Unit E1: Principles
of Differential Calculus单元 E1：微分学原理
The opening sub-unit of Topic 5 (Calculus) — the foundation everything else in Topic E rests on. You build the derivative from the limit, then learn to read it geometrically (slope of the tangent), analytically (sign tells you increasing / decreasing), and structurally (continuity and differentiability for HL). The Power Rule + tangent / normal lines bridge to Unit E2 (techniques) and E4 (problem-solving); concavity prepares the second-derivative test in E4.Topic 5（微积分）的开篇子单元 —— 整个 Topic E 都建立在它之上。本单元由极限（limit）出发构造导数（derivative），并学会三种读法：几何上（切线斜率）、分析上（符号决定单调性）、结构上（HL 的连续性与可导性）。幂法则与切线/法线衔接 Unit E2（求导技巧）和 E4（应用题）；凹凸性为 E4 的二阶导数检验做铺垫。
How to use this guide本指南使用说明
E1 is foundational. The Power Rule and tangent-line formula are the two highest-ROI cram items — they show up on almost every Paper 1 and Paper 2. The HL sections (continuity, concavity) are less computational but feed Paper 3.E1 是基础。幂法则与切线方程是性价比最高的两个考点 —— 几乎每张 Paper 1 与 Paper 2 都会出现。HL 部分（连续性、凹凸性）计算量小，但是 Paper 3 的素材。
Memorise (1) the Power Rule $\tfrac{d}{dx}(x^{n}) = n x^{n-1}$, (2) the tangent-line formula $y - f(a) = f'(a)(x - a)$, and (3) the sign convention "$f' > 0 \Rightarrow$ increasing." Skip first-principles derivations — but be able to recognise the difference quotient on multiple choice.
背熟三件事：(1) 幂法则 $\tfrac{d}{dx}(x^{n}) = n x^{n-1}$；(2) 切线方程 $y - f(a) = f'(a)(x - a)$；(3) 符号约定"$f' > 0 \Rightarrow$ 单调递增"。跳过 first-principles 推导，但要认得选择题里的差商形式。
Derive $\tfrac{d}{dx}(x^{2})$, $\tfrac{d}{dx}(x^{3})$, and $\tfrac{d}{dx}(1/x)$ from first principles in one sitting. Memorise the four canonical non-differentiable shapes (corner, cusp, vertical tangent, jump). Practise classifying stationary points by both the first-derivative sign chart and the second-derivative test.
一次性从 first principles 推出 $\tfrac{d}{dx}(x^{2})$、$\tfrac{d}{dx}(x^{3})$ 与 $\tfrac{d}{dx}(1/x)$。背熟四种典型不可导图形（角点、尖点、垂直切线、跳跃）。稳定点分类要会两种方法：一阶导符号表 + 二阶导检验。
The Idea of a Limit极限的概念 SL 5.1
- Direct substitution — if $f$ is continuous at $a$, just compute $f(a)$.
- Factor & cancel — if direct substitution gives $\tfrac{0}{0}$, factor numerator and denominator, cancel the common $(x - a)$, then substitute.
- Rationalise — when a square-root term causes $\tfrac{0}{0}$, multiply top and bottom by the conjugate.
- 直接代入 —— 若 $f$ 在 $a$ 处连续，算 $f(a)$ 即可。
- 因式分解 & 约分 —— 若直接代入得 $\tfrac{0}{0}$，把分子分母分别分解、约掉公因子 $(x - a)$，再代入。
- 有理化（
rationalise） —— 若根式造成 $\tfrac{0}{0}$，用共轭因子（conjugate）同乘分子分母。
Evaluate $\displaystyle\lim_{x \to 2} \dfrac{x^{2} - 4}{x - 2}$.求 $\displaystyle\lim_{x \to 2} \dfrac{x^{2} - 4}{x - 2}$。
Direct substitution fails. Plugging $x = 2$ gives $\tfrac{4 - 4}{2 - 2} = \tfrac{0}{0}$ — indeterminate.
