📚 Stationary Points | 驻点与极值判定
In A-Level Pure Mathematics, stationary points are key features of a curve. They occur where the gradient of a function is momentarily zero, so the tangent to the curve is horizontal. Examiners expect you to find stationary points, classify them as local maxima, local minima or points of inflection, and apply these ideas to optimisation problems.
在 A-Level 纯数学中，驻点是曲线的关键特征。它们出现在函数梯度瞬时为零的位置，因此曲线在该点的切线是水平的。考官希望你能够求出驻点，将其分类为局部最大值、局部最小值或拐点，并把这些思想应用到优化问题中。
1. What Is a Stationary Point? | 什么是驻点？
A stationary point on a curve y = f(x) is a point where the first derivative is zero: dy/dx = 0. At this point, the function is neither increasing nor decreasing at that instant, so the graph has a horizontal tangent.
曲线 y = f(x) 上的驻点是一阶导数为零的点：dy/dx = 0。在该点，函数在那一瞬间既不增加也不减少，因此图像具有水平切线。
There are three main types of stationary points: a local maximum, a local minimum, and a stationary point of inflection. The type depends on how the gradient changes as x passes through the stationary point.
驻点主要有三种类型：局部最大值、局部最小值和驻点拐点。具体类型取决于当 x 经过驻点时梯度的变化方式。
2. Finding Stationary Points | 求驻点
To find stationary points, follow a standard procedure. First differentiate the function to obtain dy/dx. Then set dy/dx = 0 and solve the resulting equation for x. Finally, substitute each x-value back into the original function y = f(x) to find the corresponding y-coordinate.
求驻点需要遵循标准步骤。首先对函数求导得到 dy/dx。然后令 dy/dx = 0，并求解所得方程得到 x 值。最后将每个 x 值代回原函数 y = f(x)，求出对应的 y 坐标。
Always give the full coordinates of a stationary point unless the question asks only for x-values. For example, if y = x² − 4x + 1, then dy/dx = 2x − 4. Setting 2x − 4 = 0 gives x = 2, so y = 2² − 4(2) + 1 = −3. The stationary point is (2, −3).
除非题目只要求 x 值，否则始终要写出驻点的完整坐标。例如，若 y = x² − 4x + 1，则 dy/dx = 2x − 4。令 2x − 4 = 0 得 x = 2，所以 y = 2² − 4(2) + 1 = −3。驻点为 (2, −3)。
3. First Derivative Test | 一阶导数判别法
The first derivative test classifies a stationary point by examining the sign of dy/dx on either side of the point. Choose test values slightly smaller and slightly larger than the stationary x-value, and substitute them into dy/dx.
一阶导数判别法通过检查驻点两侧 dy/dx 的符号来分类驻点。选择略小于和略大于驻点 x 值的测试值，并将它们代入 dy/dx。
If dy/dx changes from positive to negative as x increases, the stationary point is a local maximum. If dy/dx changes from negative to positive, it is a local minimum. If the sign does not change, the point is a stationary point of inflection.
如果随着 x 增大，dy/dx 由正变负，则驻点为局部最大值。如果 dy/dx 由负变正，则为局部最小值。如果符号不变，则该点是驻点拐点。
| Left of stationary point | At stationary point | Right of stationary point | Classification |
|---|---|---|---|
| dy/dx > 0 | dy/dx = 0 | dy/dx < 0 | Local maximum |
| dy/dx < 0 | dy/dx = 0 | dy/dx > 0 | Local minimum |
| dy/dx > 0 | dy/dx = 0 | dy/dx > 0 | Stationary inflection |
| dy/dx < 0 | dy/dx = 0 | dy/dx < 0 | Stationary inflection |
This method is especially useful when the second derivative is zero or difficult to evaluate.
当二阶导数为零或难以计算时，这种方法尤其有用。
4. Second Derivative Test | 二阶导数判别法
The second derivative test is often quicker. After finding a stationary point at x = a, compute the second derivative d²y/dx² and evaluate it at x = a.
