📚 Exercise 16B: Stationary Points and Curve Sketching | 练习16B：驻点与曲线草图
Welcome to this focused revision guide for Exercise 16B, a key practice set in the IB Mathematics course that deepens your ability to analyse functions using differentiation. Here we will walk through the theory of increasing and decreasing behaviour, stationary points, concavity, and the art of curve sketching, all supported by worked examples and exam-style thinking.
欢迎来到针对练习16B的专项复习指南，该练习是IB数学课程中利用微分深入分析函数的关键训练。本文将系统梳理函数的递增递减性、驻点、凹凸性以及曲线草图绘制技巧，并结合例题与考试思路帮你夯实核心技能。
1. Recap of Differentiation Essentials | 重温微分基础
The derivative f' the instantaneous rate of change and the fundamental tool for understanding the shape of a graph.
导数 f'(x) 表示曲线 y = f(x) 上任意一点处切线的梯度。它体现了函数的瞬时变化率，是剖析图形形状的基础工具。
sum and difference rules. For products and quotients, use the product and quotient rules, and for composite functions, the chain rule is essential.
需要牢记的基本法则：幂法则 d/dx(xⁿ) = n xⁿ⁻¹，常数乘法则，和差法则。处理乘积和商时，分别使用乘法法则和除法法则，而链式法则在处理复合函数时不可或缺。
In Exercise 16B, you will constantly rely on finding first and second derivatives quickly and accurately, so fluency in these rules is a must.
在练习16B中，你需要快速且准确地求出一阶和二阶导数，因此对这些法则的熟练掌握是必须的。
2. Increasing and Decreasing Functions | 函数的递增与递减
A function f(x) is increasing on an interval if f'(x) > 0 for all x in that interval. Conversely, if f'(x) < 0, the function is decreasing.
如果在某个区间上 f'(x) > 0，则函数在该区间上递增；反之，若 f'(x) < 0，则函数递减。
When f' the tangent is horizontal, and the function may change its increasing/decreasing behaviour – this is where stationary points occur.
当 f'(x) = 0 时，切线水平，函数可能在此处改变其递增或递减的性质——这正是驻点出现的位置。
To determine intervals of increase or decrease, solve f'(x) > 0 and f'(x) < 0, and present the results on a sign diagram.
要确定递增或递减区间，需解不等式 f'(x) > 0 和 f'(x) < 0，并将结果呈现在符号图上。
3. Stationary Points | 驻点
Stationary points occur where f'(x) = 0. They include local maxima, local minima, and horizontal points of inflection.
驻点发生在 f'(x) = 0 处，包括局部最大值、局部最小值以及水平拐点。
Not every point where f' l inflection also has zero gradient but does not change direction from increasing to decreasing (or vice versa).
并非所有使 f'(x) = 0 的点都是极值点；水平拐点处的梯度同样为零，但函数并未从递增变为递减（或相反）。
Exercise 16B will ask you to find all stationary points and then classify them, a process that requires either the first or second derivative test.
练习16B会要求找出所有驻点并对其进行分类，这一过程需要使用一阶导数检验法或二阶导数检验法。
4. Classifying Stationary Points: First Derivative Test | 分类驻点：一阶导数检验法
The first derivative test examines the sign of f'(x) immediately to the left and right of a stationary point x = a.
一阶导数检验法观察在驻点 x = a 左侧和右侧紧邻区域内 f'(x) 的符号变化。
If f' is a local maximum. If the sign changes from negative to positive, it is a local minimum.
若在 x = a 处 f'(x) 由正变负，则该点为局部最大值；若由负变正，则为局部最小值。
If the sign of f'(x) does not change, the stationary point is a horizontal point of inflection. This test always works, even when the second derivative is difficult to compute.
若 f'(x) 的符号没有发生变化，则驻点为水平拐点。即使二阶导数难以计算，该检验法也始终有效。
Use a sign diagram clearly marking the critical value x = a and the resulting sign of f'(x) on either side to justify your classification.
绘制符号图，清晰标出临界值 x = a 及其两侧 f'(x) 的符号，用于支撑你的分类结论。
5. Classifying Stationary Points: Second Derivative Test | 分类驻点：二阶导数检验法
If f'' , the second derivative test offers a quick way to classify the stationary point at x = a.
如果 f''(a) 存在，二阶导数检验法能快速对 x = a 处的驻点进行分类。
When f'(a) = 0 and f''(a) > 0, the function is concave up at that point, indicating a local minimum. If f''(a) < 0, the function is concave down, so the point is a local maximum.
当 f'(a) = 0 且 f''(a) > 0 时，函数在该点处凹向上，表明是局部最小值；若 f''(a) < 0，函数凹向下，该点为局部最大值。
Caution: if f'' stationary point could be a maximum, a minimum, or a point of inflection, and you must fall back on the first derivative test.
注意：若 f''(a) = 0，该检验法无法得出结论。驻点可能是最大值、最小值或拐点，此时必须使用一阶导数检验法。
6. Concavity and Points of Inflection | 凹凸性与拐点
Concavity describes how a curve bends. If f''(x) > 0, the curve is concave up (like a cup). If f''(x) < 0, it is concave down (like a cap).