直接代入失败。代入 $x = 2$ 得 $\tfrac{4 - 4}{2 - 2} = \tfrac{0}{0}$ —— 未定式。
Factor. $x^{2} - 4 = (x - 2)(x + 2)$, so
分解。$x^{2} - 4 = (x - 2)(x + 2)$，故
$$ \frac{x^{2} - 4}{x - 2} \;=\; \frac{(x - 2)(x + 2)}{x - 2} \;=\; x + 2 \quad (x \ne 2). $$Substitute. $\displaystyle\lim_{x \to 2} (x + 2) = 2 + 2 = 4$.
代入。$\displaystyle\lim_{x \to 2} (x + 2) = 2 + 2 = 4$。
Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x}$.求 $\displaystyle\lim_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x}$。
Direct substitution gives $\tfrac{0}{0}$. Multiply top and bottom by the conjugate $\sqrt{x + 1} + 1$:
直接代入得 $\tfrac{0}{0}$。同乘共轭 $\sqrt{x + 1} + 1$：
$$ \frac{\sqrt{x + 1} - 1}{x} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1} \;=\; \frac{(x + 1) - 1}{x (\sqrt{x + 1} + 1)} \;=\; \frac{x}{x (\sqrt{x + 1} + 1)} \;=\; \frac{1}{\sqrt{x + 1} + 1}. $$Substitute. $\displaystyle\lim_{x \to 0} \dfrac{1}{\sqrt{x + 1} + 1} = \dfrac{1}{1 + 1} = \dfrac{1}{2}$.
代入。$\displaystyle\lim_{x \to 0} \dfrac{1}{\sqrt{x + 1} + 1} = \dfrac{1}{1 + 1} = \dfrac{1}{2}$。
▸ Going deeper — what "indeterminate" actually means▸ 深入 —— "未定式"究竟是什么意思
"$\tfrac{0}{0}$" is not a number — it's a shape a limit can take that doesn't pin down the answer. The same shape can resolve to $0$, $1$, $\infty$, or any real number depending on how the numerator and denominator approach zero relative to each other. Example: $\lim_{x \to 0} \tfrac{x^{2}}{x} = 0$ but $\lim_{x \to 0} \tfrac{x}{x^{2}}$ diverges, even though both are "$\tfrac{0}{0}$" at face value. The algebraic manipulation (factor / rationalise) is how you uncover which case you're in.
"$\tfrac{0}{0}$"不是一个数，而是一种极限可能呈现的形态，本身并不锁定答案。同一形态可解出 $0$、$1$、$\infty$ 或任意实数 —— 取决于分子分母各自趋近 $0$ 的速率。例：$\lim_{x \to 0} \tfrac{x^{2}}{x} = 0$，但 $\lim_{x \to 0} \tfrac{x}{x^{2}}$ 发散，二者表面都是"$\tfrac{0}{0}$"。代数化简（分解 / 有理化）就是揭示真实情形的工具。
The Derivative from First Principles导数的定义（first principles） SL 5.1
- Notation: $f'(x), \;\tfrac{dy}{dx}, \;\tfrac{d}{dx}[f(x)], \;\dot{y}$ — all the same object.
- Units: if $y$ is in metres and $x$ in seconds, $\tfrac{dy}{dx}$ is in m/s.
difference quotient）$\tfrac{f(x+h) - f(x)}{h}$ 是过 $(x, f(x))$ 与 $(x + h, f(x + h))$ 的割线（secant line）的斜率。$h \to 0$ 时割线退化为切线（tangent），斜率即瞬时变化率 $f'(x)$。
- 记号：$f'(x), \;\tfrac{dy}{dx}, \;\tfrac{d}{dx}[f(x)], \;\dot{y}$ —— 同一对象。
- 单位：若 $y$ 以米为单位、$x$ 以秒为单位，则 $\tfrac{dy}{dx}$ 以 m/s 为单位。
Show that the derivative of $f(x) = x^{2}$ is $f'(x) = 2x$.证明 $f(x) = x^{2}$ 的导数为 $f'(x) = 2x$。
Apply the definition.
套用定义。
$$ f'(x) \;=\; \lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h}. $$Expand numerator and simplify.
