Using the Second Derivative Test to Determine Extrema · 使用二阶导数测试确定极值
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Second Derivative Test/ˈsekənd dɪˈrɪvətɪv test/ | 二阶导数判别法 | èr jiē dǎo shù pàn bié fǎ |
| inconclusive/ɪŋkənˈkluːsɪv/ | 无法判定 | wú fǎ pàn dìng |
Classify an extremum with one number
- The First Derivative Test reads sign changes; the Second Derivative Test 二阶导数判别法 is often quicker.
- At a critical point where $f'(c)=0$, just check the sign of $f''(c)$.
- Concave up at a flat spot → a valley; concave down at a flat spot → a peak.
- One evaluation of $f''$ can settle it — no sign chart needed.
用一个数判定极值
- 一阶导数检验读变号;二阶导数检验通常更快。
- 在 $f'(c)=0$ 的临界点,只需检查 $f''(c)$ 的符号。
- 平坦处上凹 → 谷;平坦处下凹 → 峰。
- 求一次 $f''$ 就能定下——无需符号表。
$f''(c) > 0$: a minimum
- If $f'(c)=0$ and $f''(c)>0$, then $f$ has a relative minimum at $c$.
- Concave up (a cup) with a horizontal tangent = the bottom of a valley.
- Intuition: the slope is zero and increasing, so the curve turns upward.
$f''(c) > 0$:最小值
- 若 $f'(c)=0$ 且 $f''(c)>0$,则 $f$ 在 $c$ 处有局部最小值。
- 上凹(杯)加水平切线 = 谷底。
- 直觉:斜率为零且增大,所以曲线向上转。
Cup at the min, cap at the max · 最小值处像杯子，最大值处像帽子
y = ax³ + cx
At the local min the curve is concave up ($f''>0$); at the local max it is concave down ($f''<0$). · 在局部最小值处曲线凹向上（$f''>0$）；在局部最大值处凹向下（$f''<0$）。
If $f'(c)=0$ and $f''(c)>0$, then $c$ is a... · 如果$f'(c)=0$且$f''(c)>0$，那么$c$是一个...
Concave up at a flat spot = minimum. · 平坦处凹向上 = 最小值。
$f''(c) < 0$: a maximum
- If $f'(c)=0$ and $f''(c)<0$, then $f$ has a relative maximum at $c$.
- Concave down (a cap) with a horizontal tangent = the top of a hill.
- The slope is zero and decreasing, so the curve turns downward.
$f''(c) < 0$:最大值
- 若 $f'(c)=0$ 且 $f''(c)<0$,则 $f$ 在 $c$ 处有局部最大值。
- 下凹(帽)加水平切线 = 山顶。
- 斜率为零且减小,所以曲线向下转。
If $f'(c)=0$ and $f''(c)<0$, then $c$ is a relative . · 如果$f'(c)=0$且$f''(c)<0$，那么$c$是一个相对。
Concave down at a flat spot = maximum. · 平坦处凹向下 = 最大值。
When it says nothing
- If $f''(c)=0$, the test is inconclusive 无法判定 — it gives no answer.
- The point could be a max, a min, or neither (like $y=x^3$ at $0$, or $y=x^4$ at $0$).
- Fall back on the First Derivative Test (sign change of $f'$) in that case.
- So: use the Second Derivative Test first if $f''$ is easy; switch to the first test if $f''(c)=0$.
当它什么都不说时
- 若 $f''(c)=0$,检验不确定——它给不出答案。
- 该点可能是最大、最小或都不是(如 $y=x^3$ 在 $0$,或 $y=x^4$ 在 $0$)。
- 此时退回到一阶导数检验($f'$ 的变号)。
- 所以:若 $f''$ 好算就先用二阶导数检验;若 $f''(c)=0$ 就换一阶检验。
If $f''(c)=0$, the Second Derivative Test proves there is no extremum at $c$. · 如果$f''(c)=0$，二阶导数测试证明在$c$处没有极值。
It is inconclusive — use the First Derivative Test instead. · 这是不确定的——改用一阶导数测试。
For · 支持 $f(x)=x^3-3x$ ($f''=6x$), classify $x=-1$. · 对于$f(x)=x^3-3x$（$f''=6x$），分类$x=-1$。
$f''(-1)=-6<0$ → relative maximum. · $f''(-1)=-6<0$ → 相对最大值。
The Second Derivative Test is applied at a critical point where... · 二阶导数测试应用于一个临界点，该点满足...
It classifies critical points where the tangent is horizontal, $f'(c)=0$. · 它对切线水平的临界点进行分类，即$f'(c)=0$。
When should you fall back on the First Derivative Test? · 何时应回退到一阶导数测试？
The first test always works; the second is just often faster. · 第一个测试总是有效；第二个通常更快。
The Second Derivative Test needs $f'(c)=0$ first — it classifies critical points where the tangent is horizontal. And $f''(c)=0$ is inconclusive, not "neither": you must then use the First Derivative Test. Don't conclude "no extremum" just because $f''(c)=0$.
二阶导数检验首先需要 $f'(c)=0$——它判定切线水平的临界点。而 $f''(c)=0$ 是不确定,不是"都不是":你必须再用一阶导数检验。别因为 $f''(c)=0$ 就断定"无极值"。
Classify the critical points of $f(x)=x^3-3x$ using $f''$.
- $f'(x)=3x^2-3=0$ at $x=\pm1$; $\;f''(x)=6x$.
- At $x=1$: $f''(1)=6>0$ → relative minimum.
- At $x=-1$: $f''(-1)=-6<0$ → relative maximum.
用 $f''$ 判定 $f(x)=x^3-3x$ 的临界点。
- $f'(x)=3x^2-3=0$ 在 $x=\pm1$;$\;f''(x)=6x$。
- 在 $x=1$:$f''(1)=6>0$ → 局部最小值。
- 在 $x=-1$:$f''(-1)=-6<0$ → 局部最大值。
The Second Derivative Test: at a critical point with $f'(c)=0$, $f''(c)>0$ gives a relative minimum and $f''(c)<0$ a relative maximum. If $f''(c)=0$ the test is inconclusive — fall back on the First Derivative Test.
二阶导数检验:在 $f'(c)=0$ 的临界点,$f''(c)>0$ 给出局部最小值,$f''(c)<0$ 给出局部最大值。若 $f''(c)=0$ 则检验不确定——退回一阶导数检验。