二阶导数判别法通常更快。求出 x = a 处的驻点后，计算二阶导数 d²y/dx²，并在 x = a 处求值。
If d²y/dx² at a is less than zero, the point is a local maximum. If d²y/dx² at a is greater than zero, the point is a local minimum. If d²y/dx² at a equals zero, the test is inconclusive and you should use the first derivative test.
如果 a 处的 d²y/dx² 小于零，则该点为局部最大值。如果 a 处的 d²y/dx² 大于零，则该点为局部最小值。如果 a 处的 d²y/dx² 等于零，则判别法无法得出结论，此时应使用一阶导数判别法。
d²y/dx² < 0 ⇒ local maximum; d²y/dx² > 0 ⇒ local minimum
d²y/dx² < 0 ⇒ 局部最大值；d²y/dx² > 0 ⇒ 局部最小值
5. Local Maxima and Local Minima | 局部最大值与局部最小值
A local maximum is a point where the function value is greater than all nearby values, but not necessarily greater than every value on the entire curve. The gradient changes from positive to negative through the point.
局部最大值是函数值大于附近所有值的点，但不一定大于整条曲线上的所有值。经过该点时，梯度由正变为负。
A local minimum is a point where the function value is less than all nearby values, and the gradient changes from negative to positive. Do not confuse local extrema with global extrema: a curve may have several local maxima and minima.
局部最小值是函数值小于附近所有值的点，梯度由负变为正。不要将局部极值与全局极值混淆：一条曲线可能有多个局部最大值和最小值。
6. Stationary Points of Inflection | 驻点拐点
A stationary point of inflection occurs when dy/dx = 0 but the sign of the gradient does not change on either side. The curve flattens but continues to increase or decrease after the point.
驻点拐点出现在 dy/dx = 0 但梯度符号在两侧不变时。曲线变平，但经过该点后继续增加或减少。
The simplest example is y = x³. Differentiating gives dy/dx = 3x², so dy/dx = 0 at x = 0. Since dy/dx is positive on both sides of x = 0, the origin is a stationary point of inflection.
最简单的例子是 y = x³。求导得 dy/dx = 3x²，因此当 x = 0 时 dy/dx = 0。由于 dy/dx 在 x = 0 两侧均为正，所以原点是驻点拐点。
For full classification, remember that a point of inflection also requires a change in concavity, meaning d²y/dx² changes sign. In y = x³, d²y/dx² = 6x, which changes from negative to positive at x = 0.
为了完整分类，请记住拐点还要求凹凸性发生变化，即 d²y/dx² 改变符号。在 y = x³ 中，d²y/dx² = 6x，它在 x = 0 处由负变为正。
7. Worked Example: Cubic Function | 例题：三次函数
Find and classify the stationary points of y = x³ − 3x² − 9x + 5.
求函数 y = x³ − 3x² − 9x + 5 的驻点并分类。
dy/dx = 3x² − 6x − 9
Set the derivative equal to zero:
令导数等于零：
3x² − 6x − 9 = 0
x² − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0
Therefore x = 3 or x = −1. Substitute these into y to find the coordinates:
因此 x = 3 或 x = −1。将这些值代入 y 求坐标：
At x = 3: y = 27 − 27 − 27 + 5 = −22
At x = −1: y = −1 − 3 + 9 + 5 = 10
The stationary points are (3, −22) and (−1, 10). Now find the second derivative:
驻点为 (3, −22) 和 (−1, 10)。现在求二阶导数：
d²y/dx² = 6x − 6
At x = 3: d²y/dx² = 12 > 0 ⇒ local minimum
At x = −1: d²y/dx² = −12 < 0 ⇒ local maximum
So the curve has a local maximum at (−1, 10) and a local minimum at (3, −22).
