📚 Exercise 5G: Stationary Points & Second Derivative Test | 练习5G：驻点与二阶导数检验
In IB Mathematics, Exercise 5G typically explores the relationship between stationary points and the second derivative. Understanding where the gradient of a function becomes zero and how the curve bends around those points is central to optimisation, graph sketching, and real-world modelling. This article breaks down the key concepts, provides worked examples, and offers exam-ready tips for tackling Exercise 5G with confidence.
在IB数学中，练习5G通常探讨驻点与二阶导数之间的关系。理解函数梯度何时为零，以及曲线在这些点周围如何弯曲，对于最优化、图形绘制和现实建模至关重要。本文分解了关键概念，提供了详细例题，并给出了备考建议，帮助你有信心地应对练习5G。
1. What Are Stationary Points? | 什么是驻点？
int, the tangent to the curve is horizontal, meaning the function stops increasing or decreasing momentarily.
驻点出现在 y = f(x) 图像上一阶导数为零的位置，即 f′(x) = 0。在这样的点上，曲线的切线是水平的，意味着函数暂时停止增大或减小。
Stationary points are classified as local maxima, local minima, or stationary points of inflection. The nature of the point can be determined by examining the sign of the gradient on either side, or more efficiently in many cases, by the second derivative test.
驻点被分为局部极大值、局部极小值或驻拐点。点的性质可以通过检查两侧梯度的符号来确定，或者在许多情况下更有效地，通过二阶导数检验来确定。
2. How to Find Stationary Points | 如何求驻点
Finding stationary points is a straightforward process: differentiate the function to obtain f′(x), set the derivative equal to zero, and solve for x. These x-values are the locations of the stationary points. Plugging them back into f(x) gives the corresponding y‑coordinates.
求驻点的过程很直接：对函数求导得到 f′(x)，令导数等于零，并解出 x。这些 x 值就是驻点的位置。将它们代回 f(x) 可得到相应的 y 坐标。
0 gives 3x(x − 2) = 0, so x = 0 or x = 2. The stationary points are at (0, 2) and (2, −2).
例如，如果 f(x) = x³ − 3x² + 2，那么 f′(x) = 3x² − 6x。令 f′(x) = 0 得到 3x(x − 2) = 0，因此 x = 0 或 x = 2。驻点为 (0, 2) 和 (2, −2)。
3. The Second Derivative Test | 二阶导数检验
The second derivative of a function, denoted f″(x) or d²y/dx², describes the rate of change of the gradient. If f″(x) is positive at a stationary point, the curve is bending upwards (concave up) and the point is a local minimum. If f″(x) is negative, the curve bends downwards (concave down) and the point is a local maximum.
函数的二阶导数，记作 f″(x) 或 d²y/dx²，描述梯度的变化率。如果驻点处的 f″(x) 为正，曲线向上弯曲（凹向上），该点为局部极小值。如果 f″(x) 为负，曲线向下弯曲（凹向下），该点为局部极大值。
If the second derivative is zero, the test is inconclusive, and you should examine the sign of the first derivative on either side of the point to determine its nature.
如果二阶导数为零，则检验无法得出结论，你应检查该点两侧一阶导数的符号来确定其性质。
4. Classifying Stationary Points with the Second Derivative | 用二阶导数对驻点分类
The classification rule can be summarised clearly:
分类规则可以清楚地总结如下：
| Sign of f″(x) | Nature of stationary point |
|---|---|
| f″(x) > 0 | Local minimum (convex) |
| f″(x) < 0 | Local maximum (concave) |
| f″(x) = 0 | Test fails; use gradient inspection |
< 0, so (0, 2) is a local maximum. At x = 2, f″(2) = 6 > 0, so (2, −2) is a local minimum.
对于前面的例子 f(x) = x³ − 3x² + 2，二阶导数为 f″(x) = 6x − 6。在 x = 0 处，f″(0) = −6 < 0，因此 (0, 2) 是局部极大值。在 x = 2 处，f″(2) = 6 > 0，因此 (2, −2) 是局部极小值。
5. Understanding Concavity and Points of Inflection | 理解凹凸性与拐点
flection occurs when f′(x) = 0 and f″(x) = 0, with a sign change in the second derivative.
