📚 Stationary Points | 驻点
A stationary point of a function is a point where the derivative is zero — that is, where the gradient of the curve is flat. At such a point, the function is neither increasing nor decreasing instantaneously.
函数的驻点是指导数为零的点，即曲线在该点处的斜率为零。在这样的点上，函数在该瞬间既不增加也不减少。
1. Definition and Geometric Meaning | 定义与几何意义
For a function f(x) , a stationary point occurs where f'(x) = 0 . Geometrically, this means the tangent to the curve at that point is horizontal — parallel to the x-axis.
对于函数 f(x) ，驻点出现在 f'(x) = 0 处。从几何角度看，这意味着曲线在该点的切线是水平的，即与 x 轴平行。
Key point: A stationary point is not necessarily a maximum or minimum — it could also be a point of inflection. The derivative being zero is a necessary condition, but not a sufficient one, for a local extremum.
关键点：驻点不一定是极大值点或极小值点，也可能是拐点。导数为零是局部极值的必要条件，但不是充分条件。
Consider f(x) = x3 . At x = 0 , we have f'(0) = 0 , yet the curve continues increasing through this point — it is a stationary point of inflection, not a turning point.
考虑 f(x) = x3 。在 x = 0 处，有 f'(0) = 0 ，但曲线在该点仍然继续上升——这是一个驻点型拐点，而不是转向点。
2. Finding Stationary Points | 求驻点
To find the coordinates of stationary points, follow these steps:
求驻点坐标，请按以下步骤进行：
- Differentiate the function to obtain f'(x) .
- Set f'(x) = 0 and solve for x .
- Substitute each x -value back into f(x) to find the corresponding y -coordinate.
- 对函数求导，得到 f'(x) 。
- 令 f'(x) = 0 ，解方程求得 x 。
- 将每个 x 值代回 f(x) ，求出对应的 y 坐标。
Example 1: Find the stationary points of f(x) = x3 - 3x2 + 2 .
例 1：求 f(x) = x3 - 3x2 + 2 的驻点。
f'(x) = 3x2 - 6x = 3x(x - 2)
Setting f'(x) = 0 gives 3x(x - 2) = 0 , so x = 0 or x = 2 .
令 f'(x) = 0 ，得 3x(x - 2) = 0 ，所以 x = 0 或 x = 2 。
Substituting: f(0) = 2 and f(2) = 8 - 12 + 2 = -2 . Therefore, the stationary points are (0, 2) and (2, -2) .
代回原函数： f(0) = 2 ， f(2) = 8 - 12 + 2 = -2 。因此，驻点为 (0, 2) 和 (2, -2) 。
3. Types of Stationary Points | 驻点的类型
There are three types of stationary points you need to recognise:
你需要识别以下三种驻点类型：
| Type | 类型 | Shape | 形状 | Example | 例子 |
| Local Maximum | 局部极大值 | n-shaped (∩) | 形如 ∩ | f(x) = -x2 |
| Local Minimum | 局部极小值 | u-shaped (∪) | 形如 ∪ | f(x) = x2 |
| Stationary Point of Inflection | 驻点型拐点 | S-shaped (S) | 形如 S | f(x) = x3 |
A local maximum is a point where the function changes from increasing to decreasing. A local minimum is where the function changes from decreasing to increasing. An inflection point is where the curve changes concavity but does not turn around.
局部极大值是函数从增加变为减少的点；局部极小值是函数从减少变为增加的点；拐点是曲线凹凸性改变但方向未反转的点。
4. The Second Derivative Test | 二阶导数判别法
The second derivative, f''(x) , measures the rate of change of f'(x) and is highly useful for classifying stationary points.
二阶导数 f''(x) 衡量 f'(x) 的变化率，对判别驻点类型非常有用。
At a stationary point where f'(x) = 0 :
在满足 f'(x) = 0 的驻点处：
- If f''(x) < 0 , the point is a local maximum.
- If f''(x) > 0 , the point is a local minimum.
- If f''(x) = 0 , the test is inconclusive — the point could be a maximum, minimum, or inflection.
- 若 f''(x) < 0 ，该点为局部极大值。
- 若 f''(x) > 0 ，该点为局部极小值。
- 若 f''(x) = 0 ，则判别法失效——该点可能是极大值、极小值或拐点。
Example 2: Classify the stationary points of f(x) = x3 - 3x2 + 2 .
例 2：判别 f(x) = x3 - 3x2 + 2 的驻点类型。
We already found stationary points at x = 0 and x = 2 . Now f''(x) = 6x - 6 .
已知驻点在 x = 0 和 x = 2 处。现在求 f''(x) = 6x - 6 。
At x = 0 : f''(0) = -6 < 0 , so (0, 2) is a local maximum.
