Stationary Points | 驻点

📚 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。

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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 \( f(x) = x^3 \). 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 \( f(x) = x^3 \)。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。

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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 \( f(x) = x^3 – 3x^2 + 2 \).

例 1：求 \( f(x) = x^3 – 3x^2 + 2 \) 的驻点。

\( f'(x) = 3x^2 – 6x = 3x(x – 2) \)

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 \( (2, -2) \)。

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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 ∩	\( f(x) = -x^2 \)
Local Minimum | 局部极小值	u-shaped (∪) | 形如 ∪	\( f(x) = x^2 \)
Stationary Point of Inflection | 驻点型拐点	S-shaped (S) | 形如 S	\( f(x) = x^3 \)

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。

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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 \( f(x) = x^3 – 3x^2 + 2 \).

例 2：判别 \( f(x) = x^3 – 3x^2 + 2 \) 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。

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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.
• Interpret: positive to 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 \( f(x) = x^3 \).

例 3：判别 \( f(x) = x^3 \) 的驻点类型。

We have \( f'(x) = 3x^2 \), so \( f'(0) = 0 \). The second derivative is \( f”(x) = 6x \), and \( f”(0) = 0 \), so the second derivative test fails.

已知 \( f'(x) = 3x^2 \)，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。

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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 \( f(x) = 2x^2 – 8x + 5 \).

题目：求 \( f(x) = 2x^2 – 8x + 5 \) 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。

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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：

\( A(x) = x(100 – 2x) = 100x – 2x^2 \)

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。

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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。

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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。

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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 \( f(x) = x^3 – 6x^2 + 9x + 1 \) and classify each using the second derivative test.

题目 1：求 \( f(x) = x^3 – 6x^2 + 9x + 1 \) 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 \( f(x) = x^4 – 4x^3 + 16 \), 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 \( f(x) = x^4 – 4x^3 + 16 \)，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。

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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。

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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。

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