Unit B1: Representation
of Functions单元 B1：函数的表示
The opening unit of Topic B. Function notation, domain and range, the equation of a straight line, inverse and composite functions, and graphical methods for solving equations. The HL extension covers algebraic properties of functions (odd, even, and self-inverse). Every later function unit (polynomials, asymptotes, transformations) inherits the vocabulary and graphical thinking introduced here.Topic B 的开篇。函数记号、定义域与值域、直线方程、反函数与复合函数，以及用图像解方程。HL 扩展涵盖函数的代数性质（奇、偶、自逆）。后续所有函数单元（多项式、渐近线、变换）都继承本单元建立的词汇与图像思维。
Function Notation, Domain, and Range函数记号、定义域与值域 SL 2.2
- A function $f: A \to B$ assigns to each $x \in A$ exactly one $y = f(x) \in B$.
- The domain of $f$ is the set $A$ of allowed inputs.
- The range of $f$ is $\{ f(x) : x \in A \}$, the set of actual outputs.
- $f$ is one-to-one (or injective) if different inputs give different outputs: $x_{1} \ne x_{2} \Rightarrow f(x_{1}) \ne f(x_{2})$. Geometrically, every horizontal line cuts the graph in at most one point.
- 函数 $f: A \to B$ 把每个 $x \in A$ 对应到唯一的 $y = f(x) \in B$。
- $f$ 的定义域（domain）是允许输入的集合 $A$。
- $f$ 的值域（range）是 $\{ f(x) : x \in A \}$，即实际输出的集合。
- $f$ 为一对一（one-to-one，injective）：不同输入给出不同输出，$x_{1} \ne x_{2} \Rightarrow f(x_{1}) \ne f(x_{2})$。几何上，任一水平线与图像至多交于一点。
Find the (largest) domain and the range of $f(x) = \sqrt{4 - x^{2}}$.求 $f(x) = \sqrt{4 - x^{2}}$ 的（最大）定义域与值域。
Domain. Need $4 - x^{2} \ge 0$, that is $x^{2} \le 4$, so $-2 \le x \le 2$. Domain: $[-2, 2]$.
定义域。需 $4 - x^{2} \ge 0$，即 $x^{2} \le 4$，故 $-2 \le x \le 2$。定义域 $[-2, 2]$。
Range. $\sqrt{4 - x^{2}}$ takes its maximum when $x = 0$, giving $\sqrt{4} = 2$, and its minimum when $x = \pm 2$, giving $0$. Range: $[0, 2]$.
值域。$\sqrt{4 - x^{2}}$ 在 $x = 0$ 取最大值 $\sqrt{4} = 2$，在 $x = \pm 2$ 取最小值 $0$。值域 $[0, 2]$。
Remark. The graph is the upper semicircle of $x^{2} + y^{2} = 4$. Reading domain and range off a sketch is often faster than algebra.
注。图像是 $x^{2} + y^{2} = 4$ 的上半圆。在草图上读定义域与值域往往比代数更快。
Is $f(x) = \dfrac{2x + 1}{x - 3}$ one-to-one on its domain?$f(x) = \dfrac{2x + 1}{x - 3}$ 在其定义域上是一对一的吗？
Set up the algebraic test. Suppose $f(x_1) = f(x_2)$ and show this forces $x_1 = x_2$.
建立代数检验。设 $f(x_1) = f(x_2)$，证明必然 $x_1 = x_2$。
$$ \frac{2x_1+1}{x_1-3} = \frac{2x_2+1}{x_2-3} $$Cross-multiply and expand.
交叉相乘并展开。
$$ (2x_1+1)(x_2-3) = (2x_2+1)(x_1-3) $$ $$ 2x_1x_2 - 6x_1 + x_2 - 3 = 2x_1x_2 - 6x_2 + x_1 - 3 $$Cancel the shared $2x_1x_2$ and $-3$, then collect terms.
消去公共的 $2x_1x_2$ 与 $-3$，再合并同类项。
$$ -6x_1 + x_2 = -6x_2 + x_1 \;\Longrightarrow\; 7x_2 = 7x_1 \;\Longrightarrow\; x_1 = x_2 $$Conclusion. Since $f(x_1) = f(x_2)$ forces $x_1 = x_2$, $f$ is one-to-one on its domain ($x \ne 3$). This algebraic test is the reliable alternative when a graph isn't available — the horizontal-line test is a visual shortcut for exactly this argument.