在 x = 0 处： f''(0) = -6 < 0 ，所以 (0, 2) 是局部极大值。
At x = 2 : f''(2) = 6 > 0 , so (2, -2) is a local minimum.
在 x = 2 处： f''(2) = 6 > 0 ，所以 (2, -2) 是局部极小值。
5. The Sign Table Method | 符号表判别法
When f''(x) = 0 at a stationary point, the second derivative test is inconclusive. In such cases, use the sign of f'(x) on either side of the stationary point.
当驻点处 f''(x) = 0 时，二阶导数判别法失效。此时应使用驻点两侧 f'(x) 的符号来判断。
Procedure:
步骤：
- Choose an x -value slightly less than the stationary point and an x -value slightly greater than it.
- Evaluate f'(x) at these test values and record the signs.
- negative = maximum; negative to positive = minimum; same sign on both sides = inflection.
- 选取略小于驻点的 x 值和略大于驻点的 x 值。
- 在这些测试点处计算 f'(x) 并记录符号。
- 判断：由正变负为极大值；由负变正为极小值；两侧符号相同为拐点。
Example 3: Classify the stationary point of f(x) = x3 .
例 3：判别 f(x) = x3 的驻点类型。
We have f'(x) = 3x2 , so f'(0) = 0 . The second derivative is f''(x) = 6x , and f''(0) = 0 , so the second derivative test fails.
已知 f'(x) = 3x2 ，所以 f'(0) = 0 。二阶导数为 f''(x) = 6x ，且 f''(0) = 0 ，因此二阶导数判别法失效。
Testing the sign of f'(x) : at x = -1 , f'(-1) = 3 > 0 ; at x = 1 , f'(1) = 3 > 0 . Since the sign does not change, (0, 0) is a stationary point of inflection.
测试 f'(x) 的符号：在 x = -1 处， f'(-1) = 3 > 0 ；在 x = 1 处， f'(1) = 3 > 0 。由于符号未改变， (0, 0) 是一个驻点型拐点。
6. Worked Example with a Quadratic | 二次函数实例
Let us apply both methods to a complete problem.
让我们将两种方法应用于一个完整的题目。
Problem: Find and classify the stationary point of f(x) = 2x2 - 8x + 5 .
题目：求 f(x) = 2x2 - 8x + 5 的驻点并判别其类型。
Solution: f'(x) = 4x - 8 . Setting f'(x) = 0 gives 4x - 8 = 0 , so x = 2 . Then f(2) = 2(4) - 16 + 5 = -3 . The stationary point is (2, -3) .
解答： f'(x) = 4x - 8 。令 f'(x) = 0 ，得 4x - 8 = 0 ，所以 x = 2 。于是 f(2) = 2(4) - 16 + 5 = -3 。驻点为 (2, -3) 。
Second derivative: f''(x) = 4 > 0 , so the point is a local minimum. The sign table confirms this: f'(1) = -4 < 0 and f'(3) = 4 > 0 , a change from negative to positive.
二阶导数： f''(x) = 4 > 0 ，所以该点是局部极小值。符号表也证实了这一点： f'(1) = -4 < 0 ， f'(3) = 4 > 0 ，符号由负变正。
7. Applications in Optimisation | 优化问题中的应用
Stationary points are central to solving optimisation problems — finding maximum or minimum values of quantities in real-world contexts.
驻点是解决优化问题的核心——即在现实情境中寻找数量的最大值或最小值。
Example 4: A rectangular enclosure is built using 100 m of fencing, with one side against a wall. Find the maximum area.
例 4：用 100 米篱笆靠墙围成一个矩形场地。求最大面积。
Let the two equal sides perpendicular to the wall each be x metres, and the side parallel to the wall be 100 - 2x metres. The area is:
设垂直于墙的两条边各为 x 米，平行于墙的边为 100 - 2x 米。面积为：
A(x) = x(100 - 2x) = 100x - 2x2
Differentiating: A'(x) = 100 - 4x . Setting equal to zero gives x = 25 . Then A''(x) = -4 < 0 , confirming a maximum. The maximum area is A(25) = 25(50) = 1250 m².
求导： A'(x) = 100 - 4x 。令其为零，得 x = 25 。又 A''(x) = -4 < 0 ，确认为极大值。最大面积为 A(25) = 25(50) = 1250 平方米。
When solving optimisation problems, always check the domain of x and verify whether the stationary value is a maximum or minimum using an appropriate test.
在解决优化问题时，务必检查 x 的定义域，并使用适当的方法验证驻值究竟是极大值还是极小值。
8. Common Exam Mistakes | 常见考试错误
Candidates often lose marks on stationary point questions for the following reasons:
考生在驻点类题目中常因以下原因丢分：
- Solving f'(x) = 0 incorrectly, especially when factorising.
- Forgetting to substitute the x -values back into f(x) to find the y -coordinates.
