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Axiom Learning·AP Precalculus基本函数与变换 — Graphs, Shifts, Composition & Inverses
| Term 术语 | Meaning 含义 | Chinese 中文 |
|---|---|---|
| Parent Function | Simplest, untransformed version of a function family | 基本函数 |
| Transformation | Change applied to a graph (shift, stretch, reflect) | 变换 |
| Translation | Vertical or horizontal shift | 平移 |
| Dilation | Stretch or shrink (vertical or horizontal) | 伸缩 |
| Reflection | Flip over an axis or line | 反射（翻转） |
| Composite Function | $f\circ g$: output of $g$ becomes input of $f$ | 复合函数 |
| One-to-One Function | Each output corresponds to exactly one input | 一一对应函数 |
| Inverse Function | $f^{-1}$: swaps input and output of $f$ | 反函数 |
Graph the piecewise function: $$f(x)=\begin{cases}x^2, & x<2 \\ x+2, & 2\leq x<4 \\ x, & x\geq 4\end{cases}$$ State the domain and range.
Evaluate $g(x)=\begin{cases}3x+1, & x\leq -1 \\ x^2-2, & -1<x\leq 3 \\ 5, & x>3\end{cases}$ at $x=-2, 0, 3, 5$. State the range.
| Transformation 变换 | Form 形式 | Effect 效果 | Chinese 中文 |
|---|---|---|---|
| Vertical Translation | $f(x)+k$ | $k>0$: shift up; $k<0$: shift down | 垂直平移：上移/下移 |
| Horizontal Translation | $f(x+h)$ | $h>0$: shift left; $h<0$: shift right | 水平平移：左移/右移（注意方向与符号相反！） |
| Vertical Dilation | $a\cdot f(x)$ | $|a|>1$: stretch; $0<|a|<1$: shrink | 垂直伸缩（乘以$a$） |
| Horizontal Dilation | $f(bx)$ | $|b|>1$: shrink horizontally; $0<|b|<1$: stretch | 水平伸缩（注意：$b$ 越大，图像越窄） |
| Reflect over x-axis | $-f(x)$ | Flip vertically (negate all $y$-values) | 关于x轴翻转（$y$ 取反） |
| Reflect over y-axis | $f(-x)$ | Flip horizontally (negate all $x$-values) | 关于y轴翻转（$x$ 取反） |
The graph of $y=|x|$ undergoes the following transformations in order:
(1) Horizontal shift 2 units to the right
(2) Vertical dilation by a factor of $\frac{1}{3}$ (a vertical compression 垂直压缩)
(3) Vertical translation 7 units up
Find the equation of the resulting graph.
The graph of $y=f(x)$ is given: it consists of line segments through $(-2,0)$, $(0,2)$, $(2,2)$, and $(4,0)$. Determine the graph of $y=2f(x+1)-3$ by showing the effect of each transformation in sequence. Also state the domain and range of the new function.
| on $f$ | $(-2,0)$ | $(0,2)$ | $(2,2)$ | $(4,0)$ |
|---|---|---|---|---|
| on $y=2f(x+1)-3$ | $(-3,-3)$ | $(-1,1)$ | $(1,1)$ | $(3,-3)$ |
Let $C_1$ be the curve $y=f(x)=3x^2-16$. Find the equation of $C_2$, which is a horizontal shrink of $C_1$ by a factor of $\frac{1}{2}$ and is reflected over the $x$-axis.
For each function, describe the parent function and sequence of transformations. State the domain and range.
(a) $y=(3x)^2-4$ (b) $y=-\sqrt[3]{x+1}$
Write the equation of $y=\sqrt{x}$ after: (1) reflect over y-axis, (2) horizontal stretch by factor 3, (3) shift up 2.
The point $(4, -3)$ is on the graph of $y=f(x)$. Find the corresponding point on $y=-2f(x-1)+5$.
Use the same $f$ as in T-3 (segments through $(-2,0)$, $(0,2)$, $(2,2)$, $(4,0)$).
(a) Rewrite $g(x)=-f(2x-4)+1$ in the form $a\,f(b(x+h))+k$ and describe the transformations in a correct order.
