A-Level Mathematics: Stationary Points and Extrema | A-Level数学：驻点与极值点判定

📚 A-Level Mathematics: Stationary Points and Extrema | A-Level数学：驻点与极值点判定

In A-Level Mathematics, one of the most important applications of differentiation is the location and classification of stationary points. These points reveal where a curve is momentarily horizontal, and they often correspond to local maxima or minima. Understanding how to distinguish between them is essential for curve sketching and optimisation problems.

在A-Level数学中，微分最重要的应用之一就是寻找并判定驻点。驻点对应曲线瞬时水平的位置，通常与局部极大值或极小值有关。掌握如何区分它们，对于绘制函数图像和解决优化问题至关重要。

1. What is a Stationary Point? | 什么是驻点？

A stationary point of a function f(x) is a point where the derivative is zero, i.e. f'(x) = 0. Geometrically, the tangent to the curve at that point is horizontal.

函数 f(x) 的驻点是导数为零的点，即 f'(x) = 0。从几何上看，曲线在该点的切线是水平的。

At a stationary point, the function is neither increasing nor decreasing instantaneously. However, not every stationary point is a maximum or minimum; it could also be a point of inflection.

在驻点处，函数在瞬间既不递增也不递减。然而，并非所有驻点都是极大值或极小值，它也可能是一个拐点。

f'(x) = 0

For example, for f(x) = x², we have f'(x) = 2x, so x = 0 is a stationary point. The graph has a minimum at the origin.

例如，对于 f(x) = x²，有 f'(x) = 2x，因此 x = 0 是驻点。图像在原点处取得极小值。

2. Stationary Points vs Critical Points | 驻点与临界点

It is important to distinguish between a stationary point and a critical point. A critical point is any point where the derivative is zero or where the derivative does not exist.

区分驻点和临界点非常重要。临界点是导数为零或导数不存在的点。

All stationary points are critical points, but not all critical points are stationary. For example, the point x = 0 on the graph of y = |x| is a critical point because the derivative is undefined there, but it is not a stationary point since f'(0) does not exist.

所有驻点都是临界点，但并非所有临界点都是驻点。例如，y = |x| 的图像在 x = 0 处是临界点，因为该处导数不存在，但它不是驻点，因为 f'(0) 不存在。

Type	Definition
Stationary point	f'(x) = 0 and f'(x) exists
Critical point	f'(x) = 0 or f'(x) does not exist

In most A-Level questions, you will only be asked to consider differentiable functions, so all critical points are stationary points.

在大多数A-Level题目中，只要求考虑可导函数，因此所有临界点都是驻点。

3. First Derivative Test | 一阶导数检验法

The first derivative test examines the sign of f'(x) on either side of a stationary point. If the derivative changes from positive to negative, the point is a local maximum; if it changes from negative to positive, the point is a local minimum.

一阶导数检验法考察驻点两侧 f'(x) 的符号。如果导数从正变为负，则该点是局部极大值；如果从负变为正，则该点是局部极小值。

If f'(x) changes from + to −: local maximum.
如果 f'(x) 从 + 变为 −：局部极大值。
If f'(x) changes from − to +: local minimum.
如果 f'(x) 从 − 变为 +：局部极小值。
If f'(x) does not change sign: neither maximum nor minimum (usually a stationary inflection).
如果 f'(x) 符号不变：既非极大值也非极小值（通常是驻点拐点）。

To apply the test, choose values of x just less than and just greater than the stationary point, and evaluate the sign of f'(x). This is especially useful when the second derivative is zero or difficult to compute.

应用该检验法时，选取略小于和略大于驻点的 x 值，并判断 f'(x) 的符号。这在二阶导数为零或难以计算时特别有用。

Stationary point at x = a: check f'(a − h) and f'(a + h) for small h > 0.

