📚 5 Maxima, minima, and other stationary points | 极大值、极小值及其他驻点
Understanding how a function behaves around points where its derivative vanishes is fundamental in calculus. These stationary points can tell us where a function reaches a peak, a valley, or simply flattens before continuing in the same direction. Mastering their classification not only helps with curve sketching but also underpins real-world optimization – from minimizing costs to maximizing efficiency.
理解函数在其导数为零的点附近的行为是微积分的基础。这些驻点能够告诉我们函数在哪里达到顶峰、谷底，或者只是暂时变平后继续沿原方向前进。掌握它们的分类不仅有助于绘制函数图像，还为现实世界中的最优化问题——从最小化成本到最大化效率——奠定了基础。
1. What are stationary points? | 什么是驻点？
int on the graph of a differentiable function y = f(x) occurs where the first derivative f' d points of inflection with zero gradient. They are sometimes called turning points when the function changes from increasing to decreasing or vice versa.
可微函数 y = f(x) 图像上的驻点出现在一阶导数 f'(x) 等于零的位置。从几何角度看，这意味着该点处的切线是水平的。驻点包括局部极大值点、局部极小值点以及梯度为零的拐点。当函数从递增变为递减或相反时，这些驻点有时也被称为转向点。
2. Using the first derivative to find critical points | 利用一阶导数求临界点
To locate stationary points, differentiate the function and set f'(x) = 0. Solve the resulting equation to obtain the x-coordinates. Be aware that the term "critical point" ction, our focus is on stationary points. For example, given f(x) = x⁴ − 4x³, we compute f' 4x³ − 12x² = 4x²(x − 3). Setting this to zero gives stationary points at x = 0 and x = 3.
为了找出驻点，先对函数求导并令 f'(x) = 0，然后解方程得到 x 坐标。需要注意，"临界点"这一术语有时也包括导数不存在的点，但对于光滑多项式或性质良好的函数，我们主要关注驻点。例如，对于 f(x) = x⁴ − 4x³，计算 f'(x) = 4x³ − 12x² = 4x²(x − 3)，令其为零得到驻点 x = 0 和 x = 3。
3. First derivative test for classification | 一阶导数判别法
The first derivative test examines the sign of f'(x) just to the left and right of a stationary point. If f'(x) changes from positive to negative, the function has a local maximum at that point. If f'(x) changes from negative to positive, there is a local minimum. If the sign does not change, the stationary point is a horizontal point of inflection.
一阶导数判别法通过检查驻点左右两侧紧邻处 f'(x) 的符号来进行分类。如果 f'(x) 由正变负，则函数在该点有局部极大值；如果由负变正，则为局部极小值；如果符号不变，则该驻点为水平拐点。
Consider f(x) = x³ − 3x, with derivative f' ionary points are x = −1 and x = 1. Testing signs around x = −1: for x < −1, say x = −2, f'(−2) = 9 > 0; for x between −1 and 1, say x = 0, f' from positive to negative indicates a local maximum at x = −1. Similarly, around x = 1, f' changes from negative to positive, giving a local minimum.
以 f(x) = x³ − 3x 为例，其导数为 f'(x) = 3x² − 3 = 3(x − 1)(x + 1)，驻点为 x = −1 和 x = 1。考察 x = −1 附近的符号：左侧取 x = −2，得 f'(−2) = 9 > 0；在 −1 与 1 之间取 x = 0，得 f'(0) = −3 < 0。符号由正变负表明在 x = −1 处为局部极大值。类似地，在 x = 1 附近，f' 由负变正，为局部极小值。
4. Second derivative test | 二阶导数判别法
When the second derivative f''(x) is easy to compute, it offers a quicker alternative. Evaluate f'' at the stationary point x = a. If f''(a) > 0, the function is concave up and the point is a local minimum. If f''(a) < 0, the function is concave down and the point is a local maximum. If f''(a) = 0, the test is inconclusive – mum, a minimum, or a point of inflection, and we must revert to the first derivative test.