(b) Find the images of the four key points, and the domain and range of $g$.
(c) A parabola has vertex $(3,-2)$ and passes through $(5,6)$. Write it as a transformation of $y=x^2$.
| Model 模型 | Use it when… 什么时候用（CED 1.13.A） |
|---|---|
| Linear 线性 | Roughly constant rate of change (first differences roughly constant) 变化率大致恒定 |
| Quadratic 二次 | Rates of change are roughly linear (second differences roughly constant); data roughly symmetric with a unique max or min; area / two-dimensional contexts 面积问题 |
| Cubic 三次 | Volume / three-dimensional contexts; third differences roughly constant 体积问题 |
| Polynomial, degree $n$ | Multiple real zeros or multiple maxima/minima; roughly constant nonzero $n$th differences. Any $n+1$ points with distinct inputs can be fit exactly by a polynomial of degree $\le n$. $n+1$ 个点一定能被次数 $\le n$ 的多项式穿过 |
| Rational 有理 | Quantities that are inversely proportional, e.g. $y=\frac{k}{x}$, or inverse-square laws $y=\frac{k}{x^2}$ (gravity, light intensity) 反比、平方反比 |
| Piecewise 分段 | The data or context behaves differently on different intervals 不同区间规律不同 |
| Exponential 指数（Unit 2） | Proportional (percent) change over equal intervals 等比增长/衰减 |
| Square root 平方根 | Output grows but more and more slowly (increasing, concave down) 增速递减 |
An engineer is designing a crash test barrier. Time after impact $t$ is in milliseconds; depression distance $d$ is in millimeters. The equation has form $d=a(t-h)^2+k$.
(a) If the maximum depression is 120 mm at $t=150$ ms, write the vertex form.
(b) If the ideal barrier is depressed 100 mm after 130 ms, find $a$.
(c) What is the maximum depression, and when does it occur?
(d) Find the average rate of change of $d$ from $t=130$ to $t=150$ ms, with units, and interpret it.
(e) Use the model to find $d(0)$. What does this say about where the model can be used?
If $x$ varies inversely with $y$ and $x=10$ when $y=4$, find the constant of variation $k$ and the equation.
Classify each scenario and name the best function model type:
A 500-liter tank drains through a valve. A student models the volume of water with $V(t)=500-20t$ liters, where $t$ is in minutes.
(a) What assumption does this model make about how $V$ and $t$ change together?
(b) State an appropriate domain and range for the model. Explain.
(c) In reality, water drains more slowly as the tank empties (less pressure). How would a better model's graph differ?
The weekly number of visitors to a new exhibit is modeled by $N(t)=-2t^2+24t+50$, where $t$ is the number of weeks after opening.
(a) Explain why a quadratic model might be reasonable here.
(b) State a domain restriction based on the model's outputs.
(c) The model gives $N(2.5)=97.5$. What range restriction should be applied to predictions?
(d) When does the model predict the most visitors, and how many?
How to choose: look at the scatterplot and the differences (1st → linear, 2nd → quadratic, 3rd → cubic). 先看散点图和差分，再选回归类型。
The height $h$ (meters) of a kicked ball $t$ seconds after it is kicked is recorded:
| $t$ (s) | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |
|---|---|---|---|---|---|---|---|
| $h$ (m) | 1.1 | 7.0 | 10.9 | 11.8 | 10.5 | 7.0 | 0.6 |
(a) Which function type is appropriate? Justify using differences.
(b) Find the regression model.
(c) Use the model to estimate when the ball hits the ground.
(d) Use the model to find the average rate of change of $h$ from $t=0.5$ to $t=1.5$, with units, and interpret it.
(e) State a reasonable domain for the model.
The points $(0,2)$, $(1,3)$, $(2,0)$, and $(3,5)$ are given.
(a) Without calculating, explain why some polynomial of degree at most 3 passes through all four points.
(b) Use CubicReg to find it. (c) Use the model to predict the output at $x=4$. Why should this prediction be treated with caution?