因此该曲线在 (−1, 10) 处有局部最大值，在 (3, −22) 处有局部最小值。
8. Worked Example: Optimisation Problem | 例题：优化问题
A square sheet of side 20 cm is used to make an open box by cutting a square of side x cm from each corner and folding up the sides. Find the value of x that maximises the volume.
一张边长为 20 cm 的正方形纸板，通过从每个角剪去边长为 x cm 的正方形并折起侧边来制作一个无盖盒子。求使体积最大的 x 值。
The volume of the box is given by V = x(20 − 2x)². Expanding gives V = 400x − 80x² + 4x³. Differentiate with respect to x:
盒子的体积为 V = x(20 − 2x)²。展开得 V = 400x − 80x² + 4x³。对 x 求导：
dV/dx = 400 − 160x + 12x²
Set dV/dx = 0:
令 dV/dx = 0：
12x² − 160x + 400 = 0
3x² − 40x + 100 = 0
x = (40 ± √(1600 − 1200)) / 6 = (40 ± 20) / 6
This gives x = 10 or x = 10/3. Since x = 10 would make the box base size 0, it is not valid in the domain 0 < x < 10. Therefore x = 10/3 cm.
得到 x = 10 或 x = 10/3。由于 x = 10 会使盒子底面尺寸为 0，在定义域 0 < x < 10 中不成立。因此 x = 10/3 cm。
Use the second derivative to confirm a maximum:
使用二阶导数确认最大值：
d²V/dx² = −160 + 24x
At x = 10/3: d²V/dx² = −160 + 80 = −80 < 0
Therefore the volume is maximised when x = 10/3 cm.
因此当 x = 10/3 cm 时体积最大。
9. When the Second Derivative Test Fails | 二阶导数判别法失效时
If d²y/dx² = 0 at a stationary point, the second derivative test does not tell you whether the point is a maximum, minimum or inflection. You must use the first derivative test instead.
如果驻点处 d²y/dx² = 0，二阶导数判别法无法告诉你该点是最大值、最小值还是拐点。此时必须改用一阶导数判别法。
Consider y = x⁴. The first derivative is dy/dx = 4x³, so dy/dx = 0 at x = 0. The second derivative is d²y/dx² = 12x², which is also zero at x = 0. However, dy/dx changes from negative to positive at x = 0, so (0, 0) is a local minimum.
考虑 y = x⁴。一阶导数为 dy/dx = 4x³，因此 x = 0 处 dy/dx = 0。二阶导数为 d²y/dx² = 12x²，在 x = 0 处也为零。然而 dy/dx 在 x = 0 处由负变正，因此 (0, 0) 是局部最小值。
Similarly, y = −x⁴ gives d²y/dx² = 0 at x = 0, but the point is a local maximum. This demonstrates why f”(a) = 0 alone is inconclusive.
类似地，y = −x⁴ 在 x = 0 处 d²y/dx² = 0，但该点是局部最大值。这说明仅凭 f”(a) = 0 无法得出结论。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
- Forgetting to substitute x back into the original equation to find the y-coordinate — 忘记将 x 代回原方程求 y 坐标。
- Stopping after solving dy/dx = 0 without classifying the stationary point — 解完 dy/dx = 0 后没有分类驻点就停止。
- Misinterpreting the sign of d²y/dx²: negative is a maximum, positive is a minimum — 误解 d²y/dx² 的符号：负值是最大值，正值是最小值。
- Assuming d²y/dx² = 0 always means a point of inflection — 认为 d²y/dx² = 0 总是意味着拐点。
- Ignoring domain restrictions in optimisation problems, such as 0 < x < 10 — 忽略优化问题中的定义域限制，例如 0 < x < 10。
- Using the first derivative test incorrectly by choosing test values on the wrong side — 一阶导数判别法使用错误，选择了错误一侧的测试值。
11. Quick Summary Checklist | 快速总结清单
| Step | Action | 中文提示 |
|---|---|---|
| 1 | Differentiate y = f(x) | 对 y = f(x) 求导 |
| 2 | Solve dy/dx = 0 | 解
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