📚 Example 7.4.3: Classifying Stationary Points | 例题 7.4.3：驻点分类
In AQA A-level Mathematics, differentiation is a core Paper 1 skill. Example 7.4.3 is a typical exam-style question: given a cubic curve, find the coordinates of its stationary points and determine whether each is a maximum or minimum. This article works through the complete method, including first and second derivative tests, common errors, and AQA mark scheme tips.
在 AQA A-level 数学中，微分是纯数 Paper 1 的核心技能。例题 7.4.3 是一道典型考题：给定一条三次曲线，求其驻点坐标，并判断每个驻点是极大值还是极小值。本文完整讲解解题方法，包括一阶导数检验、二阶导数检验、常见错误和 AQA 评分标准提示。
1. What Example 7.4.3 Asks | 例题 7.4.3 的题目要求
+ 9x + 1. It asks you to find the coordinates of the stationary points and determine their nature. This is a standard AQA differentiation question worth between 6 and 8 marks.
题目给定曲线 y = x³ - 6x² + 9x + 1。要求求出驻点的坐标，并判断其性质。这是一道标准的 AQA 微分题，分值通常在 6 到 8 分。
ature of a stationary point tells us whether the curve has a local maximum, a local minimum, or a point of inflection.
驻点出现在曲线梯度为零的位置。从数学上说，就是解 dy/dx = 0。驻点的性质告诉我们曲线在该处是局部极大值、局部极小值还是拐点。
2. Step 1: Differentiate the Function | 第 1 步：对函数求导
9x + 1, differentiate term by term using the power rule. The derivative of axⁿ is naxⁿ⁻¹.
对于 y = x³ - 6x² + 9x + 1，使用幂函数求导法则逐项求导。axⁿ 的导数是 naxⁿ⁻¹。
dy/dx = 3x² - 12x + 9
of 3, so dy/dx = 3(x² - 4x + 3). Factorising the quadratic gives dy/dx = 3(x - 1)(x - 3).
通常将导数因式分解会更方便。这里所有系数都有公因数 3，所以 dy/dx = 3(x² - 4x + 3)。继续因式分解二次式，得到 dy/dx = 3(x - 1)(x - 3)。
3. Step 2: Solve dy/dx = 0 | 第 2 步：解 dy/dx = 0
Stationary points occur when the gradient is zero. Set dy/dx = 0 and solve for x.
驻点出现在梯度为零时。令 dy/dx = 0，然后解出 x。
3(x - 1)(x - 3) = 0 → x = 1 or x = 3
-coordinates of the stationary points. These are where the curve has a horizontal tangent.
根 x = 1 和 x = 3 就是驻点的 x 坐标。在这些位置，曲线具有水平切线。
4. Step 3: Find the y-coordinates | 第 3 步：求 y 坐标
quation y = x³ - 6x² + 9x + 1. Do not substitute into dy/dx, because that would give zero.
将每个 x 值代回原方程 y = x³ - 6x² + 9x + 1。不要代入 dy/dx，否则结果会是零。
For x = 1:
当 x = 1 时：
y = (1)³ - 6(1)² + 9(1) + 1 = 1 - 6 + 9 + 1 = 5
So the first stationary point is (1, 5).
因此第一个驻点是 (1, 5)。
For x = 3:
当 x = 3 时：
y = (3)³ - 6(3)² + 9(3) + 1 = 27 - 54 + 27 + 1 = 1
So the second stationary point is (3, 1).
因此第二个驻点是 (3, 1)。
5. Step 4: First Derivative Test | 第 4 步：一阶导数检验
The first derivative test examines the sign of dy/dx immediately to the left and right of each stationary point. If the gradient changes from positive to negative, the point is a local maximum. If it changes from negative to positive, the point is a local minimum.
一阶导数检验考察每个驻点左右两侧 dy/dx 的符号。如果梯度从正变为负，该点是局部极大值；如果从负变为正，该点是局部极小值。
tervals x < 1, 1 < x < 3, and x > 3. For example, use x = 0, x = 2, and x = 4.
在区间 x < 1、1 < x < 3 和 x > 3 内选取检验值。例如使用 x = 0、x = 2 和 x = 4。
- At x = 0: dy/dx = 3(0 - 1)(0 - 3) = 9 > 0, so the gradient is positive. 在 x = 0 处，dy/dx = 3(0 - 1)(0 - 3) = 9 > 0，因此梯度为正。
- At x = 2: dy/dx = 3(2 - 1)(2 - 3) = -3 < 0, so the gradient is negative. 在 x = 2 处，dy/dx = 3(2 - 1)(2 - 3) = -3 < 0，因此梯度为负。
- At x = 4: dy/dx = 3(4 - 1)(4 - 3) = 9 > 0, so the gradient is positive. 在 x = 4 处，dy/dx = 3(4 - 1)(4 - 3) = 9 > 0，因此梯度为正。
maximum. At x = 3, the gradient changes from negative to positive, so (3, 1) is a minimum.
在 x = 1 处，梯度从正变为负，所以 (1, 5) 是极大值点。在 x = 3 处，梯度从负变为正，所以 (3, 1) 是极小值点。
6. Step 5: Second Derivative Test | 第 5 步：二阶导数检验
The second derivative test is often quicker for classifying stationary points. Differentiate dy/dx to get d²y/dx². Then substitute each x-coordinate of the stationary point into d²y/dx².
二阶导数检验通常能更快地判断驻点性质。对 dy/dx 求导得到 d²y/dx²，然后将每个驻点的 x 坐标代入 d²y/dx²。
d²y/dx² = 6x - 12
For x = 1:
当 x = 1 时：
d²y/dx² = 6(1) - 12 = -6 < 0
derivative is negative, the curve is concave down at x = 1, so (1, 5) is a local maximum.