A car starts from rest. For the first 10 seconds its speed is $0.3t^2$ m/s; after that it keeps the speed it has reached at $t=10$ until $t=60$.
(a) Write a piecewise-defined model $v(t)$ for $0\le t\le60$. (b) Find $v(5)$ and $v(20)$ with units.
(c) Find the average rate of change of $v$ on $[0,10]$ and on $[10,60]$, with units. Why is a piecewise model appropriate?
(a) A phone plan costs \$30 per month for up to 5 GB of data, plus \$8 per GB beyond 5 GB. Write a piecewise model $C(g)$, find $C(3)$ and $C(7.5)$, and find the data use that gives a \$70 bill.
(b) The intensity of light from a bulb is inversely proportional to the square of the distance: $I=\dfrac{k}{d^2}$. At $d=2$ m, $I=400$ units. Find $k$, the intensity at $d=4$ m, and the distance where $I=25$. What happens to $I$ when the distance doubles?
Given $f(x)=x^2+1$ and $g(x)=2x-3$, find:
(a) $(f\circ g)(x)$ (b) $(g\circ f)(x)$ (c) $(f\circ g)(2)$
Given $f(x)=\sqrt{x}$ and $g(x)=x-4$, find $(f\circ g)(x)$ and its domain.
Let $f(x)=\dfrac{1}{x}$ and $g(x)=x+2$.
(a) Find $(f\circ g)(x)$ and its domain.
(b) Find $(g\circ f)(x)$ and its domain.
(c) Are they equal?
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $f(x)$ | 3 | 4 | 1 | 0 | 2 |
| $g(x)$ | 2 | 0 | 4 | 1 | 3 |
(a) Find $f(g(2))$, $g(f(2))$, and $f(f(0))$. (b) Make a table of $(f\circ g)(x)$ for $x=0,1,2,3,4$. (c) Solve $f(g(x))=4$.
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $g(x)$ | 2 | 0 | 4 | 1 | 3 |
| $f(g(x))$ | 1 | 3 | 2 | 4 | 0 |
The graph of $f$ (segments through $(0,4)$, $(2,0)$, $(4,2)$, $(6,2)$) is shown, and $g(x)=x-1$.
Find: (a) $f(g(3))$ (b) $g(f(4))$ (c) $f(f(2))$ (d) $f(g(4.5))$
A stone dropped in a pond makes a circular ripple whose radius grows as $r(t)=5t$ cm, where $t$ is in seconds. The area of a circle is $A(r)=\pi r^2$.
(a) Write the area as a function of time. (b) Find the area at $t=4$ s. (c) Why is composition useful here?
Write each function as $f(g(x))$ by naming $f$ and $g$ (neither may be $x$ itself):
(a) $(x^2-5)^3$ (b) $\dfrac{2}{x+4}$ (c) $|4x-1|+2$ (d) Write $y=3\sqrt{x}$ and $y=\sqrt{x-2}$ as compositions of $\sqrt{x}$ with a linear function.
Find the inverse of $f(x)=3x+7$. Verify using the identity property.
Given $f(x)=\dfrac{2x}{x+1}$, $x\neq -1$, find its inverse $f^{-1}(x)$.
Find the inverse of $g(x)=\sqrt{x-3}+1$, and state the domain of $g^{-1}(x)$.
Determine if each function is one-to-one. If not, suggest a domain restriction that makes it invertible.
(a) $f(x)=x^2-4$ (b) $g(x)=x^3-1$ (c) $h(x)=|2x+1|$
The function $f$ is one-to-one. Selected values are shown.
| $x$ | $-1$ | 0 | 2 | 3 | 6 |
|---|---|---|---|---|---|
| $f(x)$ | 9 | 5 | 1 | 0 | $-4$ |
(a) Find $f^{-1}(5)$ and $f^{-1}(0)$. (b) Find $f\big(f^{-1}(-4)\big)$ and $f^{-1}\big(f(6)\big)$. (c) Make a table for $f^{-1}$.
| $x$ | 9 | 5 | 1 | 0 | $-4$ |
|---|---|---|---|---|---|
| $f^{-1}(x)$ | $-1$ | 0 | 2 | 3 | 6 |
Temperature in degrees Fahrenheit is $F(C)=\tfrac95C+32$, where $C$ is the temperature in degrees Celsius.
