Axiom Learning·AP PrecalculusChapter 1: Relations & Functions
函数与关系 · Domain · Rate of Change · Linearity
1 · Relations and Functions 关系与函数
1.1 Key Vocabulary 核心术语
| English Term | 中文 | Meaning / 含义 |
|---|---|---|
| relation | 关系 | A rule or connection between inputs and outputs. 输入和输出之间的对应规则。 |
| input | 输入值 | The value you put in. Usually $x$. 放入函数的值，通常用 $x$ 表示。 |
| output | 输出值 | The value produced. Usually $y$ or $f(x)$. 函数产生的值，通常用 $y$ 或 $f(x)$。 |
| domain | 定义域 | The set of all valid input values. 所有合法输入值的集合。 |
| range | 值域 | The set of all possible output values. 所有可能输出值的集合。 |
| function | 函数 | A relation where each input has exactly one output. 每个输入对应唯一输出的关系。 |
| independent variable | 自变量 | The input variable, often $x$. 输入变量，通常是 $x$。 |
| dependent variable | 因变量 | The output variable, often $y$. 输出变量，通常是 $y$。 |
| ordered pair | 有序数对 | A point written as $(x, y)$. 以 $(x, y)$ 形式表示的点。 |
1.2 What Is a Function? 什么是函数？
A function is a special type of relation where every input is paired with exactly one output.
✅ Is a Function 是函数
$\{(1,4),(2,5),(3,6)\}$
每个输入 (1, 2, 3) 只出现一次。
❌ Not a Function 不是函数
$\{(1,4),(2,5),(2,7)\}$
输入 $2$ 对应了 $5$ 和 $7$ 两个输出。
1.3 Function Notation 函数符号
$f(x)$ does NOT mean $f$ times $x$! $f(x)$ 不是 $f$ 乘以 $x$！
It means "the output of function $f$ when the input is $x$." 它表示"当输入为 $x$ 时，函数 $f$ 的输出值"。
Determine whether the relation is a function: $\{(-1,3),(0,5),(2,7),(2,9),(4,11)\}$
这不是函数，因为 $x=2$ 有两个不同的输出。
(a) If $f(x) = 10x - 2$, find $f(3)$. (b) If $g(x) = x^2 + 5$, find $g(x+2)$.
$= x^2 + 4x + 4 + 5$
$= x^2 + 4x + 9$
Let $h(x) = 3x^2 - 2x + 1$. Find: (a) $h(0)$ (b) $h(-2)$ (c) $h(x-1)$
(a) $h(0) = 3(0)^2 - 2(0) + 1 = \mathbf{1}$
(b) $h(-2) = 3(4) - 2(-2) + 1 = 12 + 4 + 1 = \mathbf{17}$
(c) $h(x-1) = 3(x-1)^2 - 2(x-1) + 1 = 3(x^2-2x+1) - 2x + 2 + 1$
$= 3x^2 - 6x + 3 - 2x + 3 = \mathbf{3x^2 - 8x + 6}$
1.4 Four Representations of a Function 函数的四种表示
The input and output values of a function vary in tandem (一起变化) according to the function rule. AP Precalculus expects you to work with that rule in four forms and to move between them.
AP 题目经常在四种表示之间切换：给表格判断凹凸、给文字选图、给图读区间。
| Representation 表示方式 | What you see 看什么 | Example: taxi fare 出租车费 |
|---|---|---|
| Verbal 文字 | A sentence describing how the quantities change together 用语言描述规则 | Starting fare \$3, plus \$2 per mile. 起步价 3 美元，每英里加 2 美元。 |
| Analytical 解析式 | An equation or formula 公式 | $C(m) = 3 + 2m$ |
| Numerical 数值（表格） | A table of input–output pairs 输入输出表 | $m$: 0, 1, 2, 3, 4 → $C$: 3, 5, 7, 9, 11 |
| Graphical 图像 | Overall shape and trend 整体形状与趋势 | See the graph below 见下图 |
Verbal: A ball is dropped from the top of a 64-foot building. Its height above the ground is $h(t)=64-16t^2$ feet, where $t$ is the time in seconds after it is dropped.
(a) Make a table of values for $t = 0,\ 0.5,\ 1,\ 1.5,\ 2$.
(b) When does the ball hit the ground?
(c) State a reasonable domain and range for $h$ in this context.
(d) Describe the graph: is $h$ increasing or decreasing? Concave up or concave down?
| $t$ (s) | 0 | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|---|
| $h(t)$ (ft) | 64 | 60 | 48 | 28 | 0 |
| First difference | $-4$ | $-12$ | $-20$ | $-28$ |
A phone plan costs \$20 per month plus \$0.10 for each text message sent.
(a) Write the monthly cost $C$ as a function of the number of texts $n$.
(b) Complete a table for $n = 0,\ 100,\ 200,\ 300$.
(c) Describe the graph of $C$ in words.
(d) What is a reasonable domain in this context?
(a) $C(n) = 20 + 0.10n$
(b) $n$: 0, 100, 200, 300 → $C$: 20, 30, 40, 50
(c) A line starting at $(0, 20)$ that increases at a constant rate of \$0.10 per text. It is neither concave up nor concave down. 从 $(0,20)$ 出发、以每条短信 0.10 美元的常数速率上升的直线。
(d) $n = 0, 1, 2, 3, \dots$ (whole numbers). You cannot send a negative or fractional number of texts. 情境中 $n$ 只能取非负整数。
2 · Domain and Range 定义域与值域
2.1 From a Graph 从图像读取
- Read the domain left → right (x-values) 从左到右读 定义域（$x$ 值）
- Read the range bottom → top (y-values) 从下到上读 值域（$y$ 值）
- ● Closed circle = endpoint included 实心圆 = 包含端点 $[\;]$
- ○ Open circle = endpoint excluded 空心圆 = 不包含端点 $(\;)$
- → Arrow = graph continues forever 箭头 = 趋向无穷 $\pm\infty$
2.2 Interval Notation 区间记号
| Inequality 不等式 | Interval 区间 | Meaning 含义 |
|---|---|---|
| $a \le x \le b$ | $[a,b]$ | Both endpoints included 两端均包含 |
| $a < x < b$ | $(a,b)$ | Both endpoints excluded 两端均不包含 |
| $a \le x < b$ | $[a,b)$ | Left included, right excluded 左包右不包 |
| $a < x \le b$ | $(a,b]$ | Left excluded, right included 左不包右包 |
| All real numbers | $(-\infty,+\infty)$ | No restriction 无限制 |
| $x \ge a$ | $[a,+\infty)$ | 从 $a$ 起到正无穷（包含 $a$） |
| $x > a$ | $(a,+\infty)$ | 从 $a$ 起到正无穷（不含 $a$） |
Infinity $\infty$ is never included with a bracket $[$ or $]$. Always use parentheses: $($ or $)$.
