📚 Example 6.6.1: Stationary Points and the Second Derivative | 例题 6.6.1：驻点与二阶导数
In AQA A-Level Mathematics, Example 6.6.1 is a classic differentiation problem. It asks you to find the stationary points of a cubic function and determine their nature using the second derivative. Mastering this type of question is essential because it appears frequently in Paper 1 and requires careful setting out of derivatives, factorisation, substitution and interpretation.
在 AQA A-Level 数学中，例题 6.6.1 是一道经典的微分问题。它要求你求出一个三次函数的驻点，并用二阶导数判断其性质。掌握这类题型至关重要，因为它在试卷一中经常出现，并且要求你规范地写出求导、因式分解、代入和解释的过程。
1. The Problem Statement | 问题陈述
x + 4. The task is to find the coordinates of its stationary points and classify each one as a local maximum or a local minimum.
考虑三次曲线 y = x³ - 6x² + 9x + 4。任务是求出它的驻点坐标，并判断每个驻点是局部极大值点还是局部极小值点。
2. Step 1 - Differentiate the Function | 第一步：对函数求导
Differentiate y with respect to x. Using the power rule, the derivative of x³ is 3x², the derivative of -6x² is -12x, the derivative of 9x is 9, and the derivative of the constant 4 is 0.
对 y 关于 x 求导。使用幂法则，x³ 的导数是 3x²，-6x² 的导数是 -12x，9x 的导数是 9，常数 4 的导数是 0。
dy/dx = 3x² - 12x + 9
3. Step 2 - Set dy/dx = 0 and Solve | 第二步：令 dy/dx = 0 并求解
rise the quadratic by first taking out the common factor 3, then factorising the remaining quadratic into (x - 1)(x - 3).
驻点出现在梯度为零处，因此令 dy/dx = 0。先提取公因式 3，再将剩余二次式因式分解为 (x - 1)(x - 3)，从而完成因式分解。
3x² - 12x + 9 = 0
3(x² - 4x + 3) = 0
3(x - 1)(x - 3) = 0
Since the product is zero, either x - 1 = 0 or x - 3 = 0. This gives x = 1 and x = 3.
由于乘积为零，要么 x - 1 = 0，要么 x - 3 = 0。解得 x = 1 和 x = 3。
4. Step 3 - Find the y-Coordinates | 第三步：求出 y 坐标
s. For x = 1, y = (1)³ - 6(1)² + 9(1) + 4 = 8. For x = 3, y = (3)³ - 6(3)² + 9(3) + 4 = 4.
将每个 x 值代回原方程，求出对应的 y 坐标。当 x = 1 时，y = 1 - 6 + 9 + 4 = 8。当 x = 3 时，y = 27 - 54 + 27 + 4 = 4。
The stationary points are therefore (1, 8) and (3, 4).
因此驻点坐标是 (1, 8) 和 (3, 4)。
5. Step 4 - Classify Using the Second Derivative | 第四步：使用二阶导数判断驻点性质
Differentiate dy/dx to get the second derivative. The derivative of 3x² is 6x, the derivative of -12x is -12, and the derivative of 9 is 0.
对 dy/dx 再求导，得到二阶导数。3x² 的导数是 6x，-12x 的导数是 -12，9 的导数是 0。
d²y/dx² = 6x - 12
= 3, d²y/dx² = 6(3) - 12 = 6. Since this is positive, the point (3, 4) is a local minimum.
在每个驻点处计算二阶导数的值。当 x = 1 时，d²y/dx² = 6 - 12 = -6。因为结果为负，所以 (1, 8) 是局部极大值点。当 x = 3 时，d²y/dx² = 18 - 12 = 6。因为结果为正，所以 (3, 4) 是局部极小值点。
6. Step 5 - Interpret the Results Graphically | 第五步：从图形角度解释结果
On the cubic curve, a local maximum at (1, 8) means the curve rises to a peak then begins to fall. A local minimum at (3, 4) means the curve falls to a trough then rises again. The graph has a typical cubic S-shape with two turning points.
在三次曲线上，(1, 8) 处的极大值意味着曲线上升到最高点后开始下降。(3, 4) 处的极小值意味着曲线下降到最低点后再次上升。图像具有典型的三次 S 形，有两个转折点。
g coefficient of x³. As x tends to -∞, y tends to -∞, and as x tends to +∞, y tends to +∞.
注意到极大值点的 x 值小于极小值点的 x 值，这与 x³ 的正首项系数一致。当 x 趋向 -∞ 时，y 趋向 -∞；当 x 趋向 +∞ 时，y 趋向 +∞。
7. Common Mistakes to Avoid | 需要避免的常见错误
Many students forget to differentiate the constant term or make sign errors in the power rule. Always write out each term separately before simplifying.
许多学生忘记对常数项求导，或在幂法则中出现符号错误。在化简之前，务必逐项写出每个导数。
Another common mistake is to find only the x-values and stop. AQA exam questions often ask for coordinates, so you must substitute back into the original equation to obtain y-values.
另一个常见错误是只求出 x 值就停止。AQA 考题通常要求坐标，因此你必须代回原方程求出 y 值。
In classification, some candidates mix up the sign test. Remember: a negative second derivative means a maximum, and a positive second derivative means a minimum.
在分类时，一些考生会混淆符号判定。记住：二阶导数为负意味着极大值，二阶导数为正意味着极小值。
8. Exam Technique for AQA | AQA 考试技巧
When answering a stationary point question under timed conditions, set out your work in clear steps: differentiate, set to zero, solve, substitute, classify. Use dy/dx and d²y/dx² notation exactly as shown.
在限时条件下回答驻点问题时，应按照清晰的步骤书写：求导、令导数为零、求解、代入、判定。使用与示例完全一致的 dy/dx 和 d²y/dx² 记号。
lgebraic manipulation has errors. Always show your factorisation, as this is frequently a specific mark.
AQA 的评分方案通常会给正确的求导和令 dy/dx = 0 步骤方法分，即使后面的代数操作有误。务必将因式分解过程写出来，因为这通常是一个独立的得分点。
If the quadratic does not factorise neatly, use the quadratic formula. However, in this example the factorisation is straightforward, so using the formula is unnecessary.
如果二次式不能整齐地因式分解，可以使用求根公式。不过在本例中，因式分解很简单，因此无需使用求根公式。
9. Alternative Classification Using the First Derivative Test | 使用一阶导数检验的替代方法
-3 (negative). Since the gradient changes from positive to negative, (1, 8) is a maximum.
有些 AQA 问题可能要求你在不使用二阶导数的情况下判断驻点性质。此时，在每个 x 值两侧选取测试值并检查 dy/dx 的符号。对于 x = 1，选择 x = 0 和 x = 2。x = 0 时 dy/dx 为 9（正），x = 2 时为 3(2-1)(2-3) = -3（负）。由于梯度由正变负，所以 (1, 8) 是极大值点。
For x = 3, choose x = 2 and x =
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图