凹凸性描述曲线的弯曲方式。在 f″(x) > 0 的区域，图形凹向上；在 f″(x) < 0 的区域，图形凹向下。凹凸性发生变化的点称为拐点。并非所有拐点都是驻点；当 f′(x) = 0 且 f″(x) = 0，并且二阶导数发生变号时，出现驻拐点。
x) changes from negative to positive, giving a stationary point of inflection at the origin.
一个经典例子是 f(x) = x³。在此 f′(0) = 0 且 f″(0) = 0，但 f″(x) 的符号由负变正，因此在原点出现驻拐点。
6. Worked Example – Polynomial Function | 例题：多项式函数
Let us work through a complete example required by Exercise 5G.
我们来完整演练一个练习5G所要求的例子。
Problem: Find and classify the stationary points of f(x) = 2x⁴ − 4x² + 1.
问题：求 f(x) = 2x⁴ − 4x² + 1 的驻点并分类。
First derivative:
f′(x) = 8x³ − 8x = 8x(x² − 1) = 8x(x − 1)(x + 1)
Setting f′(x) = 0 gives x = −1, 0, 1. The corresponding y‑values are f(−1) = −1, f(0) = 1, f(1) = −1.
令 f′(x) = 0 得到 x = −1, 0, 1。相应的 y 值为 f(−1) = −1，f(0) = 1，f(1) = −1。
Second derivative:
f″(x) = 24x² − 8
- At x = −1: f″(−1) = 24 − 8 = 16 > 0 → local minimum at (−1, −1).
- At x = 0: f″(0) = −8 < 0 → local maximum at (0, 1).
- At x = 1: f″(1) = 16 > 0 → local minimum at (1, −1).
- 在 x = −1：f″(−1) = 24 − 8 = 16 > 0 → (−1, −1) 处为局部极小值。
- 在 x = 0：f″(0) = −8 < 0 → (0, 1) 处为局部极大值。
- 在 x = 1：f″(1) = 16 > 0 → (1, −1) 处为局部极小值。
The graph is symmetric and has two minima and one maximum, a typical “W” shape.
该图形对称，具有两个极小值和一个极大值，典型的“W”形。
7. Worked Example – Rational Function | 例题：有理函数
Consider f(x) = x + 4/x for x ≠ 0. This type of function often appears in IB exercises.
考虑 f(x) = x + 4/x，其中 x ≠ 0。这类函数经常出现在IB练习题中。
First derivative:
f′(x) = 1 − 4x⁻²
ves 1 = 4/x², so x² = 4, hence x = 2 or x = −2. Stationary points are (2, 4) and (−2, −4).
令 f′(x) = 0 得 1 = 4/x²，因此 x² = 4，故 x = 2 或 x = −2。驻点为 (2, 4) 和 (−2, −4)。
Second derivative:
f″(x) = 8x⁻³
) is a local minimum. At x = −2: f″(−2) = 8/(−8) = −1 < 0, so (−2, −4) is a local maximum.
在 x = 2：f″(2) = 8/8 = 1 > 0，故 (2, 4) 为局部极小值。在 x = −2：f″(−2) = 8/(−8) = −1 < 0，故 (−2, −4) 为局部极大值。
This example highlights that the second derivative test works equally well for rational functions, provided you simplify carefully.
这个例子强调，二阶导数检验同样适用于有理函数，只要你仔细化简。
8. When the Second Derivative Test Fails | 当二阶导数检验失效时
test gives no information. In such cases, you must examine the sign of f′(x) immediately to the left and right of the point. Constructing a gradient table is a reliable method.
若在驻点处 f″(x) = 0，则检验无法提供信息。此时，你必须检查该点左右邻近的一阶导数符号。构建梯度表是一种可靠的方法。
ent changes from negative to positive, so (0, 0) is a local minimum despite the second derivative being zero.
对于 f(x) = x⁴，我们有 f′(0) = 0 且 f″(0) = 0。测试 x = −0.1 得 f′(−0.1) = −0.004（负），x = 0.1 得 f′(0.1) = 0.004（正）。梯度由负变正，因此 (0, 0) 是局部极小值，尽管二阶导数为零。
This approach ensures you can handle any function, including trigonometric or exponential ones, where multiple derivatives may vanish.