(a) Find the inverse function and explain what its input and output mean. (b) Evaluate the inverse at 98.6 and interpret.
(c) The coldest possible temperature is $-273.15$ °C. How does this restrict the domain of the inverse?
(a) The tank model $V(t)=500-20t$, $0\le t\le25$ (from 4.3). Find the inverse, state its domain and range, and find how long it takes for 140 L to remain.
(b) $h$ is one-to-one with $h(1)=4$, $h(2)=7$, $h(5)=-3$. Find $h^{-1}(7)$ and $h(h^{-1}(-3))$. Which point must lie on the graph of $h^{-1}$?
| Topic 主题 | Key Rule 关键规则 | Watch out! 注意 |
|---|---|---|
| Parent Functions 基本函数 | Know domain, range, shape of each | Square root: domain $[0,\infty)$; reciprocal: $x\neq0$ |
| Vertical shift 垂直平移 | $f(x)+k$: up if $k>0$ | Outside the function |
| Horizontal shift 水平平移 | $f(x+h)$: left if $h>0$ ⚠️ | Inside the function; direction is opposite! |
| Vertical dilation 垂直伸缩 | $a\cdot f(x)$: stretch if $|a|>1$ | $a<0$ also reflects over x-axis |
| Horizontal dilation 水平伸缩 | $f(bx)$: shrink if $|b|>1$ | $b$ and stretch/shrink are inversely related |
| Order 变换顺序 | $a\,f(b(x+h))+k$: $(x,y)\to\left(\frac{x}{b}-h,\ ay+k\right)$ | Dilate/reflect first, then translate |
| Model choice 模型选择 | Differences, shape, context (area → quadratic, volume → cubic) | $n+1$ points → degree $\le n$ fits exactly |
| Assumptions 假设与限制 | What stays constant; how quantities change; domain/range limits | Round counts; check the context domain |
| Regression 回归 | STAT → CALC: LinReg / QuadReg / CubicReg / QuartReg | 3 decimal places; beware extrapolation |
| Composition 复合 | $(f\circ g)(x)=f(g(x))$: inside-out, also from tables and graphs | Not commutative; check domain; transformations = composition with $x+k$ or $kx$ |
| Inverse 反函数 | Swap $x\leftrightarrow y$, solve for $y$; tables: reverse pairs | Must be one-to-one (HLT); reflect over $y=x$; context may limit it |
The functions $f$ and $g$ are defined for all real numbers such that $g(x)=f\big(3(x+2)\big)$. Which of the following sequences of transformations maps the graph of $f$ to the graph of $g$ in the same $xy$-plane?
答案：(C) 先水平伸缩 $\tfrac13$：$y=f(3x)$；再把 $x$ 换成 $x+2$（向左平移 2）：$y=f\big(3(x+2)\big)$ ✓。
(B) 先平移得 $f(x+2)$，再伸缩把 $x$ 换成 $3x$：$f(3x+2)$ ✗。(D) 得到 $f\big(3(x+6)\big)$ ✗。注意：$f(3x+6)$ 要先提出 3 才能读出平移量 $-2$；如果按"先平移后伸缩"，平移量应是 $-6$。
The function $f$ has a maximum at the point $(4,-1)$. The function $h$ is given by $h(x)=-2f(x-3)+5$. Which of the following must be true about the graph of $h$?
答案：(B) 水平：$x-3=4\Rightarrow x=7$。竖直：$-2(-1)+5=7$。乘以负数 $-2$ 把图像上下翻转，原来的最大值变成最小值，所以 $h$ 在 $(7,7)$ 取最小值。
(C) 把 $x-3$ 当成向左平移了；(D) 漏掉了负号。
| $x$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|
| $f(x)$ | $3$ | $2$ | $-1$ | $0$ | $1$ |
| $g(x)$ | $2$ | $-1$ | $0$ | $3$ | $-2$ |
The functions $f$ and $g$ have domain all real numbers. The table gives values of $f(x)$ and $g(x)$ at selected values of $x$. Which of the following gives a solution to $g\big(f(x)\big)=0$ with correct reasoning?
