📚 Example 6.10.2: Classifying Stationary Points with the Second Derivative | 例题 6.10.2：用二阶导数判断驻点性质
Stationary points are a central part of the AQA A-Level Mathematics specification, especially within the topic of differentiation. In Example 6.10.2, we are asked to locate the stationary points of a cubic function and determine whether each one is a local maximum or a local minimum by using the second derivative test. This article works through a complete model solution, explains the reasoning at each step, and highlights the common errors that can cost marks in an exam. The method shown here is directly transferable to other functions, including rational, trigonometric and exponential forms.
驻点是 AQA A-Level 数学考纲中微分部分的重点内容。在例题 6.10.2 中，我们需要求出一个三次函数的驻点，并利用二阶导数判断每个驻点是局部极大值还是局部极小值。本文给出完整的标准解题过程，逐步解释每一步的推理，并指出考试中容易失分的常见错误。这里展示的方法可以直接迁移到其他函数，包括分式函数、三角函数和指数函数。
1. Why Stationary Points Matter | 为什么驻点很重要
A stationary point is a point on a curve at which the gradient is zero. At such a point the tangent is horizontal, and the curve is momentarily neither rising nor falling. Stationary points are therefore the points at which a curve changes from increasing to decreasing or from decreasing to increasing.
驻点是曲线上切线斜率为零的点。在这样的点上切线是水平的，曲线瞬间既不上升也不下降。因此，驻点正是曲线由递增转为递减、或由递减转为递增的位置。
dy/dx = 0 at every stationary point
In AQA A-Level Mathematics you must be able to locate stationary points and decide whether each one is a local maximum, a local minimum, or a stationary point of inflection. The word 'local' matters: a local maximum is only the highest point in its own neighbourhood, not necessarily the highest point of the whole curve.
在 AQA A-Level 数学中，你必须能够求出驻点，并判断每个驻点是局部极大值、局部极小值，还是驻点拐点。'局部'二字很关键：局部极大值只是在它附近的小范围内最高，并不一定是整条曲线的最高点。
This skill is not examined in isolation. It underpins curve sketching, optimisation problems, and questions about tangents and normals, so a secure method pays off across the whole paper.
这项技能不会孤立地考查。它是曲线作图、最优化问题以及切线与法线问题的基础，因此掌握一套稳妥的方法会让整张试卷都受益。
2. The Question in Example 6.10.2 | 例题 6.10.2 的题目
e coordinates of each stationary point, together with a statement of whether it is a local maximum or a local minimum.
例题 6.10.2 给出三次函数 f(x) = x³ − 6x² + 9x − 2，要求求出每个驻点的坐标，并说明它是局部极大值还是局部极小值。
The plan is always the same four steps: differentiate once and solve f'(x) = 0 to find the x-coordinates; differentiate again to obtain f''(x); substitute each x-coordinate into f'' ponding y-coordinates so that the answer gives full coordinates rather than bare x-values.
解题计划始终是四步：先求一阶导数并解 f'(x) = 0，得到驻点的横坐标；再求二阶导数 f''(x)；把每个横坐标代入 f''(x)，根据符号判断性质；最后求出对应的纵坐标，使答案给出完整坐标而不是只有 x 值。
3. Step 1: Differentiate and Solve f'(x) = 0 | 第一步：求一阶导数并令其为零
Differentiate the function term by term. The derivative of x³ is 3x², the derivative of −6x² is −12x, the derivative of 9x is 9, and the constant term −2 differentiates to zero.
逐项求导。x³ 的导数是 3x²，−6x² 的导数是 −12x，9x 的导数是 9，常数项 −2 的导数为零。
f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3)
Setting f' = 1 or x = 3. Both values lie in the domain of f, so both give genuine stationary points.
令 f'(x) = 0，得 3(x − 1)(x − 3) = 0，所以 x = 1 或 x = 3。两个值都在函数定义域内，因此都对应真正的驻点。
If the quadratic does not factorise neatly, use the quadratic formula x = (−b ± √b² − 4ac) / 2a. If the discriminant b² − 4ac is negative, the equation has no real roots and the curve has no stationary points at all; you should state this explicitly rather than leaving the answer blank.
如果二次式不能整洁地因式分解，就使用求根公式 x = (−b ± √b² − 4ac) / 2a。若判别式 b² − 4ac 为负，方程没有实根，曲线根本没有驻点；此时应明确写出这一结论，而不是留空。
4. Step 2: The Second Derivative
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