结论。既然 $f(x_1) = f(x_2)$ 必然导出 $x_1 = x_2$，故 $f$ 在其定义域（$x \ne 3$）上一对一。当没有图像可用时，这个代数检验是可靠的替代方法——水平线检验只是这一论证的图像化捷径。
Straight Lines直线方程 SL 2.1
- Slope-intercept: $y = m x + c$ ($m$ slope, $c$ $y$-intercept).
- Point-slope: $y - y_{0} = m (x - x_{0})$ through $(x_{0}, y_{0})$ with slope $m$.
- General: $a x + b y + d = 0$.
Parallel and perpendicular. Lines $L_{1}$ and $L_{2}$ with slopes $m_{1}, m_{2}$ are
- parallel $\Leftrightarrow$ $m_{1} = m_{2}$ (and the lines are not identical).
- perpendicular $\Leftrightarrow$ $m_{1} m_{2} = -1$ (provided both slopes exist).
- 斜截式：$y = m x + c$（$m$ 斜率，$c$ $y$ 截距）。
- 点斜式：$y - y_{0} = m (x - x_{0})$，过点 $(x_{0}, y_{0})$，斜率 $m$。
- 一般式：$a x + b y + d = 0$。
平行与垂直。斜率分别为 $m_{1}, m_{2}$ 的两直线
- 平行 $\Leftrightarrow$ $m_{1} = m_{2}$（且不重合）。
- 垂直 $\Leftrightarrow$ $m_{1} m_{2} = -1$（且两斜率都存在）。
Find the equation of the line through $(3, -1)$ perpendicular to $2x - 5y = 10$.求过点 $(3, -1)$ 且垂直于 $2x - 5y = 10$ 的直线方程。
Slope of the given line. Rearrange: $y = \tfrac{2}{5} x - 2$, so $m_{1} = \tfrac{2}{5}$.
已知直线的斜率。整理：$y = \tfrac{2}{5} x - 2$，故 $m_{1} = \tfrac{2}{5}$。
Perpendicular slope. $m_{2} = -\tfrac{1}{m_{1}} = -\tfrac{5}{2}$.
垂直斜率。$m_{2} = -\tfrac{1}{m_{1}} = -\tfrac{5}{2}$。
Point-slope through $(3, -1)$.
点斜式（过 $(3, -1)$）。
$$ y - (-1) \;=\; -\tfrac{5}{2}(x - 3) \;\Longrightarrow\; y \;=\; -\tfrac{5}{2} x + \tfrac{13}{2}. $$Find the equation of the perpendicular bisector of the segment joining $A(2, 5)$ and $B(8, -3)$.求连接 $A(2, 5)$ 与 $B(8, -3)$ 的线段的垂直平分线方程。
Step 1 — Midpoint. The bisector passes through the midpoint of $AB$.
第 1 步 — 中点。平分线经过 $AB$ 的中点。
$$ M = \left(\frac{2+8}{2}, \frac{5+(-3)}{2}\right) = (5, 1) $$Step 2 — Slope of AB, then its negative reciprocal.
第 2 步 — AB 的斜率，再取负倒数。
$$ m_{AB} = \frac{-3-5}{8-2} = \frac{-8}{6} = -\frac{4}{3} \;\Longrightarrow\; m_{\perp} = \frac{3}{4} $$Step 3 — Point-slope through the midpoint.
第 3 步 — 过中点的点斜式。
$$ y - 1 = \tfrac{3}{4}(x - 5) \;\Longrightarrow\; y = \tfrac{3}{4}x - \tfrac{11}{4} $$A perpendicular bisector problem is really two earlier skills chained together — midpoint, then perpendicular slope — with no new formula to learn.
垂直平分线问题其实是把两个已学技能串联起来——先求中点，再求垂直斜率——不需要学任何新公式。
Inverse Functions反函数 SL 2.5
Procedure to find $f^{-1}$.
- Write $y = f(x)$.
- Swap $x$ and $y$.
- Solve for $y$.
- Rename $y$ as $f^{-1}(x)$.
- Domain of $f^{-1}$ equals the range of $f$. Range of $f^{-1}$ equals the domain of $f$.