展开分子并化简。
$$ (x + h)^{2} - x^{2} \;=\; x^{2} + 2xh + h^{2} - x^{2} \;=\; 2xh + h^{2} \;=\; h(2x + h). $$Cancel $h$ and take the limit.
约去 $h$ 后取极限。
$$ f'(x) \;=\; \lim_{h \to 0} \frac{h(2x + h)}{h} \;=\; \lim_{h \to 0} (2x + h) \;=\; 2x. \;\square $$Show that the derivative of $f(x) = 1/x$ is $f'(x) = -1/x^{2}$ (for $x \ne 0$).证明 $f(x) = 1/x$ 的导数为 $f'(x) = -1/x^{2}$（$x \ne 0$）。
Definition.
定义。
$$ f'(x) \;=\; \lim_{h \to 0} \frac{1}{h}\left[\frac{1}{x + h} - \frac{1}{x}\right]. $$Common denominator inside the brackets.
方括号内通分。
$$ \frac{1}{x + h} - \frac{1}{x} \;=\; \frac{x - (x + h)}{x(x + h)} \;=\; \frac{-h}{x(x + h)}. $$Substitute back and cancel $h$.
回代并约去 $h$。
$$ f'(x) \;=\; \lim_{h \to 0} \frac{1}{h} \cdot \frac{-h}{x(x + h)} \;=\; \lim_{h \to 0} \frac{-1}{x(x + h)} \;=\; -\frac{1}{x^{2}}. \;\square $$▸ Going deeper — the alternate form using $x = a$▸ 深入 —— 用 $x = a$ 的另一种形式
An equivalent definition computes $f'(a)$ at a specific point $a$:
在某固定点 $a$ 处，导数的等价定义为：
$$ f'(a) \;=\; \lim_{x \to a} \frac{f(x) - f(a)}{x - a}. $$This form is often cleaner when working at a named point — there's no $h$ to chase. Set $x = a + h$ to see the two are the same limit. The Paper 1 markscheme accepts either form; pick whichever produces less algebra.
在指定点处工作时该形式往往更简洁 —— 无需追踪 $h$。令 $x = a + h$ 即可见两式为同一极限。Paper 1 评分对两种形式都认可；用代数最少的那一种即可。
Power Rule & Tangent / Normal Lines幂法则与切线 / 法线 SL 5.1
- Sum rule: $\tfrac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$.
- Constant multiple: $\tfrac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$.
- Derivative of a constant: $\tfrac{d}{dx}(c) = 0$.
power rule）（对任意实指数 $n$ 成立，含负指数与分数指数）：
$$ \frac{d}{dx}(x^{n}) \;=\; n x^{n - 1}. $$
配合线性性：
- 和差法则：$\tfrac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$。
- 常数倍法则：$\tfrac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$。
- 常数的导数：$\tfrac{d}{dx}(c) = 0$。
Differentiate $f(x) = 4x^{3} - 5x^{2} + 7x - 9$.求 $f(x) = 4x^{3} - 5x^{2} + 7x - 9$ 的导数。
Apply power rule term-by-term with the constant multiple rule.
逐项使用幂法则与常数倍法则。
$$ f'(x) \;=\; 4 \cdot 3x^{2} - 5 \cdot 2x + 7 \cdot 1 - 0 \;=\; 12 x^{2} - 10 x + 7. $$Differentiate $g(x) = 3\sqrt{x} - \dfrac{2}{x^{2}}$ (rewrite as exponents first).求 $g(x) = 3\sqrt{x} - \dfrac{2}{x^{2}}$ 的导数（先化为指数形式）。
Rewrite. $g(x) = 3x^{1/2} - 2x^{-2}$.
改写。$g(x) = 3x^{1/2} - 2x^{-2}$。
Apply power rule.
应用幂法则。
$$ g'(x) \;=\; 3 \cdot \tfrac{1}{2} x^{-1/2} - 2 \cdot (-2) x^{-3} \;=\; \frac{3}{2\sqrt{x}} + \frac{4}{x^{3}}. $$Find the equations of the tangent and the normal to the curve $y = x^{3} - 2x$ at the point where $x = 1$.求曲线 $y = x^{3} - 2x$ 在 $x = 1$ 处的切线方程与法线方程。
Step 1 — Point on the curve. $f(1) = 1 - 2 = -1$, so the point is $(1, -1)$.