凹凸性描述了曲线的弯曲方向。若 f''(x) > 0，曲线凹向上（如杯形）；若 f''(x) < 0，则凹向下（如帽形）。
inflection occurs where the concavity changes sign, and it is typically found by solving f''(x) = 0 and checking the sign change of f''(x).
拐点出现在凹凸性改变符号的位置，通常通过解 f''(x) = 0 并检查 f''(x) 的符号变化来寻找。
Not every solution to f''(x) = 0 is an inflection point; you must confirm that f''(x) actually changes sign. This is critical in curve sketching problems in Exercise 16B.
并非每个满足 f''(x) = 0 的解都是拐点；必须验证 f''(x) 确实改变符号。这一点在练习16B的曲线草图问题中至关重要。
7. Curve Sketching Strategy | 曲线草图绘制策略
16B: start by examining the domain, intercepts (x and y), and any symmetry or asymptotes.
绘制 y = f(x) 草图的系统方法将助你轻松拿下练习16B：首先审查定义域、截距（x 轴和 y 轴截距），以及对称性或渐近线。
Next, compute f'(x) to find stationary points and intervals of increase/decrease. Then compute f''(x) to determine concavity and any points of inflection.
接着，计算 f'(x) 以求出驻点及递增/递减区间；然后计算 f''(x) 以确定凹凸性和拐点。
Build a table summarising the behaviour over key intervals and draw the axes, plot critical points, and smoothly connect them consistent with the sign analysis. Always label important features.
建立一个表格，总结各关键区间上的函数行为，然后绘制坐标轴、标出关键点，并根据符号分析结果光滑地连接它们。务必标注重要特征。
8. Worked Example 1: Polynomial Function | 例题1：多项式函数
tionary points, determine their nature, identify any points of inflection, and sketch the curve.
考虑函数 f(x) = x³ − 3x + 2。求出所有驻点，判断其性质，确定所有拐点，并绘制曲线草图。
Step 1: f'(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1). Setting f'(x) = 0 gives stationary points at x = 1 and x = −1.
步骤1：f'(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1)。令 f'(x) = 0 得驻点 x = 1 和 x = −1。
Corresponding y-values: f(1) = 0, f(−1) = 4. So we have points (1,0) and (−1,4).
相应的 y 值：f(1) = 0，f(−1) = 4。得到点 (1,0) 和 (−1,4)。
Second derivative: f''(x) = 6x. At x = −1, f''(−1) = −6 < 0, so a local maximum at (−1,4). At x = 1, f''(1) = 6 > 0, giving a local minimum at (1,0).
二阶导数：f''(x) = 6x。在 x = −1 处，f''(−1) = −6 < 0，故 (−1,4) 为局部最大值；在 x = 1 处，f''(1) = 6 > 0，故 (1,0) 为局部最小值。
For inflection: f''(x) = 0 ⇒ 6x = 0 ⇒ x = 0. f(0) = 2. Sign of f'' changes from negative to positive, so (0,2) is a point of inflection.
拐点：f''(x) = 0 ⇒ 6x = 0 ⇒ x = 0。f(0) = 2。f'' 的符号由负变正，因此 (0,2) 为拐点。
Increasing/decreasing intervals: f'(x) > 0 for x < −1 and x > 1; f'(x) < 0 for −1 < x < 1. Concave down for x < 0, concave up for x > 0.
递增/递减区间：当 x < −1 和 x > 1 时 f'(x) > 0；当 −1 < x < 1 时 f'(x) < 0。x < 0 时凹向下，x > 0 时凹向上。
2)=0 gives (−2,0) and (1,0) (double root). The curve passes through these points smoothly.
草绘：标出最大值点 (−1,4)、最小值点 (1,0)、拐点 (0,2) 以及 y 截距 (0,2)。x 截距：解 x³ − 3x + 2 = 0 得 (x−1)²(x+2)=0，即 (−2,0) 和 (1,0)（重根）。曲线平滑穿过这些点。
9. Worked Example 2: Rational Function | 例题2：有理函数
− 1) for Exercise 16B style reasoning: domain, asymptotes, stationary points, and sketch.
以练习16B的风格分析 f(x) = x/(x² − 1)：定义域、渐近线、驻点及草图。
d x = −1. Horizontal asymptote: as x → ±∞, f(x) → 0, so y = 0 is the horizontal asymptote.
定义域：所有实数 x 除了分母为零的 x = ±1。垂直渐近线：x = 1 和 x = −1。水平渐近线：当 x → ±∞ 时 f(x) → 0，故 y = 0 为水平渐近线。
First derivative using quotient rule: f' x²−1)(1) − x(2x) ] / (x²−1)² = (−x² − 1) / (x²−1)². The numerator is always negative, so f' s domain, and there are no stationary points because the numerator −x² − 1 ≠ 0 for real x.
