📚 Worked Example 6.4.2: Finding and Classifying Stationary Points | 例题 6.4.2：求驻点并判断其性质
In this AQA A-Level Mathematics worked example, we use differentiation to locate stationary points on a cubic curve and classify them using the second derivative test. This technique is central to the Year 1 and Year 2 pure specification and often appears in applied optimisation problems.
在这道 AQA A-Level 数学例题中，我们利用导数求三次曲线的驻点，并用二阶导数判断其性质。这一技巧是纯数 Year 1 和 Year 2 的核心内容，也经常出现在应用优化问题中。
1. The Worked Problem | 例题题目
ints on the curve y = x³ - 6x² + 9x + 2 and determine the nature of each stationary point.
求曲线 y = x³ - 6x² + 9x + 2 上驻点的坐标，并判断每个驻点的性质。
2. Step 1 - Differentiate | 第一步：求导
Differentiate term by term. The derivative of x³ is 3x², the derivative of -6x² is -12x, the derivative of 9x is 9, and the derivative of 2 is 0.
逐项求导。x³ 的导数是 3x²，-6x² 的导数是 -12x，9x 的导数是 9，常数 2 的导数是 0。
f'(x) = 3x² - 12x + 9
3. Step 2 - Stationary Points | 第二步：驻点条件
Stationary points occur when the gradient is zero, so set f' ading coefficient can have two stationary points, one local maximum and one local minimum.
驻点出现在梯度为零的位置，因此令 f'(x) = 0。首项系数为正的三次函数可能有两个驻点：一个局部极大值和一个局部极小值。
3x² - 12x + 9 = 0
4. Step 3 - Solve for x | 第三步：解方程求 x
e quadratic. The solutions x = 1 and x = 3 are the x-coordinates of the stationary points.
两边先除以 3 化简，再对二次式因式分解。解 x = 1 和 x = 3 就是驻点的 x 坐标。
x² - 4x + 3 = 0 ⇒ (x - 1)(x - 3) = 0 ⇒ x = 1, x = 3
5. Step 4 - Find y-coordinates | 第四步：求 y 坐标
quation. This gives f(1) = 6 and f(3) = 2, so the stationary points are (1, 6) and (3, 2).
将 x = 1 和 x = 3 代入原函数。得到 f(1) = 6、f(3) = 2，因此驻点为 (1, 6) 和 (3, 2)。
f(1) = 1³ - 6(1)² + 9(1) + 2 = 1 - 6 + 9 + 2 = 6
f(3) = 3³ - 6(3)² + 9(3) + 2 = 27 - 54 + 27 + 2 = 2
6. Step 5 - Second Derivative | 第五步：求二阶导数
Differentiate f'(x) to obtain f''(x). The second derivative measures the rate of change of the gradient and tells us whether the curve is concave up or concave down at a point.
对 f'(x) 再求导得到 f''(x)。二阶导数度量梯度的变化率，它告诉我们曲线在某点是凹向上还是凹向下。
f''(x) = 6x - 12
7. Step 6 - Classify the Points | 第六步：判断驻点性质
Evaluate f''(x) at each stationary point. If f''(a) < 0, the point is a local maximum; if f''(a) > 0, it is a local minimum. Here f''(1) = -6, so (1, 6) is a maximum, and f''(3) = 6, so (3, 2) is a minimum.
在每个驻点处代入二阶导数。若 f''(a) < 0，该点为局部极大值；若 f''(a) > 0，该点为局部极小值。此处 f''(1) = -6，所以 (1, 6) 是极大值点；f''(3) = 6，所以 (3, 2) 是极小值点。
| Stationary point | f''(x) | Sign | Nature |
|---|---|---|---|
| (1, 6) | -6 | Negative | Local maximum |
| (3, 2) | 6 | Positive | Local minimum |
8. Step 7 - Sketch the Curve | 第七步：曲线草图
-∞, y → -∞; as x → ∞, y → ∞. The curve rises to (1, 6), falls to (3, 2), then rises again.
标出两个驻点，并注意三次函数两端的行为：当 x → -∞ 时 y → -∞；当 x → ∞ 时 y → ∞。曲线先上升到 (1, 6)，再下降到 (3, 2)，然后再次上升。
9. Common Mistakes | 常见错误
Avoid these typical errors: forgetting to divide by the common factor before factorising; substituting into f'(x) instead of the original function for y; sign errors in the second derivative test; and stopping after finding x-values without coordinates.
避免这些典型错误：因式分解前忘记除以公因子；求 y 坐标时误代入 f'(x) 而不是原函数；二阶导数检验时符号出错；以及只求出 x 值就停止而未给出完整坐标。
- Always simplify f'(x) = 0 before factorising.
- 在因式分解前先化简 f'(x) = 0。
- Use the original y = f(x), not f'(x), to find y-coordinates.
- 求 y 坐标时使用原函数 y = f(x)，而不是 f'(x)。
- State clearly whether each point is a maximum or minimum.
- 明确说明每个驻点是极大值还是极小值。
10. Exam Technique | 考试技巧
In an exam, always show the equation f' th y-coordinate calculations, and a clear second derivative conclusion. Use a table if it helps the examiner see your classification quickly.
考试中一定要展示 f'(x) = 0、因式分解后的二次式、两个 y 坐标的计算过程，以及清晰的二阶导数结论。用表格可以帮助阅卷老师快速看清你的分类。
11. Practice Question | 练习
x² + 12x - 4 and determine their nature. Complete the working and check your answer below.
试做类似题目：求 y = 2x³ - 9x² + 12x - 4 的驻点并判断其性质。完成以下步骤并核对答案。
f'(x) = 6x² - 18x + 12 = 6(x² - 3x + 2) = 6(x - 1)(x - 2)
Therefore x = 1 and x = 2. The stationary points are (1, 1) and (2, 0).
因此 x = 1 和 x = 2。驻点为 (1, 1) 和 (2, 0)。
f''(x) = 12x - 18
At x = 1, f''(1) = -6 < 0, so (1, 1) is a local maximum. At x = 2, f''(2) = 6 > 0, so (2, 0) is a local minimum.
在 x = 1 处，f''(1) = -6 < 0，因此 (1, 1) 是局部极大值。在 x = 2 处，f''(2) = 6 >
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