📚 Example 6.9.1: Classifying Stationary Points with the Second Derivative Test | 例题 6.9.1：用二阶导数检验判定驻点性质
In this worked example, we solve a typical AQA A-level Mathematics problem involving stationary points, monotonic intervals and the second derivative test. The method is presented step by step so that you can follow the exact reasoning expected in an exam. The function chosen is a cubic polynomial, which is a very common type in AQA papers.
在本例题中，我们将解答一道典型的 AQA A-level 数学题，涉及驻点、单调区间和二阶导数检验。我们逐步展示方法，以便你能够掌握考试中要求呈现的推理过程。所选函数是一个三次多项式，这是 AQA 试卷中非常常见的类型。
1. The Question | 题目
x − 5. Find the coordinates of the stationary points on the curve and determine the nature of each stationary point.
曲线方程为 y = 2x³ − 9x² + 12x − 5。求曲线上驻点的坐标，并判断每个驻点的性质。
Also state the intervals on which the function is increasing or decreasing.
同时说明函数在哪些区间上递增或递减。
2. Key Concepts | 关键概念
ative is zero: dy/dx = 0. This means the tangent to the curve is horizontal at that point.
驻点出现在一阶导数为零的位置：dy/dx = 0。这意味着曲线在该点的切线是水平的。
The second derivative test says that if d²y/dx² < 0 at a stationary point, it is a local maximum; if d²y/dx² > 0, it is a local minimum.
二阶导数检验：若在驻点处 d²y/dx² < 0，则该点为局部最大值；若 d²y/dx² > 0，则该点为局部最小值。
If the second derivative is zero, the test is inconclusive and you should use a sign table for the first derivative instead.
如果二阶导数为零，该检验不能得出结论，此时应使用一阶导数的符号表进行判断。
For a cubic function of this form, we usually expect two stationary points: one local maximum and one local minimum.
对于这种形式的三次函数，我们通常预期会有两个驻点：一个局部最大值点和一个局部最小值点。
3. Step 1: Differentiate the Function | 第一步：求导数
We differentiate y = 2x³ − 9x² + 12x − 5 term by term using the power rule.
我们使用幂法则逐项对 y = 2x³ − 9x² + 12x − 5 求导。
For each term, multiply by the power and then reduce the power by one.
对每一项来说，先乘以指数，再将指数减一。
dy/dx = 6x² − 18x + 12
derivative is a quadratic, so solving dy/dx = 0 will give two possible stationary points.
该导数是一个二次函数，因此解 dy/dx = 0 会得到两个可能的驻点。
4. Step 2: Solve dy/dx = 0 | 第二步：解方程 dy/dx = 0
Set 6x² − 18x + 12 = 0 and factorise.
令 6x² − 18x + 12 = 0，并进行因式分解。
6(x² − 3x + 2) = 0
6(x − 1)(x − 2) = 0
Therefore x = 1 or x = 2.
因此 x = 1 或 x = 2。
If the quadratic did not factorise, you could use the quadratic formula, but AQA questions often include factorisable quadratics to keep the arithmetic manageable.
如果该二次式无法因式分解，你也可以使用求根公式，但 AQA 的题目通常会包含可因式分解的二次式，以便控制计算量。
5. Step 3: Find the y-Coordinates | 第三步：求 y 坐标
Substitute each x-value into the original equation y = 2x³ − 9x² + 12x − 5.
将每个 x 值代入原方程 y = 2x³ − 9x² + 12x − 5。
For x = 1: y = 2(1)³ − 9(1)² + 12(1) − 5 = 2 − 9 + 12 − 5 = 0.
当 x = 1 时：y = 2(1)³ − 9(1)² + 12(1) − 5 = 2 − 9 + 12 − 5 = 0。
For x = 2: y = 2(2)³ − 9(2)² + 12(2) − 5 = 16 − 36 + 24 − 5 = −1.
当 x = 2 时：y = 2(2)³ − 9(2)² + 12(2) − 5 = 16 − 36 + 24 − 5 = −1。
The stationary points are (1, 0) and (2, −1).
驻点为 (1, 0) 和 (2, −1)。
6. Step 4: Compute the Second Derivative | 第四步：求二阶导数
Differentiate dy/dx = 6x² − 18x + 12 to obtain d²y/dx².
对 dy/dx = 6x² − 18x + 12 求导，得到 d²y/dx²。
d²y/dx² = 12x − 18
Since d²y/dx² is linear, it is easy to evaluate at each stationary point.
由于 d²y/dx² 是线性的，因此可以很容易地求出它在每个驻点处的值。
7. Step 5: Apply the Second Derivative Test | 第五步：应用二阶导数检验
At x = 1, d²y/dx² = 12(1) − 18 = −6.
在 x = 1 处，d²y/dx² = 12(1) − 18 = −6。
Because d²y/dx² < 0, the stationary point (1, 0) is a local maximum. The curve is concave down here.
因为 d²y/dx² < 0，所以驻点 (1, 0) 是局部最大值点。此处曲线
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