答案：(D) 由外往里：$g(\text{?})=0$ 要求 $\text{?}=1$（$g(1)=0$）；再解 $f(x)=1$，得 $x=3$。验证：$f(3)=1$，$g(1)=0$ ✓。
(A) 算的是 $f\big(g(-1)\big)=f(2)=0$，复合顺序反了；(B)、(C) 只用了一个函数。
The function $f$ is given by $f(x)=\dfrac{2x+1}{x-3}$ for $x\ne3$. Which of the following is an expression for $f^{-1}(x)$?
答案：(A) 设 $y=\dfrac{2x+1}{x-3}$，交换 $x,y$：$x=\dfrac{2y+1}{y-3}$。去分母 $xy-3x=2y+1$，含 $y$ 的项移到一边：$y(x-2)=3x+1$，所以 $f^{-1}(x)=\dfrac{3x+1}{x-2}$。
检验：$f(4)=9$，$f^{-1}(9)=\dfrac{28}{7}=4$ ✓。(B) 是 $\dfrac{1}{f(x)}$——$f^{-1}$ 不是倒数。
The function $k$ is a transformation of the parent function $y=\sqrt{x}$. The domain of $k$ is $x\le3$ and the range of $k$ is $y\ge-1$. Which of the following could be an expression for $k(x)$?
答案：(C) $\sqrt{3-x}=\sqrt{-(x-3)}$：先关于 $y$ 轴反射，再右移 3，定义域 $3-x\ge0\iff x\le3$；根号的值 $\ge0$，减 1 后值域 $y\ge-1$ ✓。
(A) 定义域 $x\ge3$；(B) 前面的负号让值域变成 $y\le-1$；(D) $-x-3\ge0\iff x\le-3$。
The graph of the piecewise-linear function $f$ is shown for $-2\le x\le5$. Let $g$ be the inverse function of $f$. What is the domain of $g$, and what is the value of $g(4)$?
答案：(A) 反函数的定义域 = 原函数的值域：$[-4,7]$。
$g(4)$ 问的是"哪个输入让 $f$ 输出 4"。在 $[0,5]$ 上 $f(x)=x+2$，令 $x+2=4$ 得 $x=2$，所以 $g(4)=2$。
(B) 算的是 $f(4)=6$；(C) 把 $f^{-1}$ 当成了倒数；(D) 用了 $f$ 的定义域。
| $t$ (minutes) | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
|---|---|---|---|---|---|---|---|
| $s(t)$ (km/h) | $10.2$ | $14.1$ | $14.9$ | $14.3$ | $14.0$ | $15.2$ | $19.1$ |
The table gives the speed of a cyclist at selected times $t$ minutes after the start of a ride. A cubic regression is used to model these data. What is the speed, in kilometers per hour, at $t=4.5$ predicted by the cubic regression model?
答案：(B) STAT → CALC → 6: CubicReg：$s(t)\approx0.242t^3-2.869t^2+10.661t+2.214$，代入 $t=4.5$ 得 $\approx14.111$ km/h（用存进 $Y_1$ 的完整系数计算）。
其他选项来自别的回归：(A) QuartReg，(C) QuadReg，(D) LinReg。
Oil leaking from a ship forms a circular spill. The radius of the spill, in meters, $t$ hours after the leak begins is modeled by $r(t)=2+3.5\sqrt{t}$. The area of a circle with radius $r$ is $A(r)=\pi r^2$. According to the model, how many hours after the leak begins does the area of the spill reach $2000$ square meters?
