📚 Example 6.11.1: Stationary Points and Second Derivative Test | 例题 6.11.1：驻点与二阶导数判别法
x² - 9x + 5, find its stationary points, and use the second derivative to determine whether each point is a local maximum or a local minimum.
在这个 AQA A-Level 数学例题中，我们研究曲线 y = x³ - 3x² - 9x + 5，求它的驻点，并利用二阶导数判断每个驻点是局部最大值还是局部最小值。
1. What Are Stationary Points? | 什么是驻点？
to the curve is horizontal. For a function y = f(x), this means dy/dx = 0 at that x-value.
曲线上的驻点是指梯度为零的点，因此曲线在该点的切线是水平的。对于函数 y = f(x)，这意味着在该 x 值处 dy/dx = 0。
There are three main types of stationary points: local maximum points, local minimum points, and points of inflection with zero gradient. The second derivative, d²y/dx², helps us distinguish between these types when the first derivative test is less direct.
驻点主要有三种类型：局部最大值点、局部最小值点以及梯度为零的拐点。二阶导数 d²y/dx² 帮助我们区分这些类型，尤其是在一阶导数判别法不够直接时。
2. The Worked Example: Setting the Scene | 例题引入
pe of polynomial has degree 3, so its gradient function is a quadratic and can have up to two real roots, giving up to two stationary points.
在例题 6.11.1 中，我们给出三次函数 y = x³ - 3x² - 9x + 5。这种多项式次数为 3，因此它的梯度函数是二次函数，最多有两个实根，从而最多有两个驻点。
We will find where the gradient is zero, calculate the coordinates of those points, and then classify each one. The second derivative test is especially efficient here because the second derivative is linear.
我们将找到梯度为零的位置，计算这些点的坐标，然后对每个点进行分类。这里使用二阶导数判别法特别高效，因为二阶导数是一次函数。
3. Step 1 - Differentiate the Function | 第一步：求导
x is found by differentiating each term separately. For y = x³ - 3x² - 9x + 5, we obtain:
先对每一项分别求导，得到 y 关于 x 的一阶导数。对于 y = x³ - 3x² - 9x + 5，我们得到：
dy/dx = 3x² - 6x - 9
The constant term 5 differentiates to 0. This derivative represents the gradient of the curve at any point x.
常数项 5 求导后为 0。这个导数表示曲线在任意点 x 处的梯度。
4. Step 2 - Solve dy/dx = 0 | 第二步：解方程 dy/dx = 0
ivative equal to zero and solve for x. This gives the quadratic equation 3x² - 6x - 9 = 0.
为了确定驻点的位置，令一阶导数等于零并解出 x。这样得到二次方程 3x² - 6x - 9 = 0。
es it to x² - 2x - 3 = 0, and factorising gives (x - 3)(x + 1) = 0. Hence x = 3 or x = -1.
两边同时除以 3，化简为 x² - 2x - 3 = 0，因式分解得到 (x - 3)(x + 1) = 0。因此 x = 3 或 x = -1。
5. Step 3 - Find the y-Coordinates | 第三步：求 y 坐标
k into the original equation y = x³ - 3x² - 9x + 5 to find the corresponding y-coordinate.
将每个 x 值代回原方程 y = x³ - 3x² - 9x + 5，求出对应的 y 坐标。
For x = 3: y = 27 - 27 - 27 + 5 = -22. For x = -1: y = -1 - 3 + 9 + 5 = 10.
当 x = 3 时：y = 27 - 27 - 27 + 5 = -22。当 x = -1 时：y = -1 - 3 + 9 + 5 = 10。
Therefore the stationary points are (3, -22) and (-1, 10).
因此驻点为 (3, -22) 和 (-1, 10)。
6. Step 4 - Use the Second Derivative Test | 第四步：使用二阶导数判别法
Differentiate dy/dx = 3x² - 6x - 9 to obtain the second derivative: d²y/dx² = 6x - 6.
对 dy/dx = 3x² - 6x - 9 求导，得到二阶导数：d²y/dx² = 6x - 6。
Evaluate the second derivative at each x-value. If d²y/dx² > 0, the point is a local minimum; if d²y/dx² < 0, the point is a local maximum.
在每个 x 值处计算二阶导数。如果 d²y/dx² > 0，该点是局部最小值；如果 d²y/dx² < 0，该点是局部最大值。
-22) is a local minimum. At x = -1, d²y/dx² = -12 < 0, so (-1, 10) is a local maximum.
在 x = 3 处，d²y/dx² = 12 > 0，所以 (3, -22) 是局部最小值。在 x = -1 处，d²y/dx² = -12 < 0，所以 (-1, 10) 是局部最大值。
7. Step 5 - State the Nature of Each Point | 第五步：说明每个驻点性质
We can now give a complete answer: the curve has a local maximum at (-1, 10) and a local minimum at (3, -22).
现在我们可以给出完整答案：曲线在 (-1, 10) 处有局部最大值，在 (3, -22) 处有局部最小值。
It is important to state the coordinates and use the word 'local' because cubic curves do not have a global maximum or minimum over all real values.
重要的是要给出坐标并使用 "局部" 一词，因为三次曲线在所有实数上不存在全局最大值或最小值。
8. Quick Summary Table | 小结表
The table below summarises the calculations in this example. Use it as a checklist in the exam.
下表总结了本例的计算过程。考试时可以使用它作为核对清单。
| x-value | x 值 | Stationary point | 驻点 | d²y/dx² | 二阶导数 | Nature | 性质 | Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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