求 $f^{-1}$ 的步骤。
- 写 $y = f(x)$。
- 交换 $x$ 与 $y$。
- 对 $y$ 解出。
- 把 $y$ 重命名为 $f^{-1}(x)$。
- $f^{-1}$ 的定义域 $=$ $f$ 的值域；$f^{-1}$ 的值域 $=$ $f$ 的定义域。
$f(x) = 2x - 5$ for $x \in \mathbb{R}$. Find $f^{-1}(x)$ and state its domain and range.$f(x) = 2x - 5$，$x \in \mathbb{R}$。求 $f^{-1}(x)$，并写出其定义域与值域。
Set up. $y = 2x - 5$. Swap: $x = 2y - 5$. Solve: $y = \tfrac{x + 5}{2}$.
列式。$y = 2x - 5$。交换：$x = 2y - 5$。解出：$y = \tfrac{x + 5}{2}$。
$$ f^{-1}(x) \;=\; \frac{x + 5}{2}. $$Domain and range. $f$ has domain $\mathbb{R}$ and range $\mathbb{R}$ (linear, slope $\ne 0$). So $f^{-1}$ has domain $\mathbb{R}$ and range $\mathbb{R}$.
定义域与值域。$f$ 定义域 $\mathbb{R}$、值域 $\mathbb{R}$（线性、斜率非零）。故 $f^{-1}$ 定义域与值域均为 $\mathbb{R}$。
$f(x) = (x - 2)^{2} + 1$ for $x \ge 2$. Find $f^{-1}(x)$.$f(x) = (x - 2)^{2} + 1$，$x \ge 2$。求 $f^{-1}(x)$。
Step 1 — Swap and solve. $y = (x-2)^2 + 1 \Rightarrow x = (y-2)^2 + 1 \Rightarrow (y-2)^2 = x - 1$.
第 1 步 — 交换并解出。$y = (x-2)^2 + 1 \Rightarrow x = (y-2)^2 + 1 \Rightarrow (y-2)^2 = x - 1$。
$$ y - 2 = \pm\sqrt{x-1} \;\Longrightarrow\; y = 2 \pm \sqrt{x-1} $$Step 2 — Use the restricted domain to pick the sign. The original restriction was $x \ge 2$, so the range of $f^{-1}$ must also be $\ge 2$. That forces the $+$ sign:
第 2 步 — 用限制的定义域来确定符号。原函数的限制是 $x \ge 2$，故 $f^{-1}$ 的值域也必须 $\ge 2$，这就确定了要取 $+$ 号：
$$ f^{-1}(x) = 2 + \sqrt{x-1}, \quad x \ge 1 $$Without the original domain restriction, $(x-2)^2+1$ would not be one-to-one and $\pm\sqrt{x-1}$ would be genuinely ambiguous — this is exactly why Topic B1.3's "always announce the restricted domain" rule matters here.
若没有原函数的定义域限制，$(x-2)^2+1$ 就不是一对一的，$\pm\sqrt{x-1}$ 也就真的模棱两可——这正是 B1.3 节"务必声明限制定义域"这条规则在这里发挥作用的原因。
Composite Functions复合函数 SL 2.5
Domain of $f \circ g$. The set of $x$ such that (i) $x$ is in the domain of $g$, and (ii) $g(x)$ is in the domain of $f$.
The inverse pair. $f \circ f^{-1}$ and $f^{-1} \circ f$ both equal the identity function $\mathrm{id}(x) = x$, on their respective domains.
$f \circ g$ 的定义域。使 (i) $x$ 在 $g$ 的定义域中，且 (ii) $g(x)$ 在 $f$ 的定义域中的所有 $x$。
反函数对。$f \circ f^{-1}$ 与 $f^{-1} \circ f$ 在各自定义域上都等于恒等函数 $\mathrm{id}(x) = x$。
$f(x) = x^{2} + 1$ and $g(x) = 2x - 3$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$, and confirm they differ.$f(x) = x^{2} + 1$、$g(x) = 2x - 3$。求 $(f \circ g)(x)$ 与 $(g \circ f)(x)$，并验证两者不同。
$f \circ g$. Apply $g$ first, then $f$:
$f \circ g$。先用 $g$，再用 $f$：
$$ (f \circ g)(x) \;=\; f(2x - 3) \;=\; (2x - 3)^{2} + 1 \;=\; 4 x^{2} - 12 x + 10. $$$g \circ f$. Apply $f$ first:
$g \circ f$。先用 $f$：
$$ (g \circ f)(x) \;=\; g(x^{2} + 1) \;=\; 2(x^{2} + 1) - 3 \;=\; 2 x^{2} - 1. $$The two expressions are different. Composition is not commutative.
