📚 Finding Maxima and Minima | 函数极值问题：最大最小值求解方法
One of the most powerful applications of differentiation in A-level mathematics is finding the maximum and minimum values of a function. These turning points, collectively known as extrema, allow us to solve optimisation problems in physics, economics, engineering, and geometry. This article provides a complete, exam-focused guide to locating and classifying stationary points, handling endpoints, and solving practical optimisation questions.
在 A-level 数学中，微分最强大的应用之一就是求函数的最大值和最小值。这些转折点统称为极值，使我们能够解决物理、经济、工程和几何中的优化问题。本文将提供一份完整且紧扣考纲的指南，帮助大家定位并判断驻点、处理端点情形，以及解决实际优化问题。
1. What Are Extrema? | 什么是极值
An extremum (plural: extrema) is a point where a function attains a value that is either greater than or equal to (maximum) or less than or equal to (minimum) all nearby values. We distinguish between local extrema, which are extreme only within a small neighbourhood, and global extrema, which are the absolute largest or smallest values over the entire domain under consideration.
极值是指函数在某一点取得比其附近所有值都大（极大值）或都小（极小值）的数值。我们区分局部极值——仅在某个小邻域内为极值，以及全局极值——在考虑整个定义域上的绝对最大或最小值。
l (provided the function is differentiable there). This leads to the necessary condition f '(x) = 0. However, the converse is not true: f ' t guarantee an extremum, as demonstrated by points of inflection with horizontal tangents.
对于函数 y = f(x)，在局部最大值或最小值处，曲线的切线是水平的（前提是函数在该点可导）。由此得到必要条件 f '(x) = 0。但反过来不一定成立：f '(x) = 0 并不能保证是极值，水平切线的拐点就是反例。
2. Stationary Points and Critical Points | 驻点与临界点
A stationary point is a point on a curve where the derivative is zero, i.e., f ' tionary points includes local maxima, local minima, and horizontal points of inflection. A critical point is any point in the domain where the derivative is either zero or does not exist. When locating extrema on a closed interval, critical points of both types must be considered.
驻点是曲线上一阶导数为零的点，即 f '(x) = 0。驻点集合包括局部最大值、局部最小值以及水平拐点。临界点是定义域内导数为零或不存在的任何点。在闭区间上寻找极值时，两类临界点都必须考虑。
For example, for f(x) = x³ − 3x, we have f '(x) = 3x² − 3 = 3(x − 1)(x + 1). Setting f ' sts everywhere, so there are no additional critical points of the non-differentiable type.
例如，对于 f(x) = x³ − 3x，有 f '(x) = 3x² − 3 = 3(x − 1)(x + 1)。令 f '(x) = 0 得 x = 1 和 x = −1，因此这两个是驻点。导数处处存在，因此没有不可导类型的额外临界点。
3. First Derivative Test | 一阶导数判别法
The first derivative test examines the sign of f ' stationary point. Suppose x = a is a stationary point. Consider a small interval around a:
一阶导数判别法考察驻点两侧 f '(x) 的符号。设 x = a 是驻点，考虑 a 的一个小邻域：
If f '(x) changes from positive to negative as x increases through a, then f(a) is a local maximum.
如果在 x 增大经过 a 的过程中，f '(x) 由正变负，则 f(a) 是局部最大值。
If f '(x) changes from negative to positive as x increases through a, then f(a) is a local minimum.
如果在 x 增大经过 a 的过程中，f '(x) 由负变正，则 f(a) 是局部最小值。
If f '(x) does not change sign, then f(a) is neither a maximum nor a minimum — it is a horizontal point of inflection.
如果 f '(x) 的符号不发生变化，则 f(a) 既不是最大值也不是最小值——它是水平拐点。
For f(x) = x³, we have f '(x) = 3x² ≥ 0 for all x. Although f ' tive is positive on both sides of 0, so x = 0 is a horizontal inflection, not an extremum.