第 1 步 —— 曲线上的点。$f(1) = 1 - 2 = -1$，故点为 $(1, -1)$。
Step 2 — Slope of tangent. $f'(x) = 3x^{2} - 2$, so $f'(1) = 3 - 2 = 1$.
第 2 步 —— 切线斜率。$f'(x) = 3x^{2} - 2$，故 $f'(1) = 3 - 2 = 1$。
Step 3 — Tangent line. $y - (-1) = 1(x - 1)$, i.e. $y = x - 2$.
第 3 步 —— 切线方程。$y - (-1) = 1(x - 1)$，即 $y = x - 2$。
Step 4 — Normal line. Slope $= -1/f'(1) = -1/1 = -1$. So $y - (-1) = -1(x - 1)$, i.e. $y = -x$.
第 4 步 —— 法线方程。斜率 $= -1/f'(1) = -1/1 = -1$。故 $y - (-1) = -1(x - 1)$，即 $y = -x$。
▸ Going deeper — why tangent and normal slopes multiply to $-1$▸ 深入 —— 切线与法线斜率为何乘积 $= -1$
Two lines are perpendicular iff their slopes are negative reciprocals: $m_{1} m_{2} = -1$. The tangent at $(a, f(a))$ has slope $f'(a)$; the normal, being perpendicular, has slope $-1/f'(a)$. The exception is when $f'(a) = 0$ (horizontal tangent) — there the normal is vertical, with equation $x = a$ (no defined slope). The opposite exception: a vertical tangent (e.g. $y = \sqrt[3]{x}$ at $x = 0$, where $f'(x) = \tfrac{1}{3} x^{-2/3} \to \infty$) has a horizontal normal $y = f(a)$.
两直线垂直 $\Leftrightarrow$ 斜率互为负倒数：$m_{1} m_{2} = -1$。$(a, f(a))$ 处的切线斜率为 $f'(a)$，故法线斜率为 $-1/f'(a)$。例外：$f'(a) = 0$（水平切线）时法线是垂直的，方程 $x = a$（无定义斜率）。反向例外：若切线垂直（如 $y = \sqrt[3]{x}$ 在 $x = 0$ 处，$f'(x) = \tfrac{1}{3} x^{-2/3} \to \infty$），法线为水平的 $y = f(a)$。
Increasing / Decreasing Functions & Stationary Points单调性与稳定点 SL 5.2
- $f'(x) > 0$ on an interval $\Rightarrow$ $f$ is strictly increasing there.
- $f'(x) < 0$ on an interval $\Rightarrow$ $f$ is strictly decreasing there.
- $f'(x) = 0$ at a point $\Rightarrow$ that point is stationary (slope is zero; tangent is horizontal).
- $f'$ goes $+ \to -$: local maximum.
- $f'$ goes $- \to +$: local minimum.
- $f'$ keeps the same sign on either side: horizontal point of inflection (no local extremum).
- 区间上 $f'(x) > 0$ $\Rightarrow$ $f$ 在该区间严格递增（
strictly increasing）。 - 区间上 $f'(x) < 0$ $\Rightarrow$ $f$ 在该区间严格递减。
- 某点 $f'(x) = 0$ $\Rightarrow$ 该点为稳定点（
stationary point；斜率为零，切线水平）。
first-derivative test）分类 $x = a$ 处的稳定点：
- $f'$ 由 $+ \to -$：局部极大值。
- $f'$ 由 $- \to +$：局部极小值。
- $f'$ 两侧同号：水平拐点（
horizontal point of inflection），无局部极值。
Find the intervals on which $f(x) = x^{3} - 3x^{2} + 2$ is increasing / decreasing, and classify all stationary points.求 $f(x) = x^{3} - 3x^{2} + 2$ 的增减区间，并对所有稳定点分类。
Step 1 — Derivative. $f'(x) = 3x^{2} - 6x = 3x(x - 2)$.