两式不同。复合不满足交换律。
$f(x) = \sqrt{x}$ (domain $x \ge 0$) and $g(x) = x - 5$ (domain $\mathbb{R}$). Find the domain of $(f \circ g)(x)$.$f(x) = \sqrt{x}$（定义域 $x \ge 0$），$g(x) = x - 5$（定义域 $\mathbb{R}$）。求 $(f \circ g)(x)$ 的定义域。
Step 1 — Write the composite.
第 1 步 — 写出复合式。
$$ (f \circ g)(x) = f(x-5) = \sqrt{x-5} $$Step 2 — Apply the two-part domain rule. Condition (i): $x$ must be in the domain of $g$, which is all $\mathbb{R}$ — no restriction yet. Condition (ii): $g(x) = x - 5$ must be in the domain of $f$, i.e. $x - 5 \ge 0$.
第 2 步 — 套用两条定义域规则。条件 (i)：$x$ 须在 $g$ 的定义域中，即全体 $\mathbb{R}$——暂无限制。条件 (ii)：$g(x) = x - 5$ 须在 $f$ 的定义域中，即 $x - 5 \ge 0$。
$$ \text{Domain of } f \circ g: \quad x \ge 5 $$Notice the final answer came entirely from condition (ii) here — but both conditions must always be checked, since it's condition (i) that binds when the inner function $g$ itself has a restricted domain (e.g. $g(x) = \sqrt{x}$).
这里最终答案完全来自条件 (ii)——但两个条件都必须检查，因为当内函数 $g$ 本身就有受限定义域时（例如 $g(x) = \sqrt{x}$），起约束作用的就会是条件 (i)。
Graphical Solving of Equations用图像解方程 SL 2.10
GDC technique (Paper 2).
- For $f(x) = 0$: graph $y = f(x)$ and use the "zero" or "root" finder.
- For $f(x) = g(x)$: graph both, use the "intersect" feature.
- State answers to at least 3 significant figures unless the question asks for exact values.
GDC 操作（Paper 2）。
- $f(x) = 0$：画 $y = f(x)$，用 "zero / root" 功能。
- $f(x) = g(x)$：两条都画，用 "intersect" 功能。
- 非精确题答案保留至少 3 位有效数字。
Use the GDC to solve $e^{x} = 4 - x^{2}$ for $x \in [-3, 3]$.用 GDC 求 $e^{x} = 4 - x^{2}$ 在 $[-3, 3]$ 上的解。
Approach. Graph $y = e^{x}$ and $y = 4 - x^{2}$ on $[-3, 3]$ and find intersections.
方法。在 $[-3, 3]$ 上画 $y = e^{x}$ 与 $y = 4 - x^{2}$，求交点。
GDC output. Two intersections at approximately $x \approx -1.96$ (where $e^{x}$ is small and the parabola is descending) and $x \approx 1.06$ (where both are about $2.88$).
GDC 结果。两个交点约为 $x \approx -1.96$（$e^{x}$ 很小、抛物线下降处）与 $x \approx 1.06$（两边均约 $2.88$）。
Solutions to 3 sf: $x \approx -1.96$ or $x \approx 1.06$.
解（3 位有效数字）：$x \approx -1.96$ 或 $x \approx 1.06$。
Use a graph to solve $x^{2} - 2 < 2x - 3$.用图像求解 $x^{2} - 2 < 2x - 3$。
Step 1 — Set up as two graphs. Let $f(x) = x^2 - 2$ and $g(x) = 2x - 3$. The inequality asks where the graph of $f$ lies below the graph of $g$.
第 1 步 — 化为两条图像。令 $f(x) = x^2 - 2$，$g(x) = 2x - 3$。不等式问的是 $f$ 的图像在何处低于 $g$ 的图像。
Step 2 — Find the intersection points first (GDC "intersect" or algebra).