对于 f(x) = x³，有 f '(x) = 3x² ≥ 0 对所有 x 成立。虽然 f '(0) = 0，但导数在 0 的两侧均为正，因此 x = 0 是水平拐点，而非极值点。
First Derivative Test: sign change of f '(x) determines the nature of a stationary point.
一阶导数判别法：f '(x) 的符号变化决定驻点的性质。
4. Second Derivative Test | 二阶导数判别法
he second derivative test provides a faster method for classifying stationary points. If f '(a) = 0 and f ''(x) is continuous near x = a, then:
二阶导数判别法提供了一种更快捷的判断驻点性质的方法。若 f '(a) = 0 且 f ''(x) 在 x = a 附近连续，则：
If f ''(a) < 0, then f(a) is a local maximum. The curve is concave down at this point.
若 f ''(a) < 0，则 f(a) 是局部最大值，曲线在该点下凹（凸向上）。
If f ''(a) > 0, then f(a) is a local minimum. The curve is concave up at this point.
若 f ''(a) > 0，则 f(a) 是局部最小值，曲线在该点上凹（凹向上）。
If f ''(a) = 0, the test is inconclusive, and we must revert to the first derivative test.
若 f ''(a) = 0，判别法失效，此时必须回到一阶导数判别法。
Returning to f(x) = x³ − 3x, we have f '(x) = 3x² − 3 and f ''(x) = 6x. At x = 1, f ''(1) = 6 > 0, so f(1) = −2 is a local minimum. At x = −1, f ''(−1) = −6 < 0, so f(−1) = 2 is a local maximum.
回到 f(x) = x³ − 3x，有 f '(x) = 3x² − 3 和 f ''(x) = 6x。在 x = 1 处，f ''(1) = 6 > 0，所以 f(1) = −2 是局部最小值。在 x = −1 处，f ''(−1) = −6 < 0，所以 f(−1) = 2 是局部最大值。
5. When f ''(x) = 0: The Inconclusive Case | f ''(x) = 0 时的失效情形
Consider f(x) = x⁴ and g(x) = −x⁴. Both have f '(0) = 0 and f '' ond derivative test cannot distinguish between these cases. Instead, examine the sign of f ' changes from negative to positive at x = 0; for −x⁴, it changes from positive to negative.
考虑 f(x) = x⁴ 和 g(x) = −x⁴。两者都有 f '(0) = 0 且 f ''(0) = 0，但 x = 0 对 x⁴ 是最小值，对 −x⁴ 却是最大值。二阶导数判别法无法区分这两种情形。此时应直接考察 f '(x) 的符号：对 x⁴，导数在 x = 0 处由负变正；对 −x⁴，导数由正变负。
Another classic example is f(x) = x³. Here f ' ot change sign at x = 0, confirming a horizontal point of inflection. This three-way split — maximum, minimum, or inflection — is exactly why the first derivative test remains the definitive method.
另一个经典例子是 f(x) = x³。这里 f '(x) = 3x² 在 x = 0 处符号不变，证实了水平拐点。这种三分情形——最大值、最小值或拐点——正是一阶导数判别法始终作为最终判定依据的原因。
6. Global Extrema on a Closed Interval | 闭区间上的全局最值
When a function is defined on a closed interval [a, b], the Extreme Value Theorem guarantees the existence of both a global maximum and a global minimum. These must occur either at stationary points inside the interval or at the endpoints. The systematic procedure is as follows:
当函数定义在闭区间 [a, b] 上时，极值定理保证了全局最大值和全局最小值的存在。它们必然出现在区间内部的驻点或端点处。系统化的求解步骤如下：
Step 1: Find f '(x) and solve f '(x) = 0 to locate all stationary points within (a, b).
第一步：求 f '(x) 并解 f '(x) = 0，找出 (a, b) 内所有驻点。
Step 2: Evaluate f(x) at every stationary point and at both endpoints a and b.
第二步：计算函数在每一个驻点以及两个端点 a、b 处的取值。
Step 3: The largest of these values is the global maximum; the smallest is the global minimum.