当二阶导数 f''(x) 易于计算时，它提供了一种更快的替代方法。在驻点 x = a 处计算 f''。若 f''(a) > 0，函数凹向上，该点为局部极小值；若 f''(a) < 0，函数凹向下，该点为局部极大值；若 f''(a) = 0，判别法失效——该点可能是极大值、极小值或拐点，此时必须回到一阶导数判别法进行判断。
A summary table for the second derivative test:
二阶导数判别法总结表：
| f''(a) sign | Concavity | Stationary point type |
| f''(a) > 0 | concave up | local minimum |
| f''(a) < 0 | concave down | local maximum |
| f''(a) = 0 | inconclusive | use first derivative test |
5. Understanding local maxima and minima | 理解局部极大值与极小值
A local maximum at x = a means that f(a) is greater than all nearby function values – me open interval around a. A local minimum satisfies f(a) ≤ f(x) in an interval. These are "local" istinction between local and global extrema is crucial when solving optimization problems.
在 x = a 处有局部极大值，意味着 f(a) 大于所有邻近的函数值——即对于 a 周围某个开区间内的所有 x，有 f(a) ≥ f(x)。局部极小值则满足在该区间内 f(a) ≤ f(x)。这些都是"局部"性质；函数在别处可以有更大或更小的值。求解最优化问题时，区分局部极值与全局极值至关重要。
6. Horizontal points of inflection | 水平拐点
but the concavity changes sign. A classic example is f(x) = x³ at x = 0. Its derivative f'(x) = 3x² is zero at x = 0, yet f'' om concave down to concave up. The function flattens momentarily but continues increasing.
既非极大值也非极小值的驻点就是水平拐点。在该点梯度为零，但凹凸性改变符号。典型的例子是 f(x) = x³ 在 x = 0 处。其导数 f'(x) = 3x² 在 x = 0 处为零，而二阶导数 f''(x) = 6x 在 x = 0 处由负变正，表明凹凸性由向下转为向上。函数在瞬间变平后继续递增。
When using the second derivative test, if f''(a) = 0 and the first derivative test shows no sign change in f', it is a horizontal inflection point. In some cases, f''(a) = 0 with a sign change in f'' itself confirms an inflection, but careful sign analysis of f' is the most reliable method.
使用二阶导数判别法时，若 f''(a) = 0 且一阶导数检测显示 f' 符号不变，则为水平拐点。在某些情况下，f''(a) = 0 且 f'' 自身改变符号也能确认拐点，但仔细分析 f' 的符号仍是最可靠的方法。
7. Global extrema on a closed interval | 闭区间上的全局极值
To find the absolute maximum and minimum of a continuous function f on a closed interval [a, b], compute the function values at all stationary points inside the interval and at the endpoints a and b. The largest among these is the absolute maximum; the smallest is the absolute minimum. This method, often called the closed interval method, works because a continuous function on a closed interval must attain its extreme values either at stationary points or at endpoints.
要找到闭区间 [a, b] 上连续函数 f 的绝对最大值和最小值，需要计算该区间内所有驻点以及端点 a 和 b 处的函数值。其中的最大值即为绝对最大值，最小值为绝对最小值。这种方法通常被称为闭区间法，其依据是闭区间上的连续函数必然在驻点或端点处取得极值。
Example: Find the absolute extrema of f(x) = x³ − 3x + 2 on [−2, 2].
示例：求 f(x) = x³ − 3x + 2 在 [−2, 2] 上的绝对极值。
f'(x) = 3x² − 3 = 0 → x = −1, x = 1
e maximum is 4 (at x = −1 and x = 2), and the absolute minimum is 0 (at x = −2 and x = 1).
计算：f(−2) = 0，f(−1) = 4，f(1) = 0，f(2) = 4。绝对最大值为 4（在 x = −1 和 x = 2 处取得），绝对最小值为 0（在 x = −2 和 x = 1 处取得）。
8. Applications in optimization | 最优化应用
Stationary points are the heart of optimization. When a real-world quantity must be maximized or minimized, we express it as a function of one variable, differentiate, and set the derivative to zero. The resulting stationary points, together with domain endpoints, are tested to identify the optimal solution. Always verify with the second derivative or a sign chart that the stationary point indeed gives the desired extremum.