第 2 步 — 先求交点（GDC 的 "intersect" 功能或代数法）。
$$ x^2 - 2 = 2x - 3 \;\Longrightarrow\; x^2 - 2x + 1 = 0 \;\Longrightarrow\; (x-1)^2 = 0 \;\Longrightarrow\; x = 1 $$Step 3 — Interpret the single (repeated) intersection. The parabola $f$ touches the line $g$ at $x=1$ but does not cross it — since $f$ is $\cup$-shaped and $g$ is a line, $f$ lies at or above $g$ everywhere. There is no interval where $f(x) < g(x)$ strictly.
第 3 步 — 解读这个重根交点。抛物线 $f$ 在 $x=1$ 处与直线 $g$ 相切，但并未穿过它——由于 $f$ 呈 $\cup$ 形而 $g$ 是直线，$f$ 处处大于或等于 $g$。不存在使 $f(x) < g(x)$ 严格成立的区间。
Answer: no solution ($x^2-2 \ge 2x-3$ for all real $x$, with equality only at $x=1$). This is a useful reminder that "solve graphically" sometimes means confirming there's nothing to solve — a repeated root at the boundary is the signal to check, not assume, strict inequality.
答案：无解（对所有实数 $x$ 都有 $x^2-2 \ge 2x-3$，仅在 $x=1$ 处取等号）。这提醒我们："图像法求解"有时意味着确认根本无解可求——边界处出现重根，正是需要去检验而非假设严格不等式成立的信号。
Odd, Even, and Self-Inverse Functions奇函数、偶函数与自逆函数 HL AHL 2.14
- Even. $f(-x) = f(x)$ for all $x$ in the domain. Graph is symmetric about the $y$-axis. Examples: $f(x) = x^{2}$, $\cos x$, $|x|$.
- Odd. $f(-x) = -f(x)$ for all $x$. Graph has rotational symmetry about the origin. Examples: $f(x) = x^{3}$, $\sin x$, $\tan x$.
- Self-inverse. $f = f^{-1}$. Graph is symmetric about $y = x$. Examples: $f(x) = x$, $f(x) = 1/x$, $f(x) = c - x$ (for any constant $c$).
- 偶函数。定义域上 $f(-x) = f(x)$。图像关于 $y$ 轴对称。例：$f(x) = x^{2}$、$\cos x$、$|x|$。
- 奇函数。定义域上 $f(-x) = -f(x)$。图像关于原点中心对称。例：$f(x) = x^{3}$、$\sin x$、$\tan x$。
- 自逆。$f = f^{-1}$。图像关于 $y = x$ 对称。例：$f(x) = x$、$f(x) = 1/x$、$f(x) = c - x$（$c$ 为常数）。
Classify each function as odd, even, both, or neither: (a) $f(x) = x^{4} - 3 x^{2}$; (b) $g(x) = x^{3} + x$; (c) $h(x) = x^{2} + x$. Then verify $f(x) = 5 - x$ is self-inverse.分类下列函数（奇、偶、二者皆、二者非）：(a) $f(x) = x^{4} - 3 x^{2}$；(b) $g(x) = x^{3} + x$；(c) $h(x) = x^{2} + x$。再验证 $f(x) = 5 - x$ 是自逆函数。
(a) $f(-x) = (-x)^{4} - 3(-x)^{2} = x^{4} - 3 x^{2} = f(x)$. Even.
(a) $f(-x) = (-x)^{4} - 3(-x)^{2} = x^{4} - 3 x^{2} = f(x)$。偶。
(b) $g(-x) = (-x)^{3} + (-x) = -x^{3} - x = -(x^{3} + x) = -g(x)$. Odd.
(b) $g(-x) = (-x)^{3} + (-x) = -(x^{3} + x) = -g(x)$。奇。
(c) $h(-x) = x^{2} - x$. Not equal to $h(x)$ (would need the $x$ term to vanish) and not equal to $-h(x) = -x^{2} - x$ (would need the $x^{2}$ to vanish). Neither odd nor even.
(c) $h(-x) = x^{2} - x$，既不等于 $h(x)$（需 $x$ 项消失）也不等于 $-h(x) = -x^{2} - x$（需 $x^{2}$ 项消失）。非奇非偶。
Self-inverse check. $f(f(x)) = f(5 - x) = 5 - (5 - x) = x$. Hence $f \circ f = \mathrm{id}$, so $f = f^{-1}$.