无穷大 $\infty$ 和 $-\infty$ 永远用圆括号 $(\;)$，不能用方括号，因为无穷大不是一个具体的数值。
2.3 Algebraic Rules for Domain 代数求定义域规则
| # | Rule 规则 | Restriction 限制 | Example |
|---|---|---|---|
| 1 | No division by zero 分母不能为零 |
Set denominator $\neq 0$ | $\dfrac{1}{x-3}$: $x\neq 3$, domain $(-\infty,3)\cup(3,\infty)$ |
| 2 | No even root of a negative 偶次根号内不能为负数 |
Set radicand $\ge 0$ | $\sqrt{x+5}$: $x+5\ge 0\Rightarrow x\ge -5$, domain $[-5,\infty)$ |
| 3 | Log input must be positive 对数的真数必须大于零 |
Set argument $> 0$ | $\ln(x)$: $x>0$, domain $(0,\infty)$ |
Find the domain of $f(x) = \dfrac{x+2}{x^2+5x-36}$.
Set equal to zero: $(x+9)(x-4) = 0 \Rightarrow x = -9 \text{ or } x = 4$
Find the domain of $g(x) = \sqrt{2x - 8}$.
Find the domain of each function:
(a) $f(x) = \dfrac{x+2}{x^2+5x-36}$ (b) $g(x) = \sqrt{2x-8}$ (c) $h(x) = \ln(5-x)$
(a) $x^2+5x-36=(x+9)(x-4)=0 \Rightarrow x=-9,\,4$
Domain: $(-\infty,-9)\cup(-9,4)\cup(4,+\infty)$
(b) $2x-8\ge0 \Rightarrow x\ge4$ → Domain: $[4,+\infty)$
(c) $5-x>0 \Rightarrow x<5$ → Domain: $(-\infty,5)$
3 · Intercepts, Zeros, and the Vertical Line Test
截距、零点与垂直线检验
3.1 x-Intercepts and Zeros x 截距与零点
x-Intercept (x 截距)
Where the graph crosses or touches the x-axis.
图像与 $x$ 轴的交点。
At an x-intercept: $y = 0$
Zero (零点)
The x-value where $f(x) = 0$.
使 $f(x)=0$ 成立的 $x$ 值。
To find: set $f(x) = 0$ and solve.
3.2 y-Intercept y 截距
- To find zeros / x-intercepts: set $f(x) = 0$ and solve for $x$. 令 $f(x)=0$ 求解。
- To find the y-intercept: substitute $x = 0$, compute $f(0)$. 代入 $x=0$，计算 $f(0)$。
3.3 Vertical Line Test 垂直线检验
A graph represents a function if and only if no vertical line intersects the graph more than once.
Find the zeros and the $y$-intercept of $f(x) = x^2 - 48$.
$x^2 = 48$
$x = \pm\sqrt{48} = \pm 4\sqrt{3}$
y-intercept: $(0, -48)$
Find the zeros of $f(x) = (x-3)(x+5)(x+13)$.
$x - 3 = 0 \Rightarrow x = 3$
$x + 5 = 0 \Rightarrow x = -5$
$x + 13 = 0 \Rightarrow x = -13$
For each function, find all zeros and the y-intercept:
(a) $f(x) = x^2 - 9$ (b) $g(x) = 2x^3 - 8x$ (c) $h(x) = (x+1)(x-4)(x-7)$
(a) $x^2-9=0\Rightarrow x=\pm3$. Zeros: $x=3,\,-3$. y-int: $f(0)=-9$ → $(0,-9)$
(b) $2x^3-8x=2x(x^2-4)=2x(x-2)(x+2)=0$
Zeros: $x=0,\,2,\,-2$. y-int: $g(0)=0$ → $(0,0)$
(c) Zeros: $x=-1,\,4,\,7$. y-int: $h(0)=(1)(-4)(-7)=28$ → $(0,28)$
4 · Increasing, Decreasing, and Concavity
递增、递减与凹凸性
4.1 Increasing and Decreasing 递增与递减
Increasing 递增
As $x$ increases, $f(x)$ increases.
$x$ 越大，$f(x)$ 越大。
Formally: if $x_1 < x_2$, then $f(x_1) < f(x_2)$.
图像从左到右上升。
Decreasing 递减
As $x$ increases, $f(x)$ decreases.
$x$ 越大，$f(x)$ 越小。
Formally: if $x_1 < x_2$, then $f(x_1) > f(x_2)$.
图像从左到右下降。
4.2 Concavity 凹凸性
Rate of change is increasing.
变化率（斜率）越来越大。
Rate of change is decreasing.
变化率（斜率）越来越小。
• 上凹（Concave Up）：图像像碗一样开口向上（∪形），平均变化率从左到右在增大。
• 下凹（Concave Down）：图像像帽子一样开口向下（∩形），平均变化率从左到右在减小。
注意：函数可以同时是递减的且上凹（如 $y=x^2$ 在 $x<0$ 处）。
Comparing first differences to judge concavity only works when the $x$-values are equally spaced. If they are not, compute the average rate of change $\dfrac{\Delta y}{\Delta x}$ on each interval and compare those instead.
用一阶差分判断凹凸的前提是 $x$ 等间距。如果不等间距，必须先算每段的平均变化率 $\dfrac{\Delta y}{\Delta x}$，再比较大小。
4.3 Four Shape Combinations 增减 × 凹凸：四种形状
Increasing or decreasing tells you the sign of the rate of change. Concavity tells you whether the rate of change is increasing or decreasing. Put them together and you get four basic shapes.