- Stating the type of stationary point without justification — always show f''(x) or the sign table.
- Concluding "inconclusive" from f''(x) = 0 without proceeding to the sign table.
- Confusing local extrema with global extrema on a restricted domain.
- 解 f'(x) = 0 时出错，尤其是在因式分解时。
- 忘记将 x 值代回 f(x) 以求出 y 坐标。
- 未给出依据就直接说明驻点类型——必须展示 f''(x) 或符号表。
- 当 f''(x) = 0 时直接判定"失效"，而未继续使用符号表。
- 混淆限制定义域下的局部极值与全局极值。
Tip: Always write "Since f'(x) = 0 at x = a , the point is stationary" to clearly signal your reasoning to the examiner.
提示：务必写出"因为 f'(x) = 0 在 x = a 处成立，所以该点为驻点"，以便向阅卷老师清晰展示你的推理过程。
9. Summary of Key Formulas | 关键公式汇总
The table below summarises everything you need to remember:
下表总结了你需要记住的全部要点：
| Condition | 条件 | Conclusion | 结论 |
| f'(x) = 0 | Stationary point exists | 存在驻点 |
| f'(x) = 0 and f''(x) < 0 | Local maximum | 局部极大值 |
| f'(x) = 0 and f''(x) > 0 | Local minimum | 局部极小值 |
| f'(x) = 0 and f''(x) = 0 , sign of f'(x) unchanged | Point of inflection | 拐点 |
For polynomial functions, the degree of the polynomial determines the maximum number of stationary points: a degree- n polynomial has at most n - 1 stationary points.
对于多项式函数，其次数决定了驻点的最大个数： n 次多项式至多有 n - 1 个驻点。
10. Practice Questions | 练习题
Test your understanding with these IGCSE-style questions:
请用以下 IGCSE 风格题目检验你的理解：
Question 1: Find the stationary points of f(x) = x3 - 6x2 + 9x + 1 and classify each using the second derivative test.
题目 1：求 f(x) = x3 - 6x2 + 9x + 1 的驻点，并使用二阶导数判别法分类。
Question 2: A farmer has 80 m of fencing. He uses it to create a rectangular pen, but one side is along a river and needs no fencing. Find the maximum area of the pen.
题目 2：一名农夫有 80 米篱笆，他需要用其围成一个矩形围栏，其中一边沿河，无需围栏。求围栏的最大面积。
Question 3: For f(x) = x4 - 4x3 + 16 , find all stationary points and determine their nature. Note the second derivative test may fail at some points — use the sign table where necessary.
题目 3：对于 f(x) = x4 - 4x3 + 16 ，求所有驻点并判断其性质。注意：某些点处二阶导数判别法可能失效——必要时请使用符号表。
Work through each question systematically: differentiate, solve, classify, and verify your answers graphically or by substitution.
请系统地完成每道题：求导、解方程、分类，并通过图形或代入验证你的答案。
11. Final Exam Tips | 最终考试建议
In the Edexcel IGCSE Mathematics exam, stationary point questions often appear in both Paper 1 (non-calculator) and Paper 2 (calculator). They may be standalone or embedded in a longer problem involving curve sketching or optimisation.
在 Edexcel IGCSE 数学考试中，驻点问题常出现在试卷 1（不允许使用计算器）和试卷 2(允许使用计算器）中。它们可能独立出现，也可能嵌入到涉及曲线作图或优化的综合题中。
Examiner's tip: For a curve y = f(x) , solving f'(x) = 0 gives any stationary points. Then use f''(x) or a sign table to classify them. Always give coordinates as (x, y), not just the x -value.
阅卷官提示：对于曲线 y = f(x) ，解 f'(x) = 0 可得驻点。然后用 f''(x) 或符号表进行分类。务必给出完整的坐标 (x, y)，而不仅仅是 x 值。
Revise thoroughly, practise past paper questions, and always check the domain of the function before drawing conclusions about maxima and minima.
请全面复习，认真练习历年真题，并在下结论前始终检查函数的定义域。
12. Conclusion | 结论
Stationary points are one of the most examinable topics in IGCSE differentiation. Mastering the art of finding and classifying them — using both the second derivative test and the sign table method — will serve you well across multiple parts of the syllabus, particularly in curve sketching and optimisation problems.
驻点是 IGCSE 微分部分中考察频率最高的知识点之一。熟练掌握求驻点及其分类的方法——包括二阶导数判别法和符号表法——将在课程大纲中多个部分为你助力，尤其是在曲线作图和优化问题中。
Remember the golden rule: stationary means the gradient is zero; classification requires further investigation. With consistent practice, this topic will become one of your strongest areas.
请记住黄金法则：驻点意味着梯度为零；分类则需要进一步研究。通过持续练习，这一主题将成为你最擅长的领域之一。
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