Using the Candidates Test to Determine Absolute (Global) Extrema · 使用候选测试法确定绝对（全局）极值
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Candidates Test/ˈkændɪdeɪts test/ | 候选点判别法 | hòu xuǎn diǎn pàn bié fǎ |
Finding the true highest and lowest
- On a closed interval, the EVT promises an absolute max and min exist. Where are they?
- They can only be at a critical point or at an endpoint — a short list of suspects.
- The Candidates Test 候选点判别法 just evaluates $f$ at every suspect and compares.
- No sign charts needed — the biggest output wins, the smallest loses.
找出真正的最高和最低
- 在闭区间上,EVT 保证绝对最大值和最小值存在。它们在哪里?
- 它们只能在临界点或端点——一份简短的嫌疑名单。
- 候选点检验只需在每个嫌疑点求 $f$ 的值并比较。
- 不需要符号表——最大的输出胜出,最小的落败。
The three steps
- 1. Find all critical points in $[a,b]$ (where $f'=0$ or is undefined).
- 2. Evaluate $f$ at each critical point and at both endpoints $a$ and $b$.
- 3. Compare the values: the largest is the absolute maximum, the smallest the absolute minimum.
- Report both the value and its location.
三个步骤
- 1. 找出 $[a,b]$ 内所有临界点($f'=0$ 或无定义)。
- 2. 在每个临界点和两个端点 $a$、$b$ 处求 $f$ 的值。
- 3. 比较这些值:最大的是绝对最大值,最小的是绝对最小值。
- 报告值及其位置。
Endpoints and critical points · 端点和临界点
y = ax³ + cx
The absolute max/min on a closed interval hide among the endpoints and the interior critical points — evaluate and compare. · 闭区间上的绝对最大/最小值隐藏在端点和内部临界点中——进行评估并比较。
To find absolute extrema on $[a,b]$, evaluate $f$ at which points? · 为了在 $[a,b]$ 上找到绝对极值，应在 哪些 点评估 $f$？
Critical points plus both endpoints — that is the full candidate list. · 临界点加上两个端点——这就是全部候选列表。
Don't forget the endpoints
- On a closed interval, the extremes often sit at the endpoints, not a critical point.
- A monotonic function (always increasing) has its max at the right end and min at the left.
- So you must always include $f(a)$ and $f(b)$ in your candidate list.
- Missing an endpoint is the most common way to get the wrong absolute extremum.
别忘了端点
- 在闭区间上,极端点常落在端点,而非临界点。
- 单调函数(一直递增)的最大值在右端、最小值在左端。
- 所以你必须始终把 $f(a)$ 和 $f(b)$ 纳入候选名单。
- 漏掉一个端点是求错绝对极值最常见的原因。
For · 支持 $f(x)=x^3-3x$ on $[0,2]$, with $f(0)=0$, $f(1)=-2$, $f(2)=2$, what is the absolute maximum value? · 对于$f(x)=x^3-3x$在$[0,2]$上，给定$f(0)=0$、$f(1)=-2$、$f(2)=2$，绝对最大值是多少？
Largest candidate value is $2$ (at the endpoint $x=2$). · 最大的候选值是$2$（在端点$x=2$处）。
Same function and candidates: what is the absolute minimum value? · 相同的函数和候选值：绝对最小值是多少？
Smallest candidate value is $-2$ (at $x=1$). · 最小的候选值是$-2$（在$x=1$处）。
You can skip the endpoints and still be sure of the absolute extrema. · 你可以跳过端点，仍然确定绝对极值。
Extremes often sit at endpoints; skipping them can give a wrong answer. · 极值通常位于端点；跳过它们可能导致错误答案。
Compare, don't classify
- Unlike the First Derivative Test, the Candidates Test doesn't care what kind each point is.
- It simply asks: of all the candidate outputs, which is biggest and which smallest?
- One evaluation per candidate, then a comparison — clean and reliable on a closed interval.
- (It needs a closed interval; on an open interval an absolute extremum may not exist.)
比较,而非判定
- 与一阶导数检验不同,候选点检验不在意每个点是哪种。
- 它只问:在所有候选输出中,哪个最大、哪个最小?
- 每个候选求一次值,再比较——在闭区间上干净又可靠。
- (它需要闭区间;在开区间上绝对极值可能不存在。)
The Candidates Test requires a ____ interval, where the EVT guarantees extrema exist. · 候选测试需要一个____区间，其中EVT保证极值存在。
On an open interval an absolute extremum may not exist. · 在开区间上，绝对极值可能不存在。
The absolute maximum is the candidate with the largest... · 绝对最大值是具有最大...的候选值
Compare outputs $f(x)$, not the inputs. · 比较输出$f(x)$，而不是输入。
The Candidates Test requires a closed interval $[a,b]$ and must include the endpoints. Evaluating only the critical points misses extremes that live at $a$ or $b$. And compare $f$-values, not $x$-values — the winner is the largest output, wherever it occurs.
候选点检验需要闭区间 $[a,b]$,并必须包含端点。只求临界点会漏掉住在 $a$ 或 $b$ 的极端点。而且要比较 $f$ 值,而非 $x$ 值——胜者是最大的输出,无论它出现在哪里。
Find the absolute extrema of $f(x)=x^3-3x$ on $[0,2]$.
- Critical points in $[0,2]$: $f'=3(x-1)(x+1)=0$ at $x=1$ (only $x=1$ is in range).
- Evaluate candidates: $f(0)=0$, $f(1)=-2$, $f(2)=2$.
- Absolute max $=2$ at $x=2$ (endpoint); absolute min $=-2$ at $x=1$.
求 $f(x)=x^3-3x$ 在 $[0,2]$ 上的绝对极值。
- $[0,2]$ 内的临界点:$f'=3(x-1)(x+1)=0$ 在 $x=1$(只有 $x=1$ 在范围内)。
- 求候选值:$f(0)=0$,$f(1)=-2$,$f(2)=2$。
- 绝对最大值 $=2$ 在 $x=2$(端点);绝对最小值 $=-2$ 在 $x=1$。
The Candidates Test finds absolute extrema on a closed interval: list the critical points and both endpoints, evaluate $f$ at each, and compare. Largest value = absolute maximum, smallest = absolute minimum. Always include the endpoints, and compare outputs, not inputs.
候选点检验在闭区间上求绝对极值:列出临界点和两个端点,在每处求 $f$ 的值,再比较。最大值 = 绝对最大值,最小值 = 绝对最小值。永远包含端点,并比较输出而非输入。