4. Second Derivative Test | 二阶导数检验法

The second derivative test uses f”(x) to determine the concavity of the curve at the stationary point. For a stationary point x = a:

二阶导数检验法利用 f”(x) 判断曲线在驻点处的凹凸性。对于驻点 x = a：

If f”(a) > 0: the curve is concave up, giving a local minimum.
如果 f”(a) > 0：曲线凹向上，取得局部极小值。
If f”(a) < 0: the curve is concave down, giving a local maximum.
如果 f”(a) < 0：曲线凹向下，取得局部极大值。
If f”(a) = 0: the test is inconclusive; use the first derivative test instead.
如果 f”(a) = 0：检验法失效；应改用一阶导数检验法。

For example, take f(x) = x³ − 3x². Then f'(x) = 3x² − 6x = 3x(x − 2), so stationary points occur at x = 0 and x = 2. The second derivative is f”(x) = 6x − 6. At x = 0, f”(0) = −6 < 0, so it is a local maximum. At x = 2, f''(2) = 6 > 0, so it is a local minimum.

例如，设 f(x) = x³ − 3x²。则 f'(x) = 3x² − 6x = 3x(x − 2)，因此驻点出现在 x = 0 和 x = 2。二阶导数为 f”(x) = 6x − 6。在 x = 0 处，f”(0) = −6 < 0，所以是局部极大值。在 x = 2 处，f''(2) = 6 > 0，所以是局部极小值。

5. Points of Inflection | 拐点

A point of inflection is a point where the curve changes concavity. At such a point, f”(x) = 0, but this condition alone is not sufficient; the sign of f”(x) must also change on either side of the point.

拐点是曲线凹凸性发生改变的点。在拐点处 f”(x) = 0，但仅此条件并不充分；f”(x) 的符号还必须在点左右两侧发生变化。

If a point of inflection occurs at a point where f'(x) = 0 as well, it is called a stationary inflection. An example is f(x) = x³ at x = 0: f'(0) = 0 and f”(0) = 0, and f”(x) changes sign from negative to positive.

如果拐点同时满足 f'(x) = 0，则称为驻点拐点。例如 f(x) = x³ 在 x = 0 处：f'(0) = 0，f”(0) = 0，且 f”(x) 从负变为正。

Inflection requires: f”(x) = 0 and f”(x) changes sign.

Not all points where f”(x) = 0 are inflection points. For instance, f(x) = x⁴ has f”(0) = 0, but the curve is concave up on both sides of 0, so 0 is not an inflection point.

并非所有满足 f”(x) = 0 的点都是拐点。例如，f(x) = x⁴ 在 x = 0 处有 f”(0) = 0，但曲线在 0 两侧都是凹向上，因此 0 不是拐点。

6. Common Mistakes | 常见错误

Students often make several predictable mistakes when classifying stationary points. Being aware of them can save valuable marks.

学生在判定驻点时往往会犯一些可预测的错误。了解这些错误有助于避免失分。

Assuming f'(x) = 0 always means a maximum or minimum. It might be a stationary inflection, such as y = x³.
认为 f'(x) = 0 就一定对应极大值或极小值。它可能是驻点拐点，例如 y = x³。
Forgetting to check endpoints. In a closed interval, global extrema can occur at endpoints, not just at stationary points.
忘记检查端点。在闭区间内，全局极值可能出现在端点，而不仅仅是驻点。
Using f”(x) = 0 to conclude an inflection point without checking sign change.
仅凭 f”(x) = 0 就断定是拐点，而不检查符号变化。
Mixing up the first and second derivative tests. The first test looks at f'(x); the second looks at f”(x).
混淆一阶和二阶导数检验法。一阶检验看 f'(x)；二阶检验看 f”(x)。

Always write down the stationary points by solving f'(x) = 0, then apply a clear test to classify each one.

务必先解方程 f'(x) = 0 求出所有驻点，再用清晰的检验法逐一分类。

7. Worked Example | 综合例题

Let us fully analyse the function f(x) = x² + 4x + 3.

下面我们完整地分析函数 f(x) = x² + 4x + 3。

Step 1: Find stationary points.