自逆验证。$f(f(x)) = f(5 - x) = 5 - (5 - x) = x$。故 $f \circ f = \mathrm{id}$，$f = f^{-1}$。
Show that $f(x) = \dfrac{4x + 1}{x - 4}$ is self-inverse. Then decide whether $g(x) = \dfrac{4x + 1}{x - 3}$ is also self-inverse.证明 $f(x) = \dfrac{4x + 1}{x - 4}$ 是自逆函数。再判断 $g(x) = \dfrac{4x + 1}{x - 3}$ 是否也是自逆函数。
Compute $f(f(x))$. Substitute $f(x)$ into itself.
计算 $f(f(x))$。把 $f(x)$ 代入自身。
$$ f(f(x)) = \frac{4 \cdot \frac{4x+1}{x-4} + 1}{\frac{4x+1}{x-4} - 4} $$Clear the inner fraction by multiplying the numerator and the denominator by $(x - 4)$.
约去内部分式：分子分母同乘 $(x-4)$。
$$ f(f(x)) = \frac{4(4x+1) + (x-4)}{(4x+1) - 4(x-4)} = \frac{17x}{17} = x $$Conclusion for $f$. $f(f(x)) = x$ for all $x \ne 4$, so $f = f^{-1}$: $f$ is self-inverse, the same family as the cram-sheet's $f(x) = 1/x$ example.
关于 $f$ 的结论。对一切 $x \ne 4$ 都有 $f(f(x)) = x$，故 $f = f^{-1}$：$f$ 自逆，属于速记表中 $f(x) = 1/x$ 那一类函数。
Now test $g$. Only the denominator constant changed ($-4 \to -3$), so it is tempting to assume $g$ is self-inverse too. Find $g^{-1}$ directly instead of assuming.
再检验 $g$。仅分母的常数变了（$-4 \to -3$），很容易误以为 $g$ 也自逆。应直接求 $g^{-1}$，而不是想当然。
$$ y = \frac{4x+1}{x-3} \;\Longrightarrow\; x(y - 3) = 4y + 1 \;\Longrightarrow\; y(x - 4) = 3x + 1 \;\Longrightarrow\; g^{-1}(x) = \frac{3x+1}{x-4} $$Compare. $g^{-1}(x) = \dfrac{3x+1}{x-4} \ne \dfrac{4x+1}{x-3} = g(x)$. So $g$ is not self-inverse. A single constant separates "self-inverse" from "just another rational function", so always confirm with $f(f(x)) = x$ instead of pattern-matching the algebraic form.
比较。$g^{-1}(x) = \dfrac{3x+1}{x-4} \ne \dfrac{4x+1}{x-3} = g(x)$。故 $g$ 不是自逆函数。仅一个常数之差，就是"自逆"与"普通有理函数"的分界，因此务必用 $f(f(x)) = x$ 确认，切勿凭代数形式类推。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- State both whenever you write or find a function. "Find the range" is an A1 that students drop by giving a value without an interval.
- 写或求得函数时同时写出定义域与值域。"求值域"的 A1 经常因为只给数值不给区间而丢失。
- Inverse domain equals original range. Markschemes check this explicitly.
- 反函数的定义域 $=$ 原函数的值域。评分细则会专门核对。
- Convert to slope-intercept before reading the slope. $2x - 5y = 10$ does not have slope $2$.
- 先化为斜截式再读斜率。$2x - 5y = 10$ 的斜率不是 $2$。
- Perpendicular slope is the negative reciprocal. Sign and reciprocal both flip.
- 垂直斜率为负倒数。取倒数并变号。
- $f^{-1}$ is not $1/f$. The notation $f^{-1}$ always means inverse function.
- $f^{-1}$ 不是 $1/f$。$f^{-1}$ 始终指反函数。
- Verify by composition. If you suspect your inverse is wrong, compute $f(f^{-1}(x))$; it should simplify to $x$ on the appropriate domain.
- 用复合验证。怀疑反函数有误时，算 $f(f^{-1}(x))$；应在合适定义域上化简为 $x$。
- Respect the stated domain. If the question restricts $x \in [a, b]$, discard any calculator-found intersection or root outside that interval, even if it exists mathematically.