第 1 步 —— 求导。$f'(x) = 3x^{2} - 6x = 3x(x - 2)$。
Step 2 — Stationary points. $f'(x) = 0 \Rightarrow x = 0$ or $x = 2$.
第 2 步 —— 稳定点。$f'(x) = 0 \Rightarrow x = 0$ 或 $x = 2$。
Step 3 — Sign chart of $f'$. The parabola $3x(x-2)$ opens upward with roots $0$ and $2$:
第 3 步 —— $f'$ 的符号表。抛物线 $3x(x-2)$ 开口向上，零点 $0$、$2$：
| Interval区间 | $x < 0$ | $0 < x < 2$ | $x > 2$ |
|---|---|---|---|
| $f'(x)$ | $+$ | $-$ | $+$ |
| $f$ | increasing递增 | decreasing递减 | increasing递增 |
Step 4 — Classify. At $x = 0$: $f'$ goes $+ \to -$, so local max at $(0, 2)$. At $x = 2$: $f'$ goes $- \to +$, so local min at $(2, -2)$.
第 4 步 —— 分类。$x = 0$ 处：$f'$ 由 $+ \to -$，局部极大值在 $(0, 2)$。$x = 2$ 处：$f'$ 由 $- \to +$，局部极小值在 $(2, -2)$。
Continuity & Differentiability连续性与可导性 HL AHL 5.12
Differentiability at $x = a$: the limit $f'(a) = \displaystyle\lim_{h \to 0} \tfrac{f(a + h) - f(a)}{h}$ exists. Equivalently, the left-derivative equals the right-derivative.
One-way implication. Differentiable $\Rightarrow$ continuous. (The converse is false.) Four canonical shapes that are continuous but not differentiable at a point:
- Corner — left and right slopes differ. Example: $f(x) = |x|$ at $x = 0$.
- Cusp — slopes go to $+\infty$ on one side and $-\infty$ on the other. Example: $f(x) = x^{2/3}$ at $x = 0$.
- Vertical tangent — both slopes go to $+\infty$ (or both to $-\infty$). Example: $f(x) = \sqrt[3]{x}$ at $x = 0$.
- Discontinuity — not even continuous, hence not differentiable. Example: a step function at the jump.
continuous）：$\displaystyle\lim_{x \to a} f(x) = f(a)$。等价地，左极限、右极限、函数值三者一致。在 $x = a$ 处可导（
differentiable）：极限 $f'(a) = \displaystyle\lim_{h \to 0} \tfrac{f(a + h) - f(a)}{h}$ 存在。等价地，左导数（left derivative）等于右导数。单向蕴含。可导 $\Rightarrow$ 连续（反之不成立）。四种"连续但不可导"的典型形状：
- 角点（
corner） —— 左右斜率不同。例：$f(x) = |x|$ 在 $x = 0$。 - 尖点（
cusp） —— 一侧斜率趋于 $+\infty$，另一侧趋于 $-\infty$。例：$f(x) = x^{2/3}$ 在 $x = 0$。 - 垂直切线（
vertical tangent） —— 两侧斜率同趋于 $+\infty$（或同趋于 $-\infty$）。例：$f(x) = \sqrt[3]{x}$ 在 $x = 0$。 - 不连续 —— 连连续都不是，自然不可导。例：阶跃函数在跳跃处。
Show that $f(x) = |x|$ is continuous but not differentiable at $x = 0$.证明 $f(x) = |x|$ 在 $x = 0$ 处连续但不可导。
Continuity. $\displaystyle\lim_{x \to 0^{-}} |x| = 0$, $\displaystyle\lim_{x \to 0^{+}} |x| = 0$, and $f(0) = 0$. All three agree, so $f$ is continuous at $0$. ✓
连续性。$\displaystyle\lim_{x \to 0^{-}} |x| = 0$，$\displaystyle\lim_{x \to 0^{+}} |x| = 0$，$f(0) = 0$。三者一致，故 $f$ 在 $0$ 处连续。✓
Differentiability — compute one-sided derivatives.