驻点是最优化问题的核心。当需要最大化或最小化某个现实量时，我们将其表达为单一变量的函数，求导并令导数为零。所得的驻点同定义域的端点一起进行检验，从而确定最优解。务必通过二阶导数或符号表验证该驻点确实给出了所需的极值。
n 2x + y = 100, so y = 100 − 2x, and area A = x(100 − 2x) = 100x − 2x². Differentiating, A' = 100 − 4x. Setting A' = 0 gives x = 25, then y = 50, and A''(25) = −4 < 0 confirms a maximum area of 1250 m².
一个典型问题：一位农夫用 100 米围栏沿墙围成一个矩形场地，求使面积最大的尺寸。设与墙平行的边长为 y，垂直边长为 x。则有 2x + y = 100，即 y = 100 − 2x，面积 A = x(100 − 2x) = 100x − 2x²。求导得 A' = 100 − 4x，令 A' = 0 得 x = 25，于是 y = 50，A''(25) = −4 < 0 确认最大面积为 1250 m²。
9. Sketching graphs with stationary points | 利用驻点绘制图像
A function graph can be accurately sketched by combining information from stationary points, intervals of increase/decrease, and concavity. After locating stationary points, use the first derivative to determine where the function rises and falls. Compute the second derivative to see where the graph is concave up or concave down, which also helps identify non-stationary inflection points. Plot all stationary points, intercepts, and asymptotes if any, then draw smooth curves that respect all sign information.
通过结合驻点信息、增减区间以及凹凸性，可以准确地绘制函数图像。在找到驻点后，利用一阶导数确定函数何处上升、何处下降。计算二阶导数以了解图像在何处凹向上或凹向下，这也有助于识别非驻点拐点。标出所有驻点、截距以及可能存在的渐近线，然后依据所有符号信息画出平滑曲线。
For f(x) = x³ − 3x, we found a maximum at (−1, 2) and a minimum at (1, −2). f'(x) > 0 for x < −1 and x > 1 (increasing), f'(x) < 0 for −1 < x < 1 (decreasing). f'' lection at (0, 0) that is not horizontal. The graph can then be drawn with these features.
对于 f(x) = x³ − 3x，我们找到极大值点 (−1, 2) 和极小值点 (1, −2)。当 x < −1 和 x > 1 时 f'(x) > 0（递增），当 −1 < x < 1 时 f'(x) < 0（递减）。f''(x) = 6x，因此当 x < 0 时凹向下，当 x > 0 时凹向上，并在 (0, 0) 处有一个非水平拐点。随后即可根据这些特征绘制图像。
10. Points where the derivative does not exist | 导数不存在的点
Although the definition of a stationary point requires f' . These points are critical points but not stationary points. A familiar example is f(x) = |x|, which has a sharp corner at x = 0. The derivative f' earching for extrema on a closed interval, always consider endpoints and any point where f'(x) is undefined.
尽管驻点的定义要求 f'(x) = 0，但值得注意的是，函数可以在导数不存在的点取得局部极值。这些点是临界点，但不是驻点。一个熟悉的例子是 f(x) = |x|，它在 x = 0 处有一个尖角，导数 f'(0) 不存在，但 x = 0 却是全局极小值点。在闭区间上寻找极值时，务必考虑端点以及任何 f'(x) 无定义的点。
In IB questions, functions are usually differentiable everywhere on their domain, so the focus remains on stationary points. However, being aware of cusps and corners provides a more complete understanding of how extrema can arise in other related courses.
在 IB 试题中，函数在其定义域内通常处处可微，因此重点仍然是驻点。然而，了解尖点和角点有助于在其他相关课程中对极值的产生方式形成更全面的理解。
Published by TutorHao | IB 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