- 遵守题目给定的定义域。若题目限制 $x \in [a, b]$，即使计算器找到区间外的交点或根，也必须舍去。
- Find every root before stopping. A single "zero" or "intersect" call returns one solution near your cursor. A curve can cross the axis (or another curve) more than once, so scan the whole domain before reporting a final answer.
- 找出所有根再停止。一次 "zero" 或 "intersect" 只返回光标附近的一个解。曲线可能在定义域内多次穿过坐标轴（或另一条曲线），务必扫描整个定义域后再给出最终答案。
- Check the domain is symmetric about $0$ before testing odd or even. $f(x) = x^{2}$ restricted to $x \ge 0$ is neither odd nor even: $-x$ is not even in the domain, so $f(-x)$ is not defined and the identity cannot be checked at every point.
- 检验奇偶性前，先确认定义域关于 $0$ 对称。$f(x) = x^{2}$ 限制在 $x \ge 0$ 上既非奇也非偶：$-x$ 根本不在定义域中，$f(-x)$ 无定义，无法在每一点验证该恒等式。
- Self-inverse does not survive small edits. Changing a single constant, as in Worked Example B1.6b, can turn a self-inverse function into an ordinary one. Confirm with $f(f(x)) = x$, never by pattern-matching the algebraic form.
- 自逆性质经不起小改动。如例题 B1.6b 所示，改动一个常数就可能让自逆函数变成普通函数。务必用 $f(f(x)) = x$ 确认，切勿凭代数形式类推。
Flashcards闪卡
Unit B1 Practice Quiz单元 B1 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Find the largest domain of a function combining radicals, reciprocals, and logarithms求含根式、分式、对数的复合函数最大定义域
- Read range off a sketch or algebraic analysis由图像或代数分析读出值域
- Find the equation of a line from two points or from a point and slope由两点或一点加斜率求直线方程
- Find lines parallel or perpendicular to a given line through a given point求过定点平行或垂直于已知直线的直线
- Verify one-to-one using the horizontal-line test用水平线检验判断一对一
- Find $f^{-1}$ by swapping $x$ and $y$ and solving, then state its domain用"交换 $x, y$ 后解出"求 $f^{-1}$ 并写出定义域
- Restrict the domain of a non-injective function before inverting求非单射函数反函数前先限制定义域
- Compute $f \circ g$ and recognise the order matters算 $f \circ g$，并认清次序重要
- Solve $f(x) = g(x)$ graphically on the GDC, citing intersection coordinates to 3 sf用 GDC 图像解 $f(x) = g(x)$，交点保留 3 位有效数字
- Solve $f(x) > g(x)$ graphically and write the answer as an interval用图像解不等式 $f(x) > g(x)$ 并写为区间
- HL Classify a function as odd, even, both, or neither by computing $f(-x)$通过算 $f(-x)$ 把函数归类为奇、偶、二者皆、二者非
- HL Verify a function is self-inverse by checking $f(f(x)) = x$用 $f(f(x)) = x$ 验证函数自逆
IB Paper-Style PracticeIB 试卷风格练习
B1 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_B1_*.html with the bilingual built-in pattern.
B1 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_B1_*.html。
How to use this guide本指南使用说明
How to use this guide本指南使用说明
B1 is vocabulary-heavy. The computations are easy; the marks come from precise statements (domain, range, "one-to-one", inverse existence). Train the language alongside the algebra.B1 偏术语。计算不难，分数靠精确陈述（定义域、值域、一对一、反函数存在条件）。术语与代数一起练。
Memorise three formulas: gradient $m = (y_{2} - y_{1})/(x_{2} - x_{1})$; point-slope $y - y_{0} = m (x - x_{0})$; perpendicular slopes multiply to $-1$. Be able to find an inverse by swapping $x$ and $y$ and solving.
背三个公式：斜率 $m = (y_{2} - y_{1})/(x_{2} - x_{1})$；点斜式 $y - y_{0} = m (x - x_{0})$；垂直斜率乘积 $= -1$。会用"交换 $x, y$ 后解"求反函数。
Always state the domain of any function you write, and the domain of any inverse you find. Practise the horizontal-line test for "one-to-one" and explain why restricting the domain is needed when the original function fails it.
每写一个函数都要标出定义域；每求一个反函数都要写出反函数的定义域。会做水平线检验（horizontal-line test）判定一对一；当原函数不一对一时，能说明为何要限制定义域。