📚 Maxima and Minima | 极大值与极小值
In IB Mathematics, the concepts of maxima and minima are central to understanding how functions behave. Whether we are modelling the trajectory of a projectile, maximising profit in a business scenario, or minimising surface area for a given volume, the ability to locate and classify turning points is an essential skill. This article explores stationary points, the first and second derivative tests, global versus local extrema, and practical optimisation problems.
在IB数学中，极大值和极小值的概念是理解函数行为的核心。无论是模拟抛射体的运动轨迹、最大化商业场景中的利润，还是最小化给定体积的表面积，定位并分类转折点都是一项基本技能。本文探讨驻点、一阶导数和二阶导数检验、全局极值与局部极值，以及实际优化问题。
1. What Are Maxima and Minima? | 什么是极大值与极小值？
A maximum point on a curve is where the function value is greater than that of all neighbouring points; it is a 'peak'. A minimum point is where the value is lower than all adjacent points; it is a 'valley'. Together, these are called extrema or turning points. In IB, we distinguish between local (relative) extrema and global (absolute) extrema over an entire domain or closed interval.
曲线上的极大值点是指函数值大于所有邻近点的位置，是一个“峰”。极小值点是指函数值小于所有邻近点的位置，是一个“谷”。这两类点合称为极值点或转折点。在IB中，我们要区分局部（相对）极值和全局（绝对）极值，后者覆盖整个定义域或闭区间。
2. Stationary Points | 驻点
ng maxima and minima is recognising that at a turning point, the gradient of the tangent is zero. Hence, we solve f' stationary point can be a local maximum, a local minimum, or a stationary inflection point where the curve flattens but continues increasing or decreasing.
寻找极大值与极小值的关键起点是认识到，在转折点处切线的斜率为零。因此，我们解 f'(x) = 0 来找到驻点。驻点可能是局部极大点、局部极小点，或是曲线变平后继续上升或下降的驻拐点。
3. The First Derivative Test | 一阶导数检验
Once a stationary point is found at x = c, examine the sign of f'(x) on either side of c. If f'(x) changes from positive to negative, the point is a local maximum. If f' the point is a stationary inflection. Make a sign diagram to visualise this test clearly.
在 x = c 处找到驻点后，检查 c 两侧 f'(x) 的符号。若 f'(x) 由正变负，该点是局部极大点。若 f'(x) 由负变正，它是局部极小点。若符号不变，则该点是驻拐点。绘制符号图可以清晰地展示这一检验。
| Sign of f'(x) left of c | Sign of f'(x) right of c | Conclusion |
|---|---|---|
| + | − | Local maximum |
| − | + | Local minimum |
| + | + | Stationary inflection (increasing) |
| − | − | Stationary inflection (decreasing) |
4. The Second Derivative Test | 二阶导数检验
The second derivative, f''(x), gives us concavity information. At a stationary point x = c, evaluate f''(c). If f''(c) < 0, the curve is concave down, and we have a local maximum. If f''(c) > 0, the curve is concave up, indicating a local minimum. If f'' tive test. This test is often faster but requires the function to be twice differentiable.
二阶导数 f''(x) 提供凹凸性信息。在驻点 x = c 处，计算 f''(c)。若 f''(c) < 0，曲线下凹，该点是局部极大点。若 f''(c) > 0，曲线上凹，表明是局部极小点。若 f''(c) = 0，该检验无效，需使用一阶导数检验。此检验通常更快，但要求函数二次可导。
Local maximum: f'(c)=0, f''(c) < 0 | Local minimum: f'(c)=0, f''(c) > 0
5. Local vs Global Extrema | 局部极值与全局极值
A local maximum might not be the highest point on the entire curve; that title belongs to the global (or absolute) maximum. In IB, you must consider the domain. For functions defined on a closed interval [a, b], the global maximum and minimum can occur at stationary points or at the endpoints. Always evaluate f(x) at critical points and at the interval boundaries, then compare.
局部极大点可能不是整个曲线上的最高点；那个位置属于全局（绝对）极大值。在IB中，你必须考虑定义域。对于定义在闭区间 [a, b] 上的函数，全局极大值和极小值可能出现在驻点或端点处。务必计算临界点和区间端点处的 f(x)，然后进行比较。
6. Concavity and Points of Inflection | 凹凸性与拐点
Concavity is determined by f''(x). A function is concave up where f''(x) > 0 and concave down where f'' sign. Not every inflection point is stationary. A non-stationary inflection occurs where f''(x) = 0 but f'(x) ≠ 0. Understanding these helps avoid misclassifying extrema.