可导性 —— 计算单侧导数。
$$ \lim_{h \to 0^{+}} \frac{|0 + h| - 0}{h} \;=\; \lim_{h \to 0^{+}} \frac{h}{h} \;=\; 1. $$ $$ \lim_{h \to 0^{-}} \frac{|0 + h| - 0}{h} \;=\; \lim_{h \to 0^{-}} \frac{-h}{h} \;=\; -1. $$Left-derivative ($-1$) $\ne$ right-derivative ($+1$), so the two-sided limit does not exist. $f$ is not differentiable at $0$. ✗
左导数（$-1$）$\ne$ 右导数（$+1$），故双侧极限不存在。$f$ 在 $0$ 处不可导。 ✗
Geometric reading. The graph of $|x|$ has a "corner" at the origin — the tangent direction can't be determined.
几何理解。$|x|$ 的图像在原点形成"角点"，无法确定唯一的切线方向。
▸ Going deeper — why differentiable implies continuous▸ 深入 —— 为何"可导 $\Rightarrow$ 连续"
If $f'(a)$ exists, then
若 $f'(a)$ 存在，则
$$ \lim_{x \to a} \bigl[f(x) - f(a)\bigr] \;=\; \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \cdot (x - a) \;=\; f'(a) \cdot 0 \;=\; 0. $$So $\lim_{x \to a} f(x) = f(a)$ — exactly the definition of continuity at $a$. The converse fails because the difference quotient can have left/right limits that disagree even when $f(x) \to f(a)$.
即 $\lim_{x \to a} f(x) = f(a)$ —— 这正是连续性在 $a$ 处的定义。反向不成立：差商的左右极限可以不相等，即使 $f(x) \to f(a)$。
Concavity & Points of Inflection凹凸性与拐点 HL AHL 5.14
- $f''(x) > 0$ $\Rightarrow$ curve is concave up (opens like a cup; "smile"). $f'$ is increasing.
- $f''(x) < 0$ $\Rightarrow$ curve is concave down (opens like a cap; "frown"). $f'$ is decreasing.
- Point of inflection at $x = a$: $f''$ changes sign across $a$. The graph switches concavity.
- $f''(a) > 0$: local minimum.
- $f''(a) < 0$: local maximum.
- $f''(a) = 0$: inconclusive — fall back on the first-derivative sign chart.
- $f''(x) > 0$ $\Rightarrow$ 曲线凹向上（
concave up；像杯子；"微笑"）。$f'$ 递增。 - $f''(x) < 0$ $\Rightarrow$ 曲线凹向下（
concave down；像帽子；"皱眉"）。$f'$ 递减。 - 拐点（
point of inflection）$x = a$：$f''$ 在 $a$ 处变号，图像凹凸性切换。
second-derivative test）分类稳定点 $f'(a) = 0$：
- $f''(a) > 0$：局部极小值。
- $f''(a) < 0$：局部极大值。
- $f''(a) = 0$：不确定 —— 退回一阶导符号表。
Let $f(x) = x^{4} - 4x^{3}$. Find all stationary points and classify them using the second-derivative test (falling back on the first-derivative test where needed). Find all points of inflection.设 $f(x) = x^{4} - 4x^{3}$。求所有稳定点并用二阶导数检验分类（必要时退回一阶导检验）。求所有拐点。
Derivatives. $f'(x) = 4x^{3} - 12x^{2} = 4x^{2}(x - 3)$. $f''(x) = 12x^{2} - 24x = 12x(x - 2)$.
求导。$f'(x) = 4x^{3} - 12x^{2} = 4x^{2}(x - 3)$。$f''(x) = 12x^{2} - 24x = 12x(x - 2)$。
Stationary points. $f'(x) = 0 \Rightarrow x = 0$ (double root) or $x = 3$.
稳定点。$f'(x) = 0 \Rightarrow x = 0$（重根）或 $x = 3$。
Classify $x = 3$. $f''(3) = 12(3)(1) = 36 > 0$. So $(3, f(3)) = (3, 81 - 108) = (3, -27)$ is a local minimum.