第一步：求驻点。

f'(x) = 2x + 4. Setting f'(x) = 0 gives 2x + 4 = 0, so x = −2.

f'(x) = 2x + 4。令 f'(x) = 0，得 2x + 4 = 0，所以 x = −2。

Step 2: Classify using the second derivative.

第二步：用二阶导数分类。

f”(x) = 2, which is positive for all x. Therefore at x = −2, f”(−2) = 2 > 0, so this stationary point is a local minimum.

f”(x) = 2，对所有 x 都为正。因此在 x = −2 处，f”(−2) = 2 > 0，所以该驻点是局部极小值。

Step 3: Find the coordinates.

第三步：求坐标。

f(−2) = (−2)² + 4(−2) + 3 = 4 − 8 + 3 = −1. Hence the minimum point is (−2, −1).

f(−2) = (−2)² + 4(−2) + 3 = 4 − 8 + 3 = −1。因此极小值点为 (−2, −1)。

Now consider a more complex example: g(x) = 2x³ − 9x² + 12x + 1.

再考虑一个更复杂的例子：g(x) = 2x³ − 9x² + 12x + 1。

g'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2). So the stationary points are at x = 1 and x = 2.

g'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2)。因此驻点在 x = 1 和 x = 2。

g”(x) = 12x − 18. At x = 1, g”(1) = −6 < 0, so (1, 6) is a local maximum. At x = 2, g''(2) = 6 > 0, so (2, 5) is a local minimum.

g”(x) = 12x − 18。在 x = 1 处，g”(1) = −6 < 0，所以 (1, 6) 是局部极大值。在 x = 2 处，g''(2) = 6 > 0，所以 (2, 5) 是局部极小值。

8. Global Extrema and Boundaries | 全局极值与边界

A local maximum or minimum is not necessarily the greatest or least value of the function over its whole domain. The global (absolute) maximum or minimum on a closed interval [a, b] must be found by comparing the values at stationary points and at the endpoints.

局部极大值或极小值不一定是函数在整个定义域上的最大值或最小值。在闭区间 [a, b] 上的全局（绝对）最大值或最小值，必须通过比较驻点和端点的函数值来确定。

For x ∈ [a, b], check f(a), f(b), and f(x) at every stationary point in (a, b).

For example, let h(x) = x³ − 3x on the interval [−2, 3]. The stationary points are x = −1 and x = 1, with h(−1) = 2 and h(1) = −2. The endpoint values are h(−2) = −2 and h(3) = 18. Therefore the global maximum is 18 at x = 3, and the global minimum is −2, achieved at both x = 1 and x = −2.

例如，设 h(x) = x³ − 3x 在区间 [−2, 3] 上。驻点为 x = −1 和 x = 1，h(−1) = 2，h(1) = −2。端点为 h(−2) = −2，h(3) = 18。因此全局最大值为 18（在 x = 3 处），全局最小值为 −2（在 x = 1 和 x = −2 处同时取得）。

If the interval is unbounded, limits at infinity must be considered. A quadratic with a positive leading coefficient has no global maximum, only a global minimum.

如果区间无界，则需要考虑无穷远处的极限。首项系数为正的二次函数没有全局最大值，只有全局最小值。

9. Conclusion | 总结

To summarise, the classification of stationary points is a routine but essential skill. First find f'(x) and solve f'(x) = 0. Then use the first derivative test to examine sign changes, or the second derivative test to check concavity. Remember that f”(x) = 0 does not automatically imply an inflection point, and always consider endpoints for global extrema.

总结一下，判定驻点是常规但至关重要的技能。先求 f'(x) 并解方程 f'(x) = 0。然后使用一阶导数检验法考察符号变化，或使用二阶导数检验法考察凹凸性。请记住，f”(x) = 0 并不自动意味着拐点，并且求全局极值时始终要考虑端点。

With careful step-by-step working, you can confidently handle any stationary-point problem in your A-Level exam.

只要按步骤仔细演算，你就能自信地应对A-Level考试中任何关于驻点的题目。

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A-Level Mathematics: Stationary Points and Extrema | A-Level数学：驻点与极值点判定