增减性看变化率的正负，凹凸性看变化率在变大还是变小。两者组合出四种基本形状。
递增 · 上凹
变化率为正且越来越大
例：早期人口增长
递增 · 下凹
变化率为正但越来越小
例：施肥越多产量越高，但增幅递减
递减 · 上凹
变化率为负但越来越接近 0
例：热咖啡冷却
递减 · 下凹
变化率为负且越来越负
例：自由下落的球的高度
For increasing functions, "increasing at an increasing / decreasing rate" is clear. For decreasing functions, a phrase like "decreasing at a decreasing rate" is ambiguous: does "rate" mean the signed rate of change or the speed of the drop? The safest wording states the two facts separately:
The table gives values of a function $f$. Describe the behavior on $1\le x\le 6$: is $f$ increasing or decreasing? Is it concave up or down?
| $x$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $f(x)$ | 11 | 6 | −5 | −22 | −45 | −74 |
$11 \to 6 \to -5 \to -22 \to -45 \to -74$ → outputs are decreasing 递减的。
| Interval | [1,2] | [2,3] | [3,4] | [4,5] | [5,6] |
|---|---|---|---|---|---|
| Rate of change | $-5$ | $-11$ | $-17$ | $-23$ | $-29$ |
→ The graph is concave down (∩).
函数在 $[1,6]$ 上递减且下凹。
Analyze the function given by the table. State whether it is increasing or decreasing, and concave up or concave down.
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $f(x)$ | 2 | 5 | 10 | 17 | 26 |
Step 1 — Increasing/Decreasing: $2\to5\to10\to17\to26$ → outputs increase → Increasing 递增
Step 2 — First differences: $+3,\;+5,\;+7,\;+9$ → these are increasing → Concave Up 上凹 ∪
因为一阶差分（变化率）从 $3$ 增加到 $9$，所以函数上凹。
4.4 Reading Behavior from a Graph 从图像读增减与凹凸
- Scan left → right. Rising means increasing; falling means decreasing. 从左往右看，上升为递增、下降为递减。
- Turning points (where the graph switches between rising and falling) split the increasing and decreasing intervals. 转折点把增减区间分开。
- A point of inflection (where the bend switches between ∪ and ∩) splits the concavity intervals. 拐点把凹凸区间分开。
- Write intervals of $x$-values, never $y$-values. 区间写的是 $x$ 的范围，不是 $y$ 的范围。
本讲义约定：增减区间写闭区间，凹凸区间写开区间。
The graph of the function $f$ on the closed interval $-3 \le x \le 5$ is shown. The key points lie on grid points.
(a) State the domain and range of $f$.
(b) Estimate the zeros of $f$.
(c) On what intervals is $f$ increasing? Decreasing?
(d) On what intervals is the graph of $f$ concave up? Concave down?
(e) On what interval is $f$ increasing and concave down? On what interval is $f$ decreasing and concave up?
(f) Find the average rate of change of $f$ over $[-1, 3]$. What does its sign tell you?
Increasing on $[-3,-1]$ and $[3,5]$; decreasing on $[-1,3]$.
Concave down on $(-3, 1)$; concave up on $(1, 5)$.
Decreasing and concave up: $[-1,3] \cap (1,5) \to (1,3)$.
The graph of $g$ for $0 \le x \le 8$ is shown. The key points lie on grid points.
(a) On what intervals is $g$ increasing? Decreasing?
(b) On what intervals is the graph of $g$ concave up? Concave down?
(c) On what interval is $g$ increasing at a decreasing rate?
(d) At what value of $x$ is the rate of change of $g$ greatest? Explain.
(e) Which is greater: the average rate of change of $g$ over $[0,8]$ or over $[2,6]$?
(a) Increasing on $[2,6]$; decreasing on $[0,2]$ and $[6,8]$.
(b) Concave up on $(0,4)$; concave down on $(4,8)$.
(c) Increasing and concave down: $(4,6)$.
(d) $x = 4$. On $(2,4)$, $g$ is increasing and concave up, so its rate of change is increasing. On $(4,6)$, it is concave down, so its rate of change is decreasing. The rate of change is greatest at the point of inflection $x=4$.
拐点处变化率最大（先增后减）。这个结论在第 2 章多项式里还会用到。
(e) Over $[0,8]$: $\dfrac{-2-2}{8}=-0.5$. Over $[2,6]$: $\dfrac{2-(-2)}{4}=1$. The AROC over $[2,6]$ is greater.
4.5 From Words to Graphs 从文字情境到图像
AP expects you to construct a graph from a verbal description of how two quantities change together.
根据一段描述两个量如何一起变化的文字，画出或选出对应的图像。
- Name the variables: the input (often time), the output, and their units. 确定输入、输出和单位。
- Direction: in each stage, is the output going up, going down, or staying constant? 每一段输出在增、减还是不变？→ 决定增减。
- Speed: is it changing faster and faster, or slower and slower? 越来越快还是越来越慢？→ 决定凹凸。
- Special points: the starting value, zeros, maximum or minimum, and the moments where the behavior switches. 标出起点、零点、最值和转折时刻。
Water is poured at a constant rate into the flask shown. The flask is wide at the bottom and narrow at the top. Which graph best shows the height $H$ of the water as a function of time $t$?
(A) 适合圆柱形瓶子（匀速上升）；(B) 适合上宽下窄的瓶子；(D) 适合中间最宽的瓶子。
(a) Water is poured at a constant rate into the round-bottom flask shown, which has a narrow cylindrical neck. Sketch the height $H$ of the water as a function of time $t$ until the flask is full. Where does the concavity change?
(b) Lily walks from home toward school at a constant speed for 5 minutes. She stops at a café for 3 minutes. Then she gets into her mom's car, which speeds up as it drives the rest of the way to school. Sketch Lily's distance from home $D$ as a function of time $t$.
(a) $H$ is always increasing. In the lower half the sphere gets wider, so $H$ rises more and more slowly (concave down). In the upper half it gets narrower, so $H$ rises faster and faster (concave up). The concavity changes at the widest part of the sphere (point of inflection). In the neck, the width is constant, so $H$ rises at a constant rate (a line).
下半球越来越宽 → 越升越慢（下凹）；上半球越来越窄 → 越升越快（上凹）；最宽处是拐点；瓶颈是圆柱 → 直线。
(b) $0\le t\le5$: increasing line (constant speed). $5\le t\le8$: horizontal segment (distance constant). After $t=8$: increasing and concave up (the car speeds up).