第三步：这些值中的最大者即为全局最大值，最小者即为全局最小值。
Example: Find the global extrema of f(x) = x³ − 12x + 5 on [−3, 3]. We have f '(x) = 3x² − 12 = 3(x − 2)(x + 2), giving stationary points at x = ±2. Evaluating:
示例：求 f(x) = x³ − 12x + 5 在 [−3, 3] 上的全局最值。有 f '(x) = 3x² − 12 = 3(x − 2)(x + 2)，驻点为 x = ±2。代入得：
| x | −3 | −2 | 2 | 3 |
| f(x) | 14 | 21 | −11 | −4 |
Notice that the endpoint value f(3) = −4 is not the minimum, but it must still be checked.
全局最大值为 21（x = −2 处），全局最小值为 −11（x = 2 处）。注意端点值 f(3) = −4 并不是最小值，但仍然必须代入检查。
7. Extrema of Trigonometric Functions | 三角函数的极值
r R cos(x ± α). For a function of the form f(x) = a sin x + b cos x, we can rewrite it as:
三角函数经常出现在极值问题中，尤其是形如 R sin(x ± α) 或 R cos(x ± α) 的形式。对于 f(x) = a sin x + b cos x 形式的函数，可以改写为：
f(x) = R sin(x + α), where R = √a² + b² and tan α = b⁄a.
f(x) = R sin(x + α)，其中 R = √a² + b²，tan α = b⁄a。
the maximum of f is R and the minimum is −R. For example, f(x) = 3 sin x + 4 cos x has R = √9 + 16 = 5, so the maximum value is 5 and the minimum is −5.
由于正弦函数在 −1 和 1 之间振荡，f 的最大值为 R，最小值为 −R。例如，f(x) = 3 sin x + 4 cos x 的 R = √9 + 16 = 5，所以最大值为 5，最小值为 −5。
8. Optimisation Problems: Setting Up the Model | 优化问题：建立模型
In examination problems, extrema often appear in a practical context. The typical strategy is to express the quantity to be optimised as a function of a single variable, then apply the derivative tests. Consider the following classic problem:
在考试中，极值问题常以实际应用的形式出现。典型策略是将待优化的量表示为单个变量的函数，然后应用导数判别法。考虑以下经典问题：
Problem: A rectangular field is to be fenced on three sides using 100 metres of fencing, with the fourth side bounded by a river. Find the dimensions that maximise the enclosed area.
问题：用 100 米长的篱笆围一块矩形场地，其中一边靠河，只需围三边。求使围成面积最大的尺寸。
Solution: h be y metres. The fencing constraint gives x + 2y = 100, so y = (100 − x)⁄2. The area is:
解法：设平行于河边的边长为 x 米，另外两边各为 y 米。篱笆约束为 x + 2y = 100，因此 y = (100 − x)⁄2。面积为：
A(x) = x · (100 − x)⁄2 = 50x − x²⁄2, 0 < x < 100.
Differentiating, A '(x) = 50 − x. Setting A '(x) = 0 gives x = 50, and A '' ximum. The maximum area is A(50) = 50 × 25 = 1250 m², achieved when x = 50 m and y = 25 m.
求导得 A '(x) = 50 − x。令 A '(x) = 0 得 x = 50，且 A ''(x) = −1 < 0，确认为最大值。最大面积为 A(50) = 50 × 25 = 1250 平方米，此时 x = 50 米，y = 25 米。
9. Economic Application: Maximising Profit | 经济应用：最大化利润
Differentiation is widely used in economics to find optimal production levels. Suppose the profit function of a company is given by:
微分在经济学中被广泛用于确定最优产量。假设某公司的利润函数为：
P(x) = −2x³ + 9x² + 24x − 10,
where x is the number of units produced in thousands and P(x) is the profit in thousands of pounds. The derivative is:
其中 x 是以千件为单位的产量，P(x) 是以千英镑为单位的利润。其导数为：
P '(x) = −6x² + 18x + 24 = −6(x² − 3x − 4) = −6(x − 4)(x + 1).
Setting P ' = 0 gives x = 4 (since x = −1 is not feasible for production). The second derivative is P ''(x) = −12x + 18, and P '' irming a maximum. The maximum profit is P(4) = −128 + 144 + 96 − 10 = 102 thousand pounds.
