📚 Example 1.5.2: Finding Stationary Points and Their Nature | 例 1.5.2：求驻点及其性质
In this AQA A-Level Mathematics worked example, we solve a typical differentiation problem: finding the stationary points of a cubic curve and determining whether each is a local maximum or local minimum. This is Example 1.5.2, which brings together first derivatives, second derivatives, and sign inspection.
在这道 AQA A-Level 数学典型例题中，我们求解一个三次曲线的驻点，并判断每个驻点是局部极大值还是局部极小值。这是例 1.5.2，综合运用一阶导数、二阶导数与符号检验。
xam questions, where clear method marks are awarded for correct differentiation, solving f'(x) = 0, finding y-coordinates, and classifying the stationary points.
本题的完整解答与 AQA 纯数学考试题型相匹配，正确的求导、解 f'(x) = 0、求 y 坐标以及判断驻点性质都能获得清晰的方法分。
1. Understanding the Problem Statement | 理解题意
then classify each one as either a local maximum or a local minimum. A stationary point occurs where the first derivative f'(x) is zero.
我们给定三次函数 f(x) = x³ - 3x² + 2。任务是求曲线 y = f(x) 的所有驻点，并判断每个驻点是局部极大值还是局部极小值。驻点出现在一阶导数 f'(x) 等于零的地方。
The phrase "nature of a stationary point" means deciding whether the curve has a local maximum, a local minimum, or a point of inflection at that point. In this example, we will use both the second derivative test and a sign inspection of f'(x), so you can see how the two methods confirm each other.
"驻点的性质"这一说法，是指判断曲线在该点是局部极大值、局部极小值还是拐点。本例中，我们将同时使用二阶导数判别法和 f'(x) 的符号检验，以便看到两种方法如何相互验证。
2. Key Definitions: Stationary Points | 关键定义：驻点
f a differentiable function is a point on the curve where the gradient is zero, that is, f'(x) = 0. At such a point, the tangent to the curve is horizontal.
可微函数的驻点是曲线上梯度为零的点，即 f'(x) = 0。在这样的点上，曲线的切线是水平的。
There are three common types of stationary point: local maximum, local minimum, and stationary point of inflection. At a local maximum, the curve changes from increasing to decreasing. At a local minimum, it changes from decreasing to increasing. At a stationary point of inflection, the gradient does not change sign; the curve temporarily flattens but continues in the same overall direction.
常见驻点有三种：局部极大值、局部极小值和驻点拐点。在局部极大值处，曲线由递增变为递减；在局部极小值处，曲线由递减变为递增。在驻点拐点处，梯度不变号；曲线暂时变平，但总体方向保持不变。
- Local maximum: f'(x) = 0 and f''(x) < 0.
- Local minimum: f'(x) = 0 and f''(x) > 0.
- Possible inflection: f'(x) = 0 and f''(x) = 0, then check sign change of f'(x).
- 局部极大值：f'(x) = 0 且 f''(x) < 0。
- 局部极小值：f'(x) = 0 且 f''(x) > 0。
- 可能是拐点：f'(x) = 0 且 f''(x) = 0，此时需检查 f'(x) 是否变号。
3. Differentiating to Find f'(x) | 求一阶导数 f'(x)
m by term using the power rule. The derivative of x³ is 3x², the derivative of -3x² is -6x, and the derivative of the constant 2 is 0.
我们使用幂法则逐项求导 f(x) = x³ - 3x² + 2。x³ 的导数是 3x²，-3x² 的导数是 -6x，常数 2 的导数是 0。
f(x) = x³ - 3x² + 2
f'(x) = 3x² - 6x
This step carries method marks in the exam. Even if later algebra goes wrong, a correct derivative can still earn credit. Always write f'(x) clearly before setting it equal to zero.
这一步在考试中占方法分。即使后续代数出错，正确的导数仍可获得分数。在令其等于零之前，务必清楚写出 f'(x)。
4. Solving f'(x) = 0 | 解方程 f'(x) = 0
3x(x - 2) = 0. Hence x = 0 or x = 2. These are the x-coordinates of the stationary points.
令一阶导数等于零并解 x。我们有 3x² - 6x = 0。提取公因式 3x 得到 3x(x - 2) = 0。因此 x = 0 或 x = 2。这些就是驻点的 x 坐标。
3x² - 6x = 0
3x(x - 2) = 0 → x = 0 or x = 2
At this stage, we have only found the x-values. We must not stop here, because a stationary point is a point on the curve, so both coordinates are needed.
此时我们只找到了 x 值。不能就此停止，因为驻点是曲线上的点，所以需要两个坐标。