分类 $x = 3$。$f''(3) = 12(3)(1) = 36 > 0$。故 $(3, f(3)) = (3, 81 - 108) = (3, -27)$ 为局部极小值。
Classify $x = 0$. $f''(0) = 0$ — inconclusive. Fall back on the first-derivative sign chart. $f'(x) = 4x^{2}(x - 3)$: $4x^{2} \ge 0$ always; $(x - 3) < 0$ for $x < 3$. So $f'(x) \le 0$ for $x \le 3$, and the sign does not change at $x = 0$. Therefore $(0, 0)$ is a horizontal point of inflection — stationary, but not an extremum.
分类 $x = 0$。$f''(0) = 0$ —— 不确定。退回一阶导符号表。$f'(x) = 4x^{2}(x - 3)$：$4x^{2} \ge 0$ 恒成立；$(x - 3) < 0$ 当 $x < 3$。故 $x \le 3$ 时 $f'(x) \le 0$，符号在 $x = 0$ 处不变。$(0, 0)$ 是水平拐点 —— 稳定，但不是极值。
Inflection points. $f''(x) = 0 \Rightarrow x = 0$ or $x = 2$. Check sign change:
拐点。$f''(x) = 0 \Rightarrow x = 0$ 或 $x = 2$。检查变号：
| Interval区间 | $x < 0$ | $0 < x < 2$ | $x > 2$ |
|---|---|---|---|
| $f''(x)$ | $+$ | $-$ | $+$ |
| Concavity凹凸性 | up凹上 | down凹下 | up凹上 |
$f''$ changes sign at both $x = 0$ and $x = 2$. Inflections at $(0, 0)$ and $(2, f(2)) = (2, 16 - 32) = (2, -16)$. Note that $(0, 0)$ is both a horizontal inflection point and a stationary point — these aren't exclusive.
$f''$ 在 $x = 0$ 与 $x = 2$ 处都变号。拐点为 $(0, 0)$ 与 $(2, f(2)) = (2, 16 - 32) = (2, -16)$。注意 $(0, 0)$ 同时是水平拐点与稳定点 —— 二者不互斥。
▸ Going deeper — why $f''(a) = 0$ is necessary for an inflection▸ 深入 —— 为何拐点处必有 $f''(a) = 0$
A point of inflection is where $f''$ changes sign. If $f''$ is continuous (true for most IB functions you'll meet), the intermediate value theorem forces $f''(a) = 0$ at the crossing point. The converse fails: $f''(a) = 0$ without sign change gives a "flat" zero of $f''$ that's not an inflection (e.g. $f(x) = x^{4}$ at $x = 0$). On Paper 1B, you need to verify the sign change step, not just solve $f''(x) = 0$ — markschemes deduct M1 if you skip it.
拐点是 $f''$ 变号之处。若 $f''$ 连续（IB 出现的多数函数都满足），由介值定理（IVT），过零处必有 $f''(a) = 0$。反向不成立：$f''(a) = 0$ 但符号未变，是 $f''$ 的"平稳零点"，并非拐点（如 $f(x) = x^{4}$ 在 $x = 0$）。Paper 1B 必须验证变号，不能只解 $f''(x) = 0$ —— 评分细则会因省略而扣 M1。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- Cite the definition first. Open with "$f'(x) = \lim_{h \to 0} \tfrac{f(x+h) - f(x)}{h}$" — this is the M1.
- 先写定义。用 "$f'(x) = \lim_{h \to 0} \tfrac{f(x+h) - f(x)}{h}$" 开头 —— 这是 M1。
- Factor $h$ before taking the limit. Substituting $h = 0$ before cancellation costs the A1.
- 取极限之前先提因子 $h$。在约分前代 $h = 0$ 会丢 A1。
- State the limit explicitly. Write "as $h \to 0$, $(2x + h) \to 2x$" — don't just write the final answer.
- 显式写极限步骤。写出"当 $h \to 0$，$(2x + h) \to 2x$"，不要只写终值。
- Always check the point is on the curve. Compute $f(a)$ first — students forget and use the asked-for $y$-coordinate that may not satisfy the equation.