步行 → 直线上升；停留 → 水平线；汽车加速 → 上凹上升。
5 · Average Rate of Change 平均变化率
5.1 Definition and Formula 定义与公式
The average rate of change of $f$ from $x=a$ to $x=b$ is:
This equals the slope of the secant line through $(a,f(a))$ and $(b,f(b))$.
这是斜率公式 $m = \dfrac{y_2-y_1}{x_2-x_1}$ 的推广，适用于任何函数（不只是直线）。
The average rate of change over $[a,b]$ is the constant rate of change that would produce the same change in output over that interval.
例：泳池水量 $V(2)=10.6$、$V(5)=11.8$ 立方米，平均变化率为 $0.4$ m³/min。含义：如果从第 2 分钟到第 5 分钟一直以 $0.4$ m³/min 匀速注水，3 分钟内同样会增加 $1.2$ m³。
5.2 Approximate Instantaneous Rate of Change 近似瞬时变化率
- Average rate of change: over an interval $[a,b]$. 在区间上的变化率。
- Instantaneous rate of change (瞬时变化率): rate at a single point. In precalculus, we approximate it using a small interval around that point. 在一个点处的变化率。AP Precalculus 中用该点附近的小区间来近似。
- In calculus, this becomes the derivative (导数) — the slope of the tangent line (切线).
Find the average rate of change of $f(x) = 3x^2 - 2$ from $x = 2$ to $x = 6$.
$f(6) = 3(6)^2 - 2 = 108 - 2 = 106$
平均变化率为 $24$，即 $x$ 每增加1，$f(x)$ 平均增加 $24$。
The temperature of water is recorded in the table below. Approximate the rate of change at $t = 3.5$ minutes.
| minutes $t$ | 0 | 2 | 5 | 9 | 10 |
|---|---|---|---|---|---|
| $T(t)$ (°F) | 150.8 | 140 | 125.6 | 111.2 | 109.4 |
温度在 $t=3.5$ 分钟时，每分钟约下降 $4.8$°F。
(a) Find the average rate of change of $f(x) = x^3 - 4x$ from $x = 1$ to $x = 3$.
(b) A car travels 240 miles in 4 hours. What is the average speed?
(c) Using the table below, approximate the rate of change at $x = 2.5$.
| $x$ | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| $f(x)$ | 3 | 7 | 13 | 21 |
(a) $f(1)=1-4=-3$; $f(3)=27-12=15$
AROC $= \dfrac{15-(-3)}{3-1}=\dfrac{18}{2}=\mathbf{9}$
(b) Average speed $= \dfrac{240}{4} = \mathbf{60}$ mph
(c) $x=2.5$ is between $x=2$ and $x=3$. AROC $\approx \dfrac{13-7}{3-2}=\dfrac{6}{1}=\mathbf{6}$
5.3 Comparing Rates of Change at Two Points 比较两点处的变化率
- For each point, choose a small interval containing that point. 对每个点，取一个包含它的小区间。
- Compute the AROC on each small interval. These approximate the rates of change at the points. 分别算平均变化率，作为该点变化率的近似。
- Compare the two approximations. The smaller the intervals, the better the approximations. 比较两个近似值。区间越小越准确。
"Which rate of change is greater?" compares the signed values: $3 > -5$.
"Where is the output changing faster?" compares the sizes (absolute values): a rate of $-5$ is faster than $3$.
问"哪个变化率更大"，比较带符号的数值；问"哪里变化得更快"，比较绝对值。
Let $f(x) = x^3 - 6x$. Use average rates of change over the intervals $[0.9,\,1.1]$ and $[2.9,\,3.1]$ to compare the rate of change of $f$ at $x = 1$ with the rate of change at $x = 3$.
$\dfrac{-5.269-(-4.671)}{1.1-0.9} = \dfrac{-0.598}{0.2} = -2.99$
$\dfrac{11.191-6.989}{3.1-2.9} = \dfrac{4.202}{0.2} = 21.01$
$x=3$ 处的变化率更大，而且变化得更快。
The total rainfall $R(t)$, in millimeters, $t$ hours after a storm begins is shown in the table.
| $t$ (hours) | 0 | 2 | 3 | 5 | 8 | 9 |
|---|---|---|---|---|---|---|
| $R(t)$ (mm) | 0 | 15.6 | 21.0 | 28.4 | 34.7 | 35.6 |
(a) Approximate the rate of change of $R$ at $t = 2.5$ and at $t = 8.5$. At which time is rain falling faster?
(b) Compute the average rate of change on each interval between consecutive table values. What do they suggest about the concavity of the graph of $R$?
(c) Interpret the average rate of change over $3 \le t \le 5$ in context, with units.
(a) At $t=2.5$: $\dfrac{21.0-15.6}{3-2} = 5.4$ mm/hr. At $t=8.5$: $\dfrac{35.6-34.7}{9-8} = 0.9$ mm/hr. Rain is falling faster at $t = 2.5$.
(b) The $x$-values are not equally spaced, so compare the AROCs, not the raw differences:
$[0,2]: 7.8$, $[2,3]: 5.4$, $[3,5]: 3.7$, $[5,8]: 2.1$, $[8,9]: 0.9$ (mm/hr).
The AROCs are decreasing, which suggests the graph of $R$ is concave down: increasing at a decreasing rate.
(c) "Between $t = 3$ and $t = 5$ hours after the storm began, rain accumulated at an average rate of $3.7$ millimeters per hour."
5.4 The Sign of a Rate of Change & AP Language 变化率的正负与规范表述
Positive rate of change 正变化率
As one quantity increases, the other also increases. As one decreases, the other also decreases.
两个量同增同减。
Negative rate of change 负变化率
As one quantity increases, the other decreases.
一个增大，另一个减小。
Free-response answers are scored on the explanation, so write a full sentence with context and units.
简答题（FRQ）要给出理由。一句完整的回答应包含：函数名、区间或时刻、数值、单位、情境含义。
- Missing units, or wrong units (the unit of a rate is output unit per input unit). 漏写单位；变化率的单位是"输出单位 / 输入单位"。
- Not naming the interval or the time. 没写是哪个区间或哪个时刻。
- Writing "$V$ increases by 0.4" instead of "0.4 cubic meters per minute". 把"每分钟增加 0.4"说成"增加 0.4"。
- Saying only "concave up" with no reason. 只写结论，不写理由（没有列出变化率在增大）。
The volume of water in a pool is $V(t)$ cubic meters, where $t$ is measured in minutes. It is known that $V(2) = 10.6$ and $V(5) = 11.8$.