令 P '(x) = 0 得 x = 4（x = −1 在产量上无意义）。二阶导数为 P ''(x) = −12x + 18，且 P ''(4) = −30 < 0，确认为最大值。最大利润为 P(4) = −128 + 144 + 96 − 10 = 102 千英镑。
10. Common Mistakes and Exam Pitfalls | 常见错误与考试陷阱
Students frequently lose marks in extremum problems for avoidable reasons. Below are the most common pitfalls:
学生在极值问题中常因可避免的原因失分。以下是最常见的陷阱：
Forgetting endpoints: On a closed interval, the global maximum may occur at an endpoint, not at a stationary point. Always evaluate f(a) and f(b).
遗漏端点：在闭区间上，全局最大值可能出现在端点而非驻点处。务必代入计算 f(a) 和 f(b)。
Misusing the second derivative test: When f ''(x) = 0, do not guess — use the sign change table for f '(x).
误用二阶导数判别法：当 f ''(x) = 0 时不要猜测，应使用 f '(x) 的符号变化表。
Ignoring domain restrictions: In optimisation problems, variables often have natural constraints. For example, x > 0, or the quantity under a square root must be non-negative.
忽略定义域限制：在优化问题中，变量通常有自然约束，例如 x > 0，或根号下的量必须非负。
Dividing by a variable that could be zero: When solving f ' , factor rather than cancel. Cancelling x from x(x − 2) = 0 would lose the solution x = 0.
除以可能为零的变量：解 f '(x) = 0 时应因式分解而非约分。从 x(x − 2) = 0 中约去 x 会丢失 x = 0 这个解。
Confusing local and global extrema: A function may have several local maxima, only one of which is the global maximum.
混淆局部极值与全局极值：一个函数可能有多个局部最大值，但只有其中一个是全局最大值。
11. Implicit Differentiation and Extrema | 隐函数求导与极值
te extrema on such curves, differentiate both sides with respect to x and solve for dy⁄dx:
有些曲线由隐式方程定义，如 x² + y² = 25。要在这样的曲线上定位极值，可以对等式两边关于 x 求导并解出 dy⁄dx：
2x + 2y · dy⁄dx = 0 ⟹ dy⁄dx = −x⁄y.
to x² + y² = 25 yields y = ±5. These correspond to the top and bottom points of the circle — the maximum y-value 5 and the minimum y-value −5.
驻点出现在 dy⁄dx = 0 处，即 x = 0。将 x = 0 代入 x² + y² = 25 得 y = ±5。这两点正是圆的最高点和最低点——y 的最大值 5 和最小值 −5。
12. Summary: Solving Strategy | 总结：求解策略
To approach any maximum–minimum problem methodically, follow this checklist:
为了有条理地解决任何最大最小值问题，请遵循以下检查清单：
Identify the function to be optimised, and note any domain restrictions.
确定需要优化的函数，并记录定义域限制。
Compute the first derivative and solve f '(x) = 0 to find stationary points.
求一阶导数并解 f '(x) = 0 来找出驻点。
Classify each stationary point using the first or second derivative test.
使用一阶或二阶导数判别法判断每个驻点的性质。
If the domain is a closed interval, evaluate the function at all endpoints and critical points.
如果定义域是闭区间，在端点及所有临界点处计算函数值。
Compare all candidate values: the largest is the absolute maximum, the smallest is the absolute minimum.
比较所有候选值：最大者为绝对最大值，最小者为绝对最小值。
Answer the original question — in optimisation problems, state the required quantities with correct units.
回答原问题——在优化题中，带上正确单位说明要求的量。
Mastery of extremum problems requires both procedural fluency and conceptual clarity. Practise classifying stationary points without a calculator, and always justify the nature of each turning point explicitly in your exam response.
掌握极值问题既需要熟练的计算能力，也需要清晰的概念理解。练习不使用计算器判断驻点的性质，并在考试作答中始终明确说明每个转折点的判定依据。
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