由于二阶导数为负，曲线在 x = 1 处向下凹，所以 (1, 5) 是局部极大值。
For x = 3:
当 x = 3 时：
d²y/dx² = 6(3) - 12 = 6 > 0
nd derivative is positive, the curve is concave up at x = 3, so (3, 1) is a local minimum.
由于二阶导数为正，曲线在 x = 3 处向上凹，所以 (3, 1) 是局部极小值。
7. Interpreting the Results | 结果解释
The stationary points are (1, 5) as a local maximum and (3, 1) as a local minimum. On the cubic curve, the graph rises to a peak at (1, 5), then falls to a trough at (3, 1).
驻点为 (1, 5) 的局部极大值和 (3, 1) 的局部极小值。在这条三次曲线上，图像上升至 (1, 5) 的峰值，然后下降至 (3, 1) 的谷值。
Because the coefficient of x³ is positive, the curve starts low on the left and ends high on the right. The maximum occurs before the minimum, which is typical for a positive cubic with two distinct stationary points.
由于 x³ 的系数为正，曲线左侧较低，右侧较高。极大值出现在极小值之前，这是具有两个不同驻点的正三次函数的典型特征。
8. AQA Mark Scheme Technique | AQA 评分标准技巧
dx = 3(x - 1)(x - 3) is useful because it shows factorisation and makes the roots obvious.
AQA 考官期望清晰的步骤分，包括求导、解 dy/dx = 0、代回原函数求 y、以及判断驻点性质。写出 dy/dx = 3(x - 1)(x - 3) 很有帮助，因为这样展示了因式分解，并且使根一目了然。
When using the second derivative test, state the sign of d²y/dx² and link it to a maximum or minimum. A common mark scheme phrase is 'd²y/dx² < 0, so maximum' or 'd²y/dx² > 0, so minimum'.
使用二阶导数检验时，要说明 d²y/dx² 的符号，并将其与极大值或极小值联系起来。常见的评分标准表述是 d²y/dx² < 0，所以为极大值，或 d²y/dx² > 0，所以为极小值。
9. Common Mistakes to Avoid | 常见错误提醒
One common error is substituting the x-value into dy/dx instead of the original y. This always gives zero and loses the final accuracy mark. Always return to the original function for the y-coordinate.
一个常见错误是把 x 值代入 dy/dx 而不是原函数 y。这样总是得到零，并丢失最后的精度分。求 y 坐标时一定要代回原函数。
Another mistake is ignoring the factorised form of dy/dx. Expanding and then using the quadratic formula is more work and increases the chance of arithmetic errors. Factorising is faster and clearer.
另一个错误是忽略 dy/dx 的因式分解形式。展开后再使用二次公式会增加计算量，也更容易出现算术错误。因式分解更快、更清晰。
st is inconclusive, and you must use the first derivative test or check higher derivatives.
第三个错误是认为二阶导数为零就一定表示拐点。在 A-level 中，如果 d²y/dx² = 0，二阶导数检验无法得出结论，必须使用一阶导数检验或检查更高阶导数。
10. Practice Variation for AQA | AQA 变式练习
dy/dx = 6x² - 18x + 12 = 6(x - 1)(x - 2). The stationary points occur at x = 1 and x = 2.
试用相同方法处理 y = 2x³ - 9x² + 12x - 4。求导得 dy/dx = 6x² - 18x + 12 = 6(x - 1)(x - 2)。驻点出现在 x = 1 和 x = 2。
= -6 < 0, so (1, 1) is a maximum. At x = 2, d²y/dx² = 6 > 0, so (2, 0) is a minimum.
代回原方程：x = 1 时，y = 2 - 9 + 12 - 4 = 1；x = 2 时，y = 16 - 36 + 24 - 4 = 0。二阶导数为 d²y/dx² = 12x - 18。x = 1 时，d²y/dx² = -6 < 0，所以 (1, 1) 是极大值；x = 2 时，d²y/dx² = 6 > 0，所以 (2, 0) 是极小值。
11. Why AQA Emphasises This Skill | 为什么 AQA 强调这一技能
Stationary point problems combine algebraic differentiation, solving equations, and curve sketching. They appear in short questions, longer context problems, and optimisation tasks. AQA uses them to test both fluency and mathematical communication.
驻点问题综合了代数求导、解方程和曲线草图。它们出现在短题、长情境题和优化问题中。AQA 通过这类题考查计算的流畅性和数学表达的规范性。
Practising Example 7.4.3 builds the foundation for optimisation problems, where you often model a real-life quantity, differentiate, and locate maximum or minimum values such as maximum area or minimum cost.
练习例题 7.4.3 可以为优化问题打下基础。在优化题中，你通常先对现实量建模，然后求导，并找到最大值或最小值，例如最大面积或最小成本。
12. Summary | 总结
ssify each point using the first or second derivative test. The answer is maximum at (1, 5) and minimum at (3, 1).
求解例题 7.4.3 的方法是：对三次函数求导得到 dy/dx，解 dy/dx = 0，求出 y 坐标，然后使用一阶或二阶导数检验判断驻点性质。答案为 (1, 5) 处取得极大值，(3, 1) 处取得极小值。
Showing clear factorisation and linking the sign of the second derivative to maximum or minimum are essential for full AQA marks.
展示清晰的因式分解，并将二阶导数的符号与极大值或极小值联系起来，是获得 AQA 满分的必要条件。
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