(a) Find the average rate of change of $V$ over $2 \le t \le 5$.
(b) Interpret your answer in context, using correct units.
(c) What does the sign of your answer tell you?
The temperature $C(t)$, in degrees Celsius, of a cup of coffee $t$ minutes after it is poured is shown.
| $t$ (min) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| $C(t)$ (°C) | 90 | 72 | 60 | 52 | 47 |
(a) Find the average rate of change of $C$ over $0 \le t \le 20$ and interpret it in context.
(b) Is the graph of $C$ concave up or concave down on $[0, 20]$? Give a reason.
(c) Describe how the temperature changes, in words, using the language of rates.
(a) $\dfrac{47-90}{20-0} = -2.15$. "Over the first 20 minutes, the temperature of the coffee decreased at an average rate of $2.15$ degrees Celsius per minute."
(b) AROCs over consecutive 5-minute intervals: $-3.6,\ -2.4,\ -1.6,\ -1.0$ °C/min. "These average rates of change over equal-length intervals are increasing, so the graph of $C$ is concave up."
(c) "The temperature is decreasing, and the rate of change of the temperature is increasing: the coffee cools more and more slowly."
温度在下降，但变化率在增大（越来越接近 0），所以降温越来越慢。
6 · Rates of Change in Linear Functions 线性函数的变化率
6.1 Linear Function 线性函数
A linear function has the form: $f(x) = mx + b$
- $m$ = slope (斜率) = constant rate of change (常数变化率)
- $b$ = y-intercept (纵截距) = value when $x = 0$
$m > 0$：函数递增（increasing） $m < 0$：函数递减（decreasing） $m = 0$：常数函数
6.2 First Differences for Linear Functions 一阶差分识别线性函数
When $x$-values are equally spaced, compute first differences (consecutive output differences 相邻输出的差):
- If first differences are constant → linear function. 一阶差分恒定 → 线性函数。
- If first differences are not constant → not linear. 一阶差分不恒定 → 不是线性函数。
Determine whether the data represent a linear function.
| $x$ | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|
| $f(x)$ | 50 | 47.5 | 45 | 42.5 | 40 | 37.5 |
| Interval | [20,30] | [30,40] | [40,50] | [50,60] | [60,70] |
|---|---|---|---|---|---|
| Difference | $-2.5$ | $-2.5$ | $-2.5$ | $-2.5$ | $-2.5$ |
The function is linear with slope $-0.25$ (decreasing 递减).
A bakery has a fixed cost of \$120 per day and each cookie costs \$0.99 to bake. Let $x$ be the number of cookies. Write a cost model $C(x)$.
Variable cost per cookie (每块饼干成本): \$0.99 → this is the slope $m = 0.99$
(a) Determine whether the table represents a linear function. If so, find the slope.
(b) Write a linear model: A gym charges \$50 to join and \$30 per month. Find $C(t)$ where $t$ = months.
| $x$ | 0 | 3 | 6 | 9 | 12 |
|---|---|---|---|---|---|
| $f(x)$ | 10 | 19 | 28 | 37 | 46 |
(a) First differences: $19-10=9$, $28-19=9$, $37-28=9$, $46-37=9$ → all equal → Linear.
Slope $m = \dfrac{9}{3} = 3$. (Each step of $\Delta x = 3$ gives $\Delta f = 9$.)
(b) Slope $= 30$ (monthly fee), y-intercept $= 50$ (sign-up fee).
$C(t) = 30t + 50$
7 · Rates of Change in Quadratic Functions 二次函数的变化率
7.1 Quadratic Function 二次函数
$f(x) = ax^2 + bx + c$, where $a, b, c$ are real numbers and $a \neq 0$.
- Graph is a parabola (抛物线).
- Domain: $(-\infty, +\infty)$ (all real numbers 全体实数)
- $a > 0$: opens up (向上开口), concave up (上凹)
- $a < 0$: opens down (向下开口), concave down (下凹)
7.2 Second Differences — Test for Quadratic 二阶差分——识别二次函数
With equally spaced $x$-values:
- Compute first differences $\Delta_1$ (output differences 输出差).
- Compute second differences $\Delta_2$ (differences of first differences 一阶差分的差).
- If $\Delta_2$ is constant and nonzero → quadratic function. 二阶差分恒定非零 → 二次函数。
- If $\Delta_2 > 0$ → concave up ∪ | $\Delta_2 < 0$ → concave down ∩
• 一阶差分（first differences）恒定 → 线性函数（Linear）
• 二阶差分（second differences）恒定 → 二次函数（Quadratic）
• 二阶差分为正 → 上凹（concave up）；为负 → 下凹（concave down）
For $f(x)=ax^2+bx+c$ with equally spaced inputs of step $h = \Delta x$:
Only when $h = 1$ does the second difference equal $2a$.
只有当 $x$ 的间隔 $h=1$ 时，二阶差分才等于 $2a$。
例：$y=x^2$，取 $x=0,2,4,6$，得 $y=0,4,16,36$；一阶差分 $4,12,20$；二阶差分 $8 = 2(1)(2)^2$，不是 $2$。
Find the average rate of change of $f(x) = x^2$ on each interval: $[0,1]$, $[1,2]$, $[2,3]$. What do you notice?
$[1,2]$: $\dfrac{f(2)-f(1)}{2-1}=\dfrac{4-1}{1}=\mathbf{3}$
$[2,3]$: $\dfrac{f(3)-f(2)}{3-2}=\dfrac{9-4}{1}=\mathbf{5}$
Second differences: $3-1=2$, $5-3=2$ → constant! → quadratic ✓ and concave up.
Second differences = $2$ (constant, positive) → quadratic, concave up (上凹).