使用除法法则求一阶导数：f'(x) = [ (x²−1)(1) − x(2x) ] / (x²−1)² = (−x² − 1) / (x²−1)²。分子恒为负，故在定义域内 f'(x) < 0 总成立。因此函数在各分段上递减，且由于分子 −x² − 1 在实数范围内不可能为零，故无驻点。
Second derivative for concavity: f''(x) simplifies to [2x(x²+3)]/(x²−1)³. The sign depends on x and the denominator's cube. Analyse concavity intervals: concave up for x ∈ (−∞,−1)∪(0,1), concave down for x ∈ (−1,0)∪(1,∞). Inflection at x = 0 (f(0)=0).
计算凹凸性的二阶导数：f''(x) 化简为 [2x(x²+3)]/(x²−1)³。其符号取决于 x 及分母的立方。分析凹凸区间：在 (−∞,−1)∪(0,1) 上凹向上，在 (−1,0)∪(1,∞) 上凹向下。拐点位于 x = 0（f(0)=0）。
Sketch: draw asymptotes, plot origin (inflection), note decreasing trend, and confirm shape approaching asymptotes.
草图：画出渐近线，标出原点（拐点），注意递减趋势，并验证趋近渐近线的形状。
10. Common Mistakes and Tips | 常见错误与应试技巧
One common mistake is forgetting to check the domain before differentiation, especially for rational or root functions. Always state the domain first.
一个常见错误是在微分前忘记检查定义域，尤其针对有理函数或根式函数。务必先指明定义域。
Another pitfall is assuming f''(a) = 0 automatically means a point of inflection; you must prove a sign change in f''(x). Similarly, do not rely solely on the second derivative test when f''(a) = 0 or when it is hard to compute.
另一个陷阱是认为 f''(a) = 0 就自动意味着拐点；必须证明 f''(x) 改变符号。同样，当 f''(a) = 0 或计算困难时，不要仅依赖二阶导数检验法。
When sketching, many students connect points straight without considering concavity. Use f''(x) sign analysis to ensure your curve bends correctly.
绘图时，许多学生不考虑凹凸性而直接用直线连接各点。应利用 f''(x) 符号分析确保曲线弯曲方向正确。
Finally, label everything: intercepts, stationary points, inflection points, and asymptotes. Clear labelling is expected in IB exam solutions.
最后，标注所有内容：截距、驻点、拐点和渐近线。在IB考试作答中，清晰的标注是要求做到的。
11. IB Exam-Style Question | IB 考试风格题目
nary points and classify them. (b) Determine intervals of increase/decrease and concavity. (c) Sketch the graph, showing all key features.
设 f(x) = x⁴ − 4x³。(a) 求出所有驻点并分类；(b) 确定递增/递减区间及凹凸区间；(c) 绘制曲线草图，展示所有关键特征。
Solution outline: f'(x) = 4x³ − 12x² = 4x²(x − 3). Stationary points at x = 0 and x = 3. f''(x) = 12x² − 24x = 12x(x − 2). At x = 3, f''(3) = 36 > 0 → local minimum (3, −27). At x = 0, f''(0) = 0, so we use first derivative test: f'(x) does not change sign around 0, therefore (0,0) is a horizontal point of inflection.
解答要点：f'(x) = 4x³ − 12x² = 4x²(x − 3)。驻点在 x = 0 和 x = 3。f''(x) = 12x² − 24x = 12x(x − 2)。在 x = 3 处，f''(3) = 36 > 0 → 局部最小值 (3, −27)。在 x = 0 处，f''(0) = 0，故使用一阶导数检验法：f'(x) 在 0 附近符号不变，因此 (0,0) 为水平拐点。
Increase: x > 3 (and x < 0 but careful with sign? f' ositive for x > 3, negative for 0 < x < 3, and positive for x < 0). Actually f'(x) = 4x²(x−3), x² always ≥0, so sign depends on (x−3). f'(x) > 0 for x > 3, f' g for x < 3, increasing for x > 3, but note horizontal inflection at 0. Concavity: f'' concave up for x > 2. Inflection points at x = 0 and x = 2 (y=−16). Sketch accordingly.
递增区间：x > 3 时 f'(x) > 0，x < 3（x≠0）时 f'(x) < 0，因此函数在 x < 3 时递减，x > 3 时递增，注意 x=0 处为水平拐点。凹凸性：f''(x) = 0 在 x = 0 和 x = 2 处，凹向上在 x < 0，凹向下在 0 < x < 2，凹向上在 x > 2。拐点位于 x = 0 和 x = 2（y=−16）。据此绘图。
12. Summary and Key Takeaways | 总结与要点归纳
Stationary points (f'(x) = 0) are the gateway to understanding a function' irst or second derivative test, and never forget to check concavity and inflection points.
驻点（f'(x) = 0）是理解函数极值行为及形状的入口。务必使用一阶或二阶导数检验法加以分类，并始终检查凹凸性和拐点。
Mastering Exercise 16B means you can confidently move from f(x) to its derivatives, construct sign charts, and produce labelled, accurate sketches that meet IB assessment standards. Practise with a variety of polynomial, rational, and radical functions to consolidate your skills.
掌握练习16B意味着你能自信地从 f(x) 出发求得其导数，构建符号表，并绘制出符合IB评分标准的、标注清晰的准确草图。通过练习各类多项式、有理函数和根式函数来巩固技能。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图