- 务必核对点在曲线上。先算 $f(a)$ —— 学生常忘记，用题给的 $y$ 坐标当点斜式的 $y_{0}$，但该坐标未必在曲线上。
- Use point-slope, not slope-intercept. $y - f(a) = m(x - a)$ is faster and avoids re-deriving the $y$-intercept.
- 用点斜式而非斜截式。$y - f(a) = m(x - a)$ 更快，无需再求 $y$ 截距。
- Normal: $m_{\text{normal}} = -1/m_{\text{tangent}}$. Sign error here is the #1 deduction on normal-line problems.
- 法线：$m_{\text{normal}} = -1/m_{\text{tangent}}$。这里的符号错误是法线题第一大扣分项。
- Second-derivative test is faster when it works. Use it unless $f''(a) = 0$.
- 二阶导检验能用时更快。除非 $f''(a) = 0$，否则优先用它。
- If $f''(a) = 0$, switch to the first-derivative sign chart. Don't declare "inconclusive" and stop — that loses the A1.
- 若 $f''(a) = 0$，改用一阶导符号表。不要写"无法判断"就停 —— 那会丢 A1。
- "Stationary" $\ne$ "extremum." Horizontal inflections are stationary but not extrema. Always verify sign change.
- "稳定"$\ne$"极值"。水平拐点是稳定的但非极值。务必验证变号。
Flashcards闪卡
Unit E1 — Practice Quiz单元 E1——练习测验
Readiness Checklist备考清单
Tick each one when you can do it cold — without notes, without the formula box, on your first attempt.
每一条都要"裸做"做对（不看笔记、不看公式框、一次过）才打勾。
- Evaluate a $\tfrac{0}{0}$ limit by factoring and cancelling a common $(x - a)$用"分解 + 约 $(x - a)$"求 $\tfrac{0}{0}$ 型极限
- Rationalise a square-root expression to remove $\tfrac{0}{0}$用有理化消除根式中的 $\tfrac{0}{0}$
- Write down the first-principles definition of $f'(x)$ from memory凭记忆写出 $f'(x)$ 的 first principles 定义
- Derive $\tfrac{d}{dx}(x^{2})$ and $\tfrac{d}{dx}(1/x)$ from first principles用定义推 $\tfrac{d}{dx}(x^{2})$ 与 $\tfrac{d}{dx}(1/x)$
- Apply the power rule term-by-term, including negative and fractional exponents逐项使用幂法则（含负指数与分数指数）
- Convert $\sqrt{x}$ and $1/x^{n}$ into exponent form before differentiating求导前把 $\sqrt{x}$、$1/x^{n}$ 化为指数形式
- Find the equation of the tangent and the normal at a given $x$-value求曲线在给定 $x$ 处的切线与法线方程
- Find stationary points by solving $f'(x) = 0$ and classify with the sign chart解 $f'(x) = 0$ 求稳定点，并用符号表分类
- Identify a horizontal point of inflection (stationary but no sign change in $f'$)识别水平拐点（稳定但 $f'$ 不变号）
- HL Show a function is continuous but not differentiable at a point via one-sided limits用单侧极限证明函数在某点连续但不可导
- HL Recognise the four canonical non-differentiable shapes (corner, cusp, vertical tangent, jump)识别四种典型不可导图形（角点、尖点、垂直切线、跳跃）
- HL Apply the second-derivative test to classify stationary points用二阶导数检验对稳定点分类
- HL Find points of inflection by solving $f''(x) = 0$ and verifying sign change由 $f''(x) = 0$ 并验证变号求拐点
IB Paper-Style PracticeIB 试卷风格练习
E1 Practice + Solutions are on the roadmap (sibling of the A1/A2/A3/A4/A5/D1/D2/D3 sets). They will ship under Practice Questions/Unit_E1_*.html with the bilingual built-in pattern.
E1 配套的 Practice + Solutions 已在排期（与 A1/A2/A3/A4/A5/D1/D2/D3 的练习集同级）。上线后将位于 Practice Questions/Unit_E1_*.html，采用双语内嵌格式。