当 $x$ 间隔为 1 时，$f(x)=ax^2+bx+c$ 的二阶差分等于 $2a$（一般情况为 $2ah^2$，见上方说明）。
Determine whether the table represents a quadratic function. If it does, state the concavity.
| $x$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|---|
| $f(x)$ | 76 | 55 | 36 | 19 | 4 | $-9$ |
| Interval | [-3,-2] | [-2,-1] | [-1,0] | [0,1] | [1,2] |
|---|---|---|---|---|---|
| $\Delta_1$ | $-21$ | $-19$ | $-17$ | $-15$ | $-13$ |
| $\Delta_2$ ①→② | $\Delta_2$ ②→③ | $\Delta_2$ ③→④ | $\Delta_2$ ④→⑤ | |
|---|---|---|---|---|
| $-19-(-21)=+2$ | $-17-(-19)=+2$ | $-15-(-17)=+2$ | $-13-(-15)=+2$ |
(Step $h=1$, so second difference $= 2ah^2 = 2a = 2 \Rightarrow a = 1 > 0$ → confirms concave up.)
For each table, determine if the function is linear, quadratic, or neither. If quadratic, state concavity.
Table A:
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $f(x)$ | 5 | 2 | $-1$ | $-4$ | $-7$ |
Table B:
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $g(x)$ | 0 | $-3$ | $-8$ | $-15$ | $-24$ |
Table A: First differences: $-3,-3,-3,-3$ → all equal → Linear. Slope $= -3$.
Table B: First differences: $-3,-5,-7,-9$ → not constant → not linear.
Second differences: $-5-(-3)=-2$; $-7-(-5)=-2$; $-9-(-7)=-2$ → constant $-2$ → Quadratic, concave down (∩).
Let $f(x) = -2x^2 + 4x + 1$.
(a) Find the average rate of change on $[0, 3]$.
(b) Is $f$ concave up or concave down? How do you know from the equation?
(c) Is the average rate of change increasing or decreasing as we move right? Why?
(a) $f(0)=1$; $f(3)=-18+12+1=-5$
AROC $= \dfrac{-5-1}{3-0}=\dfrac{-6}{3}=\mathbf{-2}$
(b) $a=-2 < 0$ → parabola opens down → Concave down (∩).
开口向下，下凹。
(c) Since $f$ is concave down, the rate of change is decreasing as $x$ increases.
下凹意味着变化率（斜率）从左到右在减小。
7.3 How the Average Rates of Change Change 平均变化率本身如何变化
| Linear 线性 | Quadratic 二次 | |
|---|---|---|
| AROCs over consecutive equal-length intervals 连续等长区间上的平均变化率 | Constant 常数 | Given by a linear function 构成一次函数（等差） |
| How those AROCs change 平均变化率的变化速度 | At a rate of zero 变化率为 0 | At a constant rate 以常数速率变化 |
| Differences 差分 | 1st differences constant | 2nd differences constant |
| Concavity 凹凸 | Neither 无 | Up if $a>0$, down if $a<0$ |
For $f(x)=ax^2+bx+c$, the AROC on $[x,\ x+h]$ is
$$\frac{f(x+h)-f(x)}{h} = \frac{a(2xh+h^2)+bh}{h} = 2a\,x + (ah+b),$$which is a linear function of $x$ with slope $2a$. Each time you move one interval to the right ($x \to x+h$), the AROC changes by exactly $2ah$.
所以二次函数在连续等长区间上的平均变化率是等差数列，公差为 $2ah$；它们以每单位 $x$ 变化 $2a$ 的常数速率在变。
| $x$ | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| $f(x)$ | 1 | 5 | 17 | 37 | 65 |
(a) Find the average rate of change of $f$ on $[0,2]$, $[2,4]$, $[4,6]$, $[6,8]$.
(b) How do these average rates of change change? What type of function could model $f$?
(c) If $f(x)=ax^2+bx+c$, find $a$. Is the graph concave up or concave down?
(a) $\dfrac{5-1}{2}=2$, $\dfrac{17-5}{2}=6$, $\dfrac{37-17}{2}=10$, $\dfrac{65-37}{2}=14$.
(b) They increase by $4$ each time, a constant amount. The AROCs change at a constant rate, so $f$ is quadratic.
(c) First differences: $4, 12, 20, 28$; second differences: $8$. With $h=2$: $a = \dfrac{8}{2(2)^2} = 1$ (not $4$!). Since $a>0$, the graph is concave up. (Indeed $f(x)=x^2+1$.)
也可以用 AROC 的公差：$2ah = 4 \Rightarrow a = 1$。
8 · Summary Table 总结表格
| Concept 概念 | Key Test / Formula 关键检验/公式 | 中文要点 |
|---|---|---|
| Function 函数 | Each input has exactly one output. | 每个输入唯一对应一个输出 |
| Four Representations 四种表示 |
Verbal, analytical, numerical, graphical | 文字、解析式、表格、图像；情境会限制定义域 |
| Vertical Line Test | A vertical line hits the graph at most once. | 竖线与图像最多交于一点 |
| Domain 定义域 | Valid $x$-values: exclude denom=0, even root of neg, log≤0 | 排除分母为0、偶次根号内负数、对数真数≤0 |
| Zero / x-intercept 零点 | Set $f(x)=0$, solve for $x$ | 令 $f(x)=0$ 求 $x$ |
| y-intercept y截距 | Find $f(0)$ | 令 $x=0$，计算 $f(0)$ |
| Average Rate of Change 平均变化率 |
$\dfrac{f(b)-f(a)}{b-a}$ = slope of secant line | 输出变化量 ÷ 输入变化量 = 割线斜率 |
| Linear Function 线性函数 | Constant first differences; $f(x)=mx+b$ | 一阶差分恒定，变化率为常数 |
| Quadratic Function 二次函数 | Constant second differences; AROCs form a linear function; $f(x)=ax^2+bx+c$ | 二阶差分恒定，图像为抛物线 |
| Concave Up 上凹 | Rates of change increasing; $a>0$; ∪ | 变化率递增，开口向上 |
| Concave Down 下凹 | Rates of change decreasing; $a<0$; ∩ | 变化率递减，开口向下 |
| Four Shapes 四种形状 | Inc/dec = sign of rate; concavity = whether rate is increasing or decreasing | 增减看变化率正负，凹凸看变化率增减 |
| Words → Graph 文字 → 图像 |
Variables → direction → speed → special points | 四步法：变量、方向、快慢、特殊点 |
| Rate at a Point 某点的变化率 |
Approximate with AROC on a small interval containing the point; compare two points this way | 用包含该点的小区间近似；比较两点同理 |
| Sign of Rate 变化率正负 | Positive: vary together in the same direction; negative: opposite directions | 正：同增同减；负：一增一减 |
| Interpretation 解释 | Interval + value + output unit per input unit + context | 区间、数值、单位、情境缺一不可 |
| Second Difference 二阶差分 | $\Delta_2 = 2ah^2$ (equals $2a$ only when $h=1$) | $x$ 间隔为 $h$ 时 $a=\Delta_2/(2h^2)$ |
核心要点回顾：
- 函数（function）：每个输入唯一对应一个输出。用 VLT 检验图像。
- 定义域（domain）：三条规则——分母≠0、偶次根≥0、对数真数>0。
- 零点（zero）令 $f(x)=0$；y截距令 $x=0$。
- 平均变化率（AROC）= $\dfrac{\Delta f}{\Delta x}$ = 割线斜率。近似瞬时变化率用最近区间。
- 线性 vs 二次：一阶差分恒定→线性；二阶差分恒定→二次。二阶差分的正负决定凹凸性。
- 四种表示（文字、解析式、表格、图像）要能互相转换；读图时区间写 $x$ 的范围。
- 增减 × 凹凸 = 四种形状；文字情境画图用四步法：变量、方向、快慢、特殊点。
- 比较两点的变化率：各取包含该点的小区间算 AROC 再比较；"更大"看符号，"更快"看绝对值。
- 回答要写完整句子：区间、数值、单位（输出/输入）、情境含义。
- 二阶差分 $=2ah^2$，只有 $x$ 间隔为 1 时才等于 $2a$。
Chapter 1 Review 章末习题
Which of the following relations defines $y$ as a function of $x$?
