📚 Example 6.4.3: Classifying Stationary Points with the Second Derivative | 例题 6.4.3：用二阶导数判断驻点性质
In AQA A Level Mathematics, Chapter 6 on differentiation brings together first and second derivatives through worked examples. Example 6.4.3 is a standard exam-style question: you are given a cubic function and asked to locate its stationary points, then classify each one as a maximum or minimum. This article works through the complete method, explains the second derivative test, and highlights the details examiners expect to see in your written solution.
在 AQA A Level 数学中，第六章微分通过例题将一阶导数和二阶导数结合起来。例题 6.4.3 是一道标准的考试题型：给出一个三次函数，要求找出驻点，并判断每个驻点是极大值还是极小值。本文将完整讲解解题方法、解释二阶导数判别法，并强调考官希望在你的书面解答中看到的细节。
1. The Problem Statement | 题目重述
Find the coordinates of the stationary points on C and determine the nature of each stationary point.
曲线 C 的方程为 y = x³ − 3x² + 2。求曲线 C 上驻点的坐标，并判断每个驻点的性质。
y = x³ − 3x² + 2
This type of question usually appears in the pure mathematics component of AQA papers. It tests differentiation, solving polynomial equations, substitution, and interpretation of the second derivative.
这类题目通常出现在 AQA 试卷的纯数学部分。它考查微分、多项式方程求解、代入运算以及对二阶导数的理解。
2. What Is a Stationary Point? | 什么是驻点？
he gradient of the curve is zero. If f(x) is the function, the stationary points satisfy f'(x) = 0. At these points the tangent to the curve is horizontal.
驻点出现在曲线的梯度为零的位置。若函数为 f(x)，则驻点满足 f'(x) = 0。在这些点处，曲线的切线是水平的。
Stationary points can be local maxima, local minima, or points of inflection with zero gradient. The second derivative test helps us distinguish between a maximum and a minimum when the derivative has two distinct real roots.
驻点可能是局部极大值、局部极小值，或梯度为零的拐点。当导数有两个不同实根时，二阶导数判别法可以帮助我们区分极大值和极小值。
For a cubic curve such as the one in this example, there are usually two stationary points: one maximum and one minimum, provided the derivative has two distinct real roots. This is exactly the case here.
对于本题这样的三次曲线，通常有两个驻点：一个极大值和一个极小值，只要导数有两个不同的实根。本题恰好就是这种情况。
3. Step 1: Differentiate to Find f'(x) | 第一步：求一阶导数
Differentiate the given function term by term. The derivative of x³ is 3x², the derivative of −3x² is −6x, and the derivative of the constant 2 is 0.
对给定函数逐项求导。x³ 的导数是 3x²，−3x² 的导数是 −6x，常数 2 的导数是 0。
f'(x) = 3x² − 6x
Always write f'(x) clearly before solving. This step carries method marks, even if a later substitution error occurs.
在求解之前务必清楚地写出 f'(x)。即使后续代入出现错误，这一步也能获得方法分。
4. Step 2: Solve f'(x) = 0 | 第二步：令一阶导数为零
Set the first derivative equal to zero and solve for x. Factorising is usually the quickest method for a polynomial derivative, but you may also use the quadratic formula if necessary.
令一阶导数等于零并解出 x。对于多项式导数，因式分解通常是最快的方法，但如有必要，也可以使用二次公式。
3x² − 6x = 0
3x(x − 2) = 0
ives two solutions: x = 0 and x = 2. These are the x-coordinates of the stationary points.
这给出两个解：x = 0 和 x = 2。它们就是驻点的横坐标。
nswers. The question asks for coordinates, so you must find the corresponding y-values next.
不要到此就停下。一个常见错误是以为 x = 0 和 x = 2 就是最终答案。题目要求的是坐标，因此接下来还必须求出对应的 y 值。
5. Step 3: Find the y-Coordinates | 第三步：求纵坐标
to the original equation y = x³ − 3x² + 2. Never substitute into the derivative, because f'(x) only gives gradient and not the y-coordinate on the curve.
将每个 x 值代回原方程 y = x³ − 3x² + 2。切勿代入导数，因为 f'(x) 只表示梯度，而不表示曲线上的 y 坐标。
For x = 0:
