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。
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。
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。
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 $(3, -1)$）。
$$ y - (-1) \;=\; -\tfrac{5}{2}(x - 3) \;\Longrightarrow\; y \;=\; -\tfrac{5}{2} x + \tfrac{13}{2}. $$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 $\mathbb{R}$。
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。
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 $x \approx 1.06$。
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 $f \circ f = \mathrm{id}$，$f = f^{-1}$。
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 $x$。
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。