答案：(D)
(A) $x=y^2-4$：$x=0$ 时 $y=\pm2$，一个输入两个输出。(B) 输入 $3$ 对应 $1$ 和 $7$。(C) 圆 $x^2+y^2=25$ 不过垂直线检验。(D) 每个 $x$ 只算出一个 $y$，图像是 V 形，过垂直线检验 ✓
The function $f$ is given by $f(x)=|x-1|$. What is the average rate of change of $f$ on the interval $-2\le x\le 3$, if it exists?
答案：(A) $\dfrac{f(3)-f(-2)}{3-(-2)}=\dfrac{2-3}{5}=-\dfrac15$。
平均变化率只看两个端点的函数值，图像中间有"拐角"不影响它存在。
What is the domain of $g(x)=\dfrac{\sqrt{6-2x}}{x+1}$ ?
答案：(B) 根号：$6-2x\ge0\Rightarrow x\le3$；分母：$x\ne-1$。两个条件同时满足：$(-\infty,-1)\cup(-1,3]$。$x=3$ 时 $g(3)=0$，可以取到，所以用方括号。
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $f(x)$ | 4 | 7 | 9 | 10 | 10.5 | 10.75 |
The table gives values of the function $f$ at selected values of $x$. Which of the following conclusions with reason is consistent with the values in the table?
答案：(C) 一阶差分（等间距，所以就是平均变化率）：$3,\ 2,\ 1,\ 0.5,\ 0.25$。都为正 → 递增；越来越小 → 图像下凹（increasing at a decreasing rate）。
(A) 只看输出变大，不能判断凹凸。(B) 变化率在减小，不是增大。(D) 一阶差分不恒定，不是线性。
The graph of the function $g$ is shown. The relative maximum of $g$ occurs at point $P$, and the only point of inflection of the graph of $g$ is $R$. At which of the labeled points is the rate of change of $g$ negative and increasing?
答案：(D) "变化率为负" = 递减；"变化率在增大" = 上凹。
$P$ 是相对极大值，变化率为 $0$；$Q$ 在拐点左边，递减且下凹（变化率为负且在减小）；$R$ 是拐点，递减最快（变化率最小）；$S$ 在拐点右边，递减且上凹 → 变化率为负且在增大 ✓
A ball is thrown upward from the edge of a rooftop. At $t=0$ seconds it is $6$ meters above the ground, at $t=1$ it is $21.1$ meters above the ground, and at $t=2$ it is $26.4$ meters above the ground. A quadratic model $h(t)$ is fit to these three data points. According to the model, at what time $t$ does the ball hit the ground?
答案：(B) 设 $h(t)=at^2+bt+c$：$c=6$，$a+b+c=21.1$，$4a+2b+c=26.4$，解得 $h(t)=-4.9t^2+20t+6$。
令 $h(t)=0$，取正根：$t=\dfrac{20+\sqrt{517.6}}{9.8}\approx 4.362$ 秒。（$2.041$ 是最高点的时刻；$4.082$ 是回到 $6$ 米高、也就是屋顶高度的时刻。）
| $t$ (minutes) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| $T(t)$ (°C) | 92.0 | 76.2 | 63.9 | 54.4 | 47.1 | 41.4 | 37.0 |
A bowl of soup cools on a table. The temperature of the soup $t$ minutes after it is served is $T(t)$ degrees Celsius. Using the data in the table, which of the following is the best estimate of the rate of change of $T$ at $t=12$ minutes, with a correct interpretation?
答案：(A) 用包含 $t=12$ 的最短区间 $[10,15]$：$\dfrac{54.4-63.9}{15-10}=-1.9$ °C/min。负号表示温度在下降。
(B) 用了更宽的区间 $[10,20]$，估计不如 (A) 好；(C) 忘了除以 $\Delta t$；(D) 符号和含义都错。
Water drains from a tank. The volume of water in the tank, in liters, $t$ minutes after a valve is opened is modeled by
$V(t)=\begin{cases}300-0.5t^2, & 0\le t<10\\[2pt] 400-1.5t^2, & 10\le t\le 14\end{cases}$
Which of the following is a correct statement about the average rates of change of $V$?
答案：(B) $V(0)=300,\ V(4)=292,\ V(8)=268,\ V(12)=184,\ V(14)=106$。
平均变化率：$[0,4]$ 为 $-2$，$[4,8]$ 为 $-6$，$[8,12]$ 为 $-21$，$[12,14]$ 为 $-39$（L/min）；$[0,8]$ 为 $-4$，$[8,14]$ 为 $-27$。只有 (B) 成立：$-21<-6$。
注意 $[8,12]$ 跨过了分段点 $t=10$，两个端点要分别代入对应的那一段。
| $x$ | 0 | 2 | 3 | 5 | 8 |
|---|---|---|---|---|---|
| $f(x)$ | 1 | 9 | 12 | 15 | 16.5 |
The function $f$ is increasing on $0\le x\le 8$. The table gives values of $f(x)$ at selected values of $x$.