这种方法确保你能处理任何函数，包括三角函数或指数函数，其中高阶导数可能为零。
9. Optimisation Applications in IB Problems | IB题目中的最优化应用
Exercise 5G often includes real-world optimisation: maximising area, minimising surface area for a given volume, or finding least material cost. The stationary point naturally gives the optimal solution, and the second derivative confirms whether it is a maximum or minimum.
练习5G通常包含现实世界中的最优化问题：最大化面积、在给定体积下最小化表面积，或寻找最少材料成本。驻点自然而然地给出最优解，二阶导数则确认它是极大值还是极小值。
and checking with the second derivative, you prove the optimal radius is r = (V/(2π))¹/³.
例如，一个经典问题要求确定在固定体积V下使表面积最小的圆柱体尺寸。通过将表面积表示为半径的函数，求导找到驻点，再用二阶导数检验，可以证明最优半径为 r = (V/(2π))¹/³。
In IB exams, always verify that the critical value gives the desired extreme (max or min) by using either the second derivative or a sign table – units and a concluding sentence are essential for full marks.
在IB考试中，务必通过二阶导数或符号表验证临界值给出所需的极值（最大或最小）——单位以及一句结论性语句对于获得满分至关重要。
10. Common Mistakes and How to Avoid Them | 常见错误及如何避免
Even students who understand the theory can lose marks on Exercise 5G. Below are the most frequent pitfalls:
即使理解理论的学生也可能在练习5G中失分。以下是最常遇到的陷阱：
- Forgetting to check whether f″(x) = 0: Many apply the test without realising it fails, leading to an incorrect classification.
- 忽略检查 f″(x) = 0 的情况：许多人在没有意识到检验失效的情况下就用它，导致错误分类。
- Arithmetic errors when solving f′(x) = 0: Always factorise carefully or use the quadratic formula. Double-check by substituting back.
- 解 f′(x) = 0 时的代数错误：务必要仔细因式分解或使用二次公式。通过代回进行双重检查。
- Misinterpreting concavity: Positive second derivative means concave up (minimum), not maximum. Drawing a quick sketch helps prevent this mix-up.
- 误解凹凸性：正二阶导数意味着凹向上（极小值），而非极大值。画一个快速草图有助于防止混淆。
- Omitting the units in optimisation problems: If the question asks for dimensions or cost, include the relevant unit in your final answer.
- 在最优化问题中遗漏单位：如果题目要求尺寸或成本，最终答案要包含相关单位。
11. Summary and Exam Tips | 总结与考试技巧
To master Exercise 5G, follow this routine: differentiate to find stationary points, compute the second derivative, evaluate its sign at each point, and classify. When the second derivative is zero, fall back on a gradient table. Practise with polynomials, rational functions, and simple trigonometric functions until the steps become automatic.
要掌握练习5G，请遵循以下步骤：求导找到驻点，计算二阶导数，在每个点处求值判断符号，并进行分类。当二阶导数为零时，转而使用梯度表。不断练习多项式、有理函数和简单三角函数，直到这些步骤变得自动。
In the IB exam, working is often rewarded. Show every derivative clearly, label your stationary points, and explicitly state the final classification. Efficient use of the second derivative test saves time compared to a full gradient sign check, but you must know both methods to handle any function.
在IB考试中，解题过程通常会得分。清晰地写出每一个导数，标出驻点，并明确说明最终分类。与完整的梯度符号检查相比，有效运用二阶导数检验可节省时间，但你必须两种方法都掌握，以应对任何函数。
12. Practice Questions to Cement Understanding | 巩固理解的练习题
Try these problems before moving on from Exercise 5G:
在离开练习5G之前，试试这些习题：
- Find and classify the stationary points of f(x) = x³ − 6x² + 9x − 4.
- 求 f(x) = x³ − 6x² + 9x − 4 的驻点并分类。
- Determine the nature of the stationary point of g(x) = x − 2 sin x for 0 < x < π.
- 确定 g(x) = x − 2 sin x 在区间 0 < x < π 内驻点的性质。
- A rectangle is inscribed in a circle of radius r. Express the area as a function of one variable, find the maximum area, and confirm with the second derivative.
- 一个矩形内接于半径为 r 的圆。将面积表达为单变量函数，求最大面积，并用二阶导数确认。
Working through a variety of examples builds confidence and prepares you for the mixed nature of IB questions.
通过演练各种例子来建立信心，为应对IB考试中题目的多样性做好准备。
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