📚 Worked Example 6.12.1: Classifying Stationary Points Using the Second Derivative | 例题 6.12.1：利用二阶导数对驻点进行分类
This AQA A-level Pure Mathematics worked example focuses on using the second derivative test to classify stationary points. It is a high-frequency exam skill that combines differentiation, solving equations and interpreting the sign of a derivative.
这道 AQA A-level 纯数例题重点讲解如何利用二阶导数检验法对驻点进行分类。这是考试中的高频技能，综合考查求导、解方程以及对导数符号的理解。
1. The Exam-Style Question | 考试风格题目
d the coordinates of the stationary points on the curve and use the second derivative to determine whether each point is a maximum or a minimum.
一条曲线的方程为 y = x³ - 3x² + 2。求曲线上驻点的坐标，并利用二阶导数判断每个驻点是极大值点还是极小值点。
This question is worth approximately 6 marks in an AQA AS or A-level Pure paper.
本题在 AQA AS 或 A-level 纯数试卷中约值 6 分。
2. What the Examiner Is Testing | 考官考查内容
The examiner wants to see that you can differentiate a polynomial correctly, solve the resulting equation, find the second derivative, and use its sign at each stationary point to classify the turning point.
考官希望看到你能够正确对多项式求导，解出相应方程，求出二阶导数，并利用二阶导数在每个驻点处的符号判断转折点的性质。
Clear method is essential because AQA awards method marks for correct differentiation and valid substitution, even if the final conclusion is slightly imperfect.
清晰的过程至关重要，因为 AQA 会给正确求导和有效代入的方法分，即使最终结论有轻微瑕疵。
3. Step 1: Differentiate the Function | 第1步：对函数求导
Differentiate term by term. The derivative of x³ is 3x², the derivative of -3x² is -6x, and the derivative of the constant 2 is 0.
逐项求导。x³ 的导数是 3x²，-3x² 的导数是 -6x，常数 2 的导数为 0。
dy/dx = 3x² - 6x
Always write the first derivative in a clear form before solving, as this is the key equation for locating stationary points.
在求解之前，务必先写出一阶导数的清晰形式，这是确定驻点位置的关键方程。
4. Step 2: Solve dy/dx = 0 | 第2步：解一阶导数为零的方程
Stationary points occur where the gradient is zero, so set dy/dx equal to zero.
驻点出现在梯度为零的位置，因此令 dy/dx 等于零。
3x² - 6x = 0
Factorise the left-hand side by taking out the common factor 3x.
提取公因式 3x，对左边进行因式分解。
3x(x - 2) = 0
This gives x = 0 or x = 2. These are the x-coordinates of the stationary points.
由此得到 x = 0 或 x = 2。这就是驻点的 x 坐标。
5. Step 3: Find the Second Derivative | 第3步：求二阶导数
Differentiate dy/dx to obtain the second derivative. The derivative of 3x² is 6x, and the derivative of -6x is -6.
对 dy/dx 再求导，得到二阶导数。3x² 的导数是 6x，-6x 的导数是 -6。
d²y/dx² = 6x - 6
The sign of d²y/dx² at a stationary point tells us whether the curve is concave up or concave down at that point, which determines the nature of the turning point.
驻点处 d²y/dx² 的符号告诉我们曲线在该点是上凹还是下凹，这决定了转折点的性质。
6. Step 4: Test the First Stationary Point | 第4步：检验第一个驻点
Substitute x = 0 into the second derivative.
将 x = 0 代入二阶导数。
d²y/dx² = 6(0) - 6 = -6
Since the second derivative is negative, the stationary point at x = 0 is a local maximum.
由于二阶导数为负，x = 0 处的驻点是局部极大值点。
Find the y-coordinate by substituting x = 0 into the original equation.
将 x = 0 代入原方程求出 y 坐标。
y = 0³ - 3(0)² + 2 = 2
Therefore the first stationary point is (0, 2) and it is a maximum.
因此第一个驻点为 (0, 2)，是极大值点。
7. Step 5: Test the Second Stationary Point | 第5步：检验第二个驻点
Substitute x = 2 into the second derivative.
将 x = 2 代入二阶导数。
d²y/dx² = 6(2) - 6 = 6
Since the second derivative is positive, the stationary point at x = 2 is a local minimum.
由于二阶导数为正，x = 2 处的驻点是局部极小值点。
Find the y-coordinate by substituting x = 2 into the original equation.
将 x = 2 代入原方程求出 y 坐标。
y = 2³ - 3(2)² + 2 = 8 - 12 + 2 = -2
Therefore the second stationary point is (2, -2) and it is a minimum.
因此第二个驻点为 (2, -2)，是极小值点。
8. Step 6: Present the Final Answer | 第6步：呈现最终答案
State the conclusion clearly, using a table or complete sentences.
用表格或完整的句子清晰地呈现结论。
| Stationary point / 驻点 | d²y/dx² | Nature / 性质 |
|---|---|---|
| (0, 2) | -6 | Maximum / 极大值 |
| (2, -2) | 6 | Minimum / 极小值 |
The final answer can be written as: (0, 2) is a maximum point and (2, -2) is a minimum point.
最终答案可以写成：(0, 2) 是极大值点，(2, -2) 是极小值点。
9. Common Errors and How to Avoid Them | 常见错误及避免方法
A common
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