📚 The Extreme Value Theorem and Its Typical Applications | 极值定理及其典型应用
The Extreme Value Theorem (EVT) is one of the most elegant and useful results in calculus. It guarantees that every continuous function defined on a closed interval must attain both an absolute maximum and an absolute minimum somewhere within that interval. This simple yet powerful conclusion underpins a huge range of optimisation problems across mathematics, economics, physics and engineering.
极值定理（EVT）是微积分中最优美、最实用的结论之一。它保证了定义在闭区间上的任何连续函数，必定在该区间内某处取得绝对最大值和绝对最小值。这一简洁而有力的结论，支撑着数学、经济学、物理学和工程学中大量的优化问题。
1. What Is the Extreme Value Theorem? | 什么是极值定理？
Formally, the theorem states: if a function f is continuous on a closed interval [a, b], then there exist points c₁ and c₂ in [a, b] such that for every x in [a, b],
f(c₁) ≤ f(x) ≤ f(c₂)
The value f(c₁) is the absolute minimum of f on [a, b], and f(c₂) is the absolute maximum. In other words, the function cannot "escape" its interval without reaching its extreme values.
该定理的正式表述为：若函数 f 在闭区间 [a, b] 上连续，则存在点 c₁ 和 c₂ ∈ [a, b]，使得对任意 x ∈ [a, b]，都有
f(c₁) ≤ f(x) ≤ f(c₂)
其中 f(c₁) 是 f 在 [a, b] 上的绝对最小值，f(c₂) 是绝对最大值。换句话说，函数不可能在不取得极值的情况下"逃出"它的定义区间。
Two key words deserve special attention: "continuous" and "closed interval". Both conditions are essential; if either one is missing, the conclusion may fail completely.
两个关键词值得我们特别注意："连续"和"闭区间"。这两个条件缺一不可；只要缺失任何一个，结论就可能完全不成立。
2. Conditions: Continuity and the Closed Interval | 条件：连续性与闭区间
Why do we need continuity? Consider the function f(x) = 1/x for x ≠ 0, with f(0) = 0, defined on the closed interval [−1, 1]. This function is discontinuous at x = 0. As x approaches 0 from the right, f(x) grows without bound, so no maximum exists, even though the interval is closed and bounded.
为什么需要连续性？考虑函数 f(x) = 1/x（当 x ≠ 0 时），并令 f(0) = 0，定义在闭区间 [−1, 1] 上。这个函数在 x = 0 处不连续。当 x 从右侧趋近于 0 时，f(x) 无限增大，因此不存在最大值，尽管区间是闭且有界的。
Why do we need a closed interval? Take f(x) = x² on the open interval (0, 1). Here f is perfectly continuous, but it never attains an absolute maximum: the values approach 1 as x approaches 1, but x = 1 is not included. Likewise, the values approach 0 but never actually equal 0, so there is no absolute minimum either.
为什么需要闭区间？考察 f(x) = x² 在开区间 (0, 1) 上。此时 f 完全连续，但它永远达不到绝对最大值：当 x 趋近 1 时函数值趋近于 1，但 x = 1 并不在区间内。同样地，函数值趋近于 0 却永远不等于 0，因此也不存在绝对最小值。
These two examples show that the EVT is not an empty statement: both hypotheses are genuinely required. In your exam, always check continuity and the closed interval before applying the theorem.
这两个例子表明，极值定理并非空洞的陈述：两个条件都确实必要。在考试中，应用该定理之前务必先检查连续性和闭区间这两项前提。
3. Critical Points and the EVT Procedure | 临界点与极值定理求解步骤
Once the EVT guarantees that absolute extrema exist, we still need a method to find them. The candidates for extrema come from two sources: critical points inside the interval and the two endpoints of the interval.
一旦极值定理保证了绝对极值的存在，我们仍需要一种方法来求出它们。极值的候选点来自两个来源：区间内部的临界点，以及区间的两个端点。
A number c in the interior (a, b) is called a critical point of f if f '(c) = 0 or f '(c) does not exist, and c is in the domain of f. The complete procedure for locating absolute extrema is as follows:
若 c ∈ (a, b) 且满足 f '(c) = 0 或 f '(c) 不存在，同时 c 在 f 的定义域内，则称 c 为 f 的临界点。求绝对极值的完整步骤如下：
Step 1: Differentiate f (x) to obtain f '(x).
第 1 步：对 f (x) 求导，得到 f '(x)。
Step 2: Solve f '(x) = 0 and also identify points in (a, b) where f '(x) does not exist. Keep only those that lie inside the open interval.
第 2 步：解方程 f '(x) = 0，同时找出 (a, b) 内 f '(x) 不存在的点。只保留位于开区间内部的点。
Step 3: Evaluate f at every critical point and at both endpoints a and b.
第 3 步：计算 f 在所有临界点以及两个端点 a 和 b 处的函数值。
Step 4: The largest of these values is the absolute maximum, and the smallest is the absolute minimum.
第 4 步：这些值中最大的就是绝对最大值，最小的就是绝对最小值。
This procedure works for any continuous function on a closed interval, whether the extrema occur at endpoints, at corners, or at smooth peaks and valleys.
这个步骤适用于闭区间上的任何连续函数，无论极值是出现在端点、尖角处，还是在光滑的峰谷处。
4. Worked Example: A Step-by-Step Calculation | 实例精讲：逐步计算
Let us apply the procedure to a classic exam problem. Find the absolute maximum and absolute minimum of
f(x) = x³ − 3x + 1 on [0, 2
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