(a) Find the average rate of change of $f$ over $0\le x\le 8$. Show the computation.
(b) Based on the table, is the graph of $f$ concave up or concave down on $0\le x\le 8$? Give a reason. Hint: the input values are not equally spaced.
(c) Use the table to estimate the rate of change of $f$ at $x=4$.
(d) Use your answer from (c) to estimate $f(4)$ with a line through $(3,\,f(3))$. Is this estimate likely to be greater or less than the actual value of $f(4)$? Explain.
(a) $\dfrac{f(8)-f(0)}{8-0}=\dfrac{16.5-1}{8}=1.9375$。
(b) 下凹。各段平均变化率：$[0,2]$ 为 $4$，$[2,3]$ 为 $3$，$[3,5]$ 为 $1.5$，$[5,8]$ 为 $0.5$，依次减小，所以变化率在减小 → 下凹。
注意：$x$ 不等间距，不能直接比较输出差 $8,\,3,\,3,\,1.5$（中间两个相等会误判）。
(c) 用包含 $x=4$ 的区间 $[3,5]$：$\dfrac{15-12}{2}=1.5$。
(d) $f(4)\approx 12+1.5(4-3)=13.5$。估计值偏小：图像下凹，$[3,5]$ 上的割线在曲线下方，所以割线上 $x=4$ 处的值小于 $f(4)$。
A small shop sells hand-made candles. When the price is $x$ dollars per candle, the weekly profit is $P(x)$ dollars. The shop recorded a weekly profit of $\$420$ at a price of $\$10$, $\$540$ at $\$14$, and $\$480$ at $\$20$. The profit is modeled by a quadratic function $P(x)=ax^2+bx+c$.
(a) (i) Use the given data to write three equations that can be used to find $a$, $b$, and $c$. (ii) Find the values of $a$, $b$, and $c$.
(b) (i) Use the given data to find the average rate of change of the profit from $x=10$ to $x=14$. Interpret your answer, with units. (ii) Use the average rate of change to estimate $P(12)$. Compare this estimate with $P(12)$ from the model, and use the concavity of the model to explain the difference.
(c) For what prices does the model predict a positive profit? What is the greatest weekly profit the model predicts, and at what price?
(a) $100a+10b+c=420$，$196a+14b+c=540$，$400a+20b+c=480$；解得 $a=-4,\ b=126,\ c=-440$，即 $P(x)=-4x^2+126x-440$。
(b)(i) $\dfrac{540-420}{14-10}=30$ dollars per dollar：价格从 \$10 提到 \$14，每提价 1 美元，周利润平均增加 30 美元。
(ii) 线性估计 $P(12)\approx420+30\cdot2=480$；模型 $P(12)=-576+1512-440=496$。$a<0$，模型图像下凹，割线在曲线下方，所以用平均变化率的估计偏小。
(c) $P(x)=0\Rightarrow x=\dfrac{126\pm\sqrt{8836}}{8}=\dfrac{126\pm94}{8}$，即 $x=4$ 或 $x=27.5$。利润为正：$4<x<27.5$（美元）。顶点 $x=\dfrac{126}{8}=15.75$，最大周利润 $P(15.75)=552.25$ 美元。
A drone takes off from the ground at $t=0$ seconds. It rises quickly at first and then more and more slowly, reaching a height of $60$ meters at $t=20$. It hovers at $60$ meters until $t=30$. Then it descends, slowly at first and then faster and faster, and lands at $t=50$. Let $h(t)$ be the height of the drone, in meters.
(a) Sketch a possible graph of $h$ for $0\le t\le 50$.
(b) On which intervals is $h$ increasing? decreasing? On which intervals is the graph of $h$ concave up? concave down?
(c) Write one complete sentence that describes $h$ and its rate of change on $30<t<50$.
(d) Find the average rate of change of $h$ over $0\le t\le 50$. Does this value mean the drone did not move? Explain.
(a) 参考图：
(b) 递增：$(0,20)$；递减：$(30,50)$；$(20,30)$ 上为常数。下凹：$(0,20)$ 和 $(30,50)$（两段的变化率都在减小）；没有上凹的区间。
(c) "On $30<t<50$, $h$ is decreasing, and the rate of change of $h$ is decreasing (it becomes more and more negative), so the drone descends faster and faster."
(d) $\dfrac{h(50)-h(0)}{50}=\dfrac{0-0}{50}=0$ m/s。平均变化率只反映起点和终点的净变化：无人机回到了地面，但中途上升到 60 米又落下，并不是没动。
| $x$ | $-3$ | 0 | 3 | 6 | 9 |
|---|---|---|---|---|---|
| $f(x)$ | 34 | 1 | 4 | 43 | 118 |
The table gives values of a function $f$ for equally spaced values of $x$.
(a) Use differences to show that the data are consistent with a quadratic function.
(b) Assume $f(x)=ax^2+bx+c$. Find $a$. A student claims $a=18$ because "$a$ is half the second difference." Explain the error.
(c) Find $b$ and $c$.
(d) Find the average rates of change of $f$ over $[0,3]$ and $[3,6]$. By how much does the average rate of change increase from one interval of length $3$ to the next? Use this pattern to find the average rate of change over $[9,12]$.
(a) 一阶差分：$-33,\ 3,\ 39,\ 75$；二阶差分：$36,\ 36,\ 36$，恒定 → 与二次函数一致。
(b) 间隔 $h=3$ 时 $\Delta_2=2ah^2=18a$，所以 $36=18a$，$a=2$。"$a$ 等于二阶差分的一半"只在 $h=1$ 时成立，这里 $h=3$。
(c) $f(0)=c=1$；$f(3)=18+3b+1=4\Rightarrow b=-5$。所以 $f(x)=2x^2-5x+1$。
(d) $[0,3]$：$\dfrac{4-1}{3}=1$；$[3,6]$：$\dfrac{43-4}{3}=13$。每往后一个长度为 3 的区间，平均变化率增加 $\dfrac{\Delta_2}{h}=\dfrac{36}{3}=12$。所以 $[6,9]$ 为 $25$，$[9,12]$ 为 $37$（验证：$f(12)=229$，$\dfrac{229-118}{3}=37$ ✓）。
