + x²·eˣ = eˣ·x(x + 2)。
7. Tangents and Normals | 切线与法线
the slope m of the tangent line. The equation of the tangent at (a, f(a)) is y − f(a) = f' ar to the tangent; its slope is −1/m (provided m ≠ 0), and its equation is y − f(a) = −1/f'(a) (x − a).
函数在某点的导数给出了切线的斜率 m。在点 (a, f(a)) 处的切线方程为 y − f(a) = f'(a)(x − a)。法线垂直于切线，其斜率为 −1/m（若 m ≠ 0），方程为 y − f(a) = −1/f'(a) (x − a)。
Example: Find the equations of the tangent and normal to y = x³ − x at x = 1.
例题：求曲线 y = x³ − x 在 x = 1 处的切线和法线方程。
Solution: f(1) = 0. f'(x) = 3x² − 1, f'(1) = 2. Tangent: y − 0 = 2(x − 1) ⇒ y = 2x − 2. Normal: slope = −1/2, equation y = −½(x − 1) ⇒ y = −½ x + ½.
解答：f(1) = 0。f'(x) = 3x² − 1，f'(1) = 2。切线：y − 0 = 2(x − 1) ⇒ y = 2x − 2。法线：斜率 = −1/2，方程 y = −½(x − 1) ⇒ y = −½ x + ½。
8. Stationary Points and Optimization | 驻点与优化问题
Stationary points occur where f'(x) = 0. Their nature is determined by the second derivative: if f''(x) > 0 the point is a local minimum; if f''(x) < 0 it is a local maximum; if f'' ms often require finding the global maximum or minimum of a function on a closed interval.
驻点出现在 f'(x) = 0 处。驻点的性质由二阶导数判断：若 f''(x) > 0，为局部极小值点；若 f''(x) < 0，为局部极大值点；若 f''(x) = 0，则需进一步分析。优化问题通常要求找出函数在闭区间上的全局最大值或最小值。
Example: Find and classify the stationary points of f(x) = 2x³ − 9x² + 12x + 1.
例题：求 f(x) = 2x³ − 9x² + 12x + 1 的驻点并分类。
Solution: f'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2). Set f'(x) = 0 ⇒ x = 1, 2. f''(x) = 12x − 18. f''(1) = −6 < 0 ⇒ local max at (1, 6). f''(2) = 6 > 0 ⇒ local min at (2, 5).
解答：f'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2)。令 f'(x) = 0 得 x = 1, 2。f''(x) = 12x − 18。f''(1) = −6 < 0 ⇒ (1, 6) 为局部极大点。f''(2) = 6 > 0 ⇒ (2, 5) 为局部极小点。
9. Antiderivatives and Integration | 反导数与积分
−1), ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C. Always remember to include the constant of integration, C.
积分是微分的逆运算。不定积分 ∫ f(x) dx 得到导数为 f(x) 的一族函数。对于幂函数（n ≠ −1），∫ xⁿ dx = xⁿ⁺¹/(n+1) + C。切记要加上积分常数 C。
Example: Find ∫ (4x³ − 2/x² + e²ˣ) dx.
例题：求 ∫ (4x³ − 2/x² + e²ˣ) dx。
Solution: Rewrite 2/x² as 2x⁻². Then ∫ 4x³ dx = x⁴, ∫ −2x⁻² dx = −2·(x⁻¹/(−1)) = 2/x, ∫ e²ˣ dx = ½ e²ˣ. Therefore the integral is x⁴ + 2/x + ½ e²ˣ + C.
解答：将 2/x² 写为 2x⁻²。∫ 4x³ dx = x⁴，∫ −2x⁻² dx = −2·(x⁻¹/(−1)) = 2/x，∫ e²ˣ dx = ½ e²ˣ。因此积分为 x⁴ + 2/x + ½ e²ˣ + C。
10. Area Under a Curve | 曲线下的面积
The definite integral ∫ₐᵇ f(x) dx gives the signed area between the curve y = f(x), the x‑axis and the vertical lines x = a and x = b. Areas below the x‑axis are negative, so to find the total enclosed area you must split the integral at x‑intercepts and take absolute values.
定积分 ∫ₐᵇ f(x) dx 表示曲线 y = f(x)、x 轴以及直线 x = a 和 x = b 之间的有号面积。x 轴下方的面积为负，因此求总面积时必须在曲线与 x 轴的交点处分割积分，并对各段取绝对值。
Example: Find the total area enclosed by y = x² − 4 and the x‑axis from x = 0 to x = 3.
例题：求曲线 y = x² − 4 与 x 轴在 x = 0 到 x = 3 范围内所围成的总面积。
Solution: x‑intercepts: x² − 4 = 0 ⇒ x = 2 (ignore x = −2 as it is outside [0,3]). On [0,2] the curve lies below the x‑ (9 − 12) − (8/3 − 8) = (−3) − (−16/3) = 7/3. Total area = 16/3 + 7/3 = 23/3 square units.
解答：与 x 轴交点：x² − 4 = 0 ⇒ x = 2（x = −2 不在区间 [0,3] 内，舍去）。在 [0,2] 上曲线在 x 轴下方；在 [2,3] 上在上方。总面积 = −∫₀² (x²−4) dx + ∫₂³ (x²−4) dx。计算：∫ (x²−4) dx = x³/3 − 4x。从 0 到 2：(8/3 − 8) − 0 = −16/3，故 −∫ 部分为 16/3。从 2 到 3：(9 − 12) − (8/3 − 8) = (−3) − (−16/3) = 7/3。总面积 = 16/3 + 7/3 = 23/3 平方单位。
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