凹凸性由 f''(x) 决定。当 f''(x) > 0 时函数上凹；当 f''(x) < 0 时函数下凹。拐点是凹凸性改变符号的位置。并非所有拐点都是驻点。非驻拐点出现在 f''(x) = 0 但 f'(x) ≠ 0 处。理解这些有助于避免错误分类极值。
7. Optimisation Problems | 优化问题
Maxima and minima are powerful tools for solving real-world problems. The typical IB question provides a quantity to maximise or minimise, such as area, volume, cost, or distance. You must:
- Express the quantity as a function of one variable, often using a constraint equation.
- Find the stationary points by differentiation.
- Determine the nature of the stationary points and check endpoints if a domain is specified.
- Interpret the result in context, ensuring it is a practical maximum or minimum.
极大值与极小值是解决实际问题的重要工具。典型的IB题目会给出一个需要最大化或最小化的量，如面积、体积、成本或距离。你必须：
- 将其表示为单一变量的函数，通常借助约束方程。
- 通过求导找到驻点。
- 判断驻点的性质，若给定定义域还需检查端点。
- 在具体情境中解释结果，确保它是符合实际的极大值或极小值。
8. Worked Example 1: Polynomial Function | 例题1：多项式函数
Find and classify the stationary points of f(x) = 2x³ − 3x² − 12x + 5.
找出 f(x) = 2x³ − 3x² − 12x + 5 的驻点并进行分类。
First, differentiate: f'(x) = 6x² − 6x − 12. Set f' 0 → 6(x² − x − 2) = 0 → (x − 2)(x + 1) = 0, giving stationary points at x = 2 and x = −1.
首先求导：f'(x) = 6x² − 6x − 12。令 f'(x) = 0 → 6(x² − x − 2) = 0 → (x − 2)(x + 1) = 0，得到驻点 x = 2 和 x = −1。
Use the second derivative f''(x) = 12x − 6. At x = 2, f''(2) = 18 > 0 ⇒ local minimum. At x = −1, f''(−1) = −18 < 0 ⇒ local maximum.
使用二阶导数 f''(x) = 12x − 6。x = 2 时，f''(2) = 18 > 0 ⇒ 局部极小点；x = −1 时，f''(−1) = −18 < 0 ⇒ 局部极大点。
Local minimum at (2, f(2)) = (2, −17) | Local maximum at (−1, f(−1)) = (−1, 14)
9. Worked Example 2: Optimisation | 例题2：优化问题
0, for 10 ≤ x ≤ 60. Find the number of units that maximises profit and the maximum profit.
某公司生产 x 件产品，利润函数为 P(x) = −0.5x² + 40x − 300，定义域 10 ≤ x ≤ 60。求使利润最大化的产品数量及最大利润。
P'(x) = −x + 40. Set P'(x) = 0 → x = 40. This is a stationary point. P'' = −0.5(1600) + 1600 − 300 = −800 + 1600 − 300 = 500. The global maximum is 500 at x = 40.
P'(x) = −x + 40。令 P'(x) = 0 得 x = 40，这是一个驻点。P''(x) = −1 < 0，所以是极大点。检查端点：P(10) = −0.5(100) + 400 − 300 = 50；P(60) = −0.5(3600) + 2400 − 300 = 300；P(40) = 500。全局最大利润为500，在 x = 40 时取得。
10. Common Mistakes | 常见错误
e domain when identifying global extrema, or they misclassify stationary points by relying solely on f'' n. Another error is confusing a stationary point with a point of inflection. Carefully read the question: sometimes you are asked for the coordinates of a turning point, not just its x-value. Always provide x and y coordinates.
学生经常忘记在识别全局极值时考虑定义域，或者仅凭 f''(x) = 0 而未进一步调查就错误分类驻点。另一个错误是将驻点与拐点混淆。仔细读题：有时要求给出转折点的坐标，而不仅仅是 x 值。务必提供 x 和 y 坐标。
11. Tips for IB Exam Success | IB考试成功秘诀
- Memorise the two derivative tests and know when each is advantageous.
- Practise setting up constraint equations for optimisation problems.
- Always sketch a quick sign diagram for f'(x) when the second derivative test fails.
- Check endpoints on closed intervals—global extrema often hide there.
- 熟记两种导数检验方法，并了解各自的优势场景。
- 练习针对优化问题建立约束方程。
- 当二阶导数检验失效时，始终快速绘制 f'(x) 的符号图。
- 检查闭区间的端点——全局极值常隐藏于此。
12. Summary | 总结
Maxima and minima analysis is a cornerstone of IB calculus, connecting differentiation with graph behaviour and practical applications. By mastering the identification of stationary points, the first and second derivative tests, and distinguishing between local and global extrema, you equip yourself with the tools to solve a wide range of problems. Remember to always interpret your mathematical findings in the context of the question.
极大值与极小值分析是IB微积分的基石，它将求导与图像行为及实际应用联系起来。通过掌握驻点的识别、一阶和二阶导数检验，以及区分局部与全局极值，你就具备了解决各类问题的工具。切记始终在题目所给的语境中解释你的数学发现。
Published by TutorHao | 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