📚 Example 3.6.1: Stationary Points | 示例3.6.1：驻点
In this article, we work through Example 3.6.1 from the AQA A-Level Mathematics specification, focusing on finding and classifying stationary points. This is a core skill in differentiation that appears frequently in Pure Mathematics exams.
本文我们将详细讲解 AQA A-Level 数学考纲中的例 3.6.1，主题是求驻点并判断其性质。这是微分运算中的核心技能，在纯数学考试中经常出现。
1. Problem Statement | 问题陈述
Consider the curve given by the equation y = x3 - 6x2 + 9x + 1. Find the coordinates of all stationary points and determine whether each one is a local maximum, a local minimum, or a point of inflection.
考虑曲线方程 y = x3 - 6x2 + 9x + 1。求所有驻点的坐标，并判断每个驻点是局部极大值、局部极小值还是拐点。
- We must use calculus to find the first derivative and set it equal to zero.
- 我们需要使用微积分求出一阶导数，并令其等于零。
2. First Derivative | 一阶导数
To locate stationary points, we differentiate the function with respect to x. Applying the power rule term by term gives:
为了找到驻点，我们需要对函数关于 x 求导。逐项应用幂法则得到：
f'(x) = 3x² - 12x + 9
Notice that each coefficient is divisible by 3, which will simplify the next step.
注意每一项的系数都能被 3 整除，这将简化下一步计算。
3. Setting f'(x) = 0 | 令 f'(x) = 0
Stationary points occur where the gradient is zero. Hence we solve:
驻点出现在梯度为零的地方。因此我们解方程：
3x² - 12x + 9 = 0
Divide both sides by 3 to obtain a simpler quadratic:
两边同时除以 3，得到一个更简单的二次方程：
x² - 4x + 3 = 0
This factorises neatly as (x - 1)(x - 3) = 0, so x = 1 or x = 3.
该式可以整洁地因式分解为 (x - 1)(x - 3) = 0，因此 x = 1 或 x = 3。
4. y-coordinates | 纵坐标
Substitute x = 1 into the original equation y = x³ - 6x² + 9x + 1:
将 x = 1 代入原方程 y = x³ - 6x² + 9x + 1：
y = 1 - 6 + 9 + 1 = 5
So the first stationary point is (1, 5).
因此第一个驻点是 (1, 5)。
Now substitute x = 3:
再代入 x = 3：
y = 27 - 54 + 27 + 1 = 1
Thus the second stationary point is (3, 1).
因此第二个驻点是 (3, 1)。
5. Second Derivative | 二阶导数
To classify these stationary points, we calculate the second derivative f''(x). Differentiating f'(x) = 3x² - 12x + 9 gives:
为了判断这些驻点的性质，我们需要计算二阶导数 f''(x)。对 f'(x) = 3x² - 12x + 9 求导得到：
f''(x) = 6x - 12
The sign of f''(x) at each stationary point tells us about its nature.
f''(x) 在每个驻点处的正负号将告诉我们该点的性质。
6. Classifying Using f''(x) | 用二阶导数判断
Evaluate f''(x) at x = 1:
在 x = 1 处求 f''(x)：
f''(1) = 6(1) - 12 = -6 < 0
Since f''(1) < 0, the point (1, 5) is a local maximum.
因为 f''(1) < 0，所以点 (1, 5) 是局部极大值。
Now evaluate at x = 3:
现在求 x = 3 处的值：
f''(3) = 6(3) - 12 = 6 > 0
Since f''(3) > 0, the point (3, 1) is a local minimum.
因为 f''(3) > 0，所以点 (3, 1) 是局部极小值。
7. Sign Test for Verification | 符号检验以作验证
hem. Choose values slightly less than and greater than x = 1, for example x = 0 and x = 2.
如果你不想使用二阶导数，也可以在每个驻点附近做符号检验来判断。例如在 x = 1 左右分别取 x = 0 和 x = 2。
- f'(0) = 9 > 0, so gradient is positive before x = 1.
- f'(0) = 9 > 0，说明 x = 1 之前梯度为正。
- f'(2) = 3(4) - 24 + 9 = -3 < 0, so gradient is negative after x = 1.
- f'(2) = 3(4) - 24 + 9 = -3 < 0，说明 x = 1 之后梯度为负。
A positive-to-negative change indicates a local maximum, confirming our earlier result.
由正变负说明该点是局部极大值，与之前的结果一致。
For x = 3, test x = 2 and x = 4:
对于 x = 3，测试 x = 2 和 x = 4：
- f'(2) = -3 < 0 before x = 3.
- f'(2) = -3 < 0，说明 x = 3 之前梯度为负。
- f'(4) = 48 - 48 + 9 = 9 > 0 after x = 3.
- f'(4) = 48 - 48 + 9 = 9 > 0，说明 x = 3 之后梯度为正。
A negative-to-positive change indicates a local minimum.
由负变正说明该点是局部极小值。
8. Curve Sketching | 曲线绘制
Combining the information we have, we can sketch the cubic curve. The y-intercept is at (0, 1), and the stationary points are (1, 5) and (3, 1). Since the leading coefficient is positive, the curve rises to the right and falls to the left.
综合已有信息，我们可以画出这条三次曲线的草图。y 截距为 (0, 1)，驻点为 (1, 5) 和 (3, 1)。由于最高次项系数为正，曲线向右上方延伸，向左下方延伸。
- At (1, 5) the curve reaches a local peak.
- 在 (1, 5) 处曲线达到局部峰值。
- At (3, 1) the curve reaches a local trough.
- 在 (3, 1) 处曲线达到局部谷值。
- The curve has no asymptotes and is continuous.
- 曲线没有渐近线且连续。
9. Common Mistakes | 常见错误
g to more complicated algebra. Another common error is substituting x = 1 and x = 3 into f'(x) instead of the original function when finding y-coordinates.
学生在解二次方程时常常忘记约去公因子，导致代数运算更复杂。另一个常见错误是求纵坐标时把 x = 1 和 x = 3 代入 f'(x) 而不是原函数。
- Always check that the y-coordinate comes from the original equation.
- 始终记住纵坐标要代入原方程求得。
- Remember that f''(x) = 0 does not always mean a point of inflection; the sign must also change.
- 记住 f''(x) = 0 并不一定就是拐点，还必须看符号是否改变。
10. Practice Question | 练习
ionary points of y = x³ - 3x² - 9x + 2 and classify them using the second derivative test.
请自己尝试这个相关题目：求 y = x³ - 3x² - 9x + 2 的驻点，并用二阶导数判别法判断它们的性质。
Answer: f'(x) = 3x² - 6x - 9 = 3(x - 3)(x + 1), so x = 3 or x = -1. f''(x) = 6x - 6, so f''(3) = 12 > 0 (local minimum at (3, -25)) and f''(-1) = -12 < 0 (local maximum at (-1, 7)).
答案：f'(x) = 3x² - 6x - 9 = 3(x - 3)(x + 1)，故 x = 3 或 x = -1。f''(x) = 6x - 6，所以 f''(3) = 12 > 0（在 (3, -25) 处为局部极小值），f''(-1) = -12 < 0（在 (-1, 7) 处为局部极大值）。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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