答案：(C) $A\big(r(t)\big)=\pi\big(2+3.5\sqrt t\big)^2$ 是面积关于时间的函数。令它等于 2000：$r=\sqrt{2000/\pi}\approx25.231$，$3.5\sqrt t\approx23.231$，$\sqrt t\approx6.638$，$t\approx44.057$ 小时。（也可以直接用计算器求 $Y_1=\pi(2+3.5\sqrt x)^2$ 与 $Y_2=2000$ 的交点。）
(A) 停在了 $\sqrt t$；(B) 是半径，不是时间；(D) 漏掉了 $r(t)$ 里的常数 2。
| $x$ | $-3$ | $-1$ | $0$ | $2$ | $4$ |
|---|---|---|---|---|---|
| $f(x)$ | $12$ | $4$ | $-1$ | $0$ | $24$ |
The domain of $f$ consists of the five numbers in the table, and $f$ is one-to-one. The function $g$ is given by $g(x)=\sqrt{2x+1}$. Let $h$ be the function given by $h(x)=g\big(f(x)\big)$.
(a) Find $h(-1)$. Then find all values of $x$ in the domain of $h$, and explain why $x=0$ is not in the domain of $h$.
(b) Find $f^{-1}(0)$. Then find the value of $x$ for which $h(x)=7$.
(c) Find an expression for $g^{-1}(x)$, and state the domain of $g^{-1}$.
(d) Find the value of $f^{-1}\big(g^{-1}(5)\big)$. Show the work that leads to your answer.
(a) $h(-1)=g\big(f(-1)\big)=g(4)=\sqrt9=3$。
$h$ 的定义域：$x$ 在 $f$ 的定义域里，且 $f(x)$ 在 $g$ 的定义域里（$2f(x)+1\ge0$，即 $f(x)\ge-\tfrac12$）。$f(0)=-1$，$g(-1)=\sqrt{-1}$ 无定义，所以 $x=0$ 不在定义域里。定义域：$\{-3,\,-1,\,2,\,4\}$。
(b) $f(2)=0$，所以 $f^{-1}(0)=2$。
$h(x)=7\iff\sqrt{2f(x)+1}=7\iff f(x)=24$，由表 $f(4)=24$，所以 $x=4$。
(c) $y=\sqrt{2x+1}$，交换：$x=\sqrt{2y+1}$，$x^2=2y+1$，$g^{-1}(x)=\dfrac{x^2-1}{2}$。$g^{-1}$ 的定义域 = $g$ 的值域：$x\ge0$（不写这个限制，$\dfrac{x^2-1}{2}$ 就不是一一对应的）。
(d) $g^{-1}(5)=\dfrac{25-1}{2}=12$，$f^{-1}(12)=-3$（因为 $f(-3)=12$）。所以 $f^{-1}\big(g^{-1}(5)\big)=-3$。
The graph of the piecewise-linear function $f$ is shown. Its domain is $-4\le x\le4$, and the graph consists of four line segments with endpoints $(-4,2)$, $(-2,-2)$, $(0,0)$, $(2,4)$, and $(4,4)$.
(a) The function $g$ is given by $g(x)=-2f(x-1)+3$. Find $g(3)$, and describe a sequence of transformations that maps the graph of $f$ to the graph of $g$.
(b) Find the domain and the range of $g$.
(c) The function $k$ is given by $k(x)=f(2x)$. Find the domain of $k$, the range of $k$, and all zeros of $k$.
(d) Sketch the graph of $g$, and label the coordinates of each endpoint of its segments.
(a) $g(3)=-2f(2)+3=-2(4)+3=-5$。
变换：向右平移 1 个单位；竖直伸缩因子 2 并关于 $x$ 轴反射（即竖直伸缩因子 $-2$）；最后向上平移 3 个单位。（竖直方向必须先乘后加。）
(b) 定义域：$-4\le x-1\le4\Rightarrow[-3,5]$。$f$ 的值域 $[-2,4]$，乘 $-2$ 后变成 $[-8,4]$（翻转后大小对调），再加 3：值域 $[-5,7]$。
(c) 定义域：$-4\le2x\le4\Rightarrow[-2,2]$。水平伸缩不改变输出，值域仍是 $[-2,4]$。$f$ 的零点是 $x=-3$ 和 $x=0$，所以 $2x=-3$ 或 $2x=0$：$k$ 的零点是 $x=-\tfrac32$ 和 $x=0$。
(d) 每个端点 $(a,b)\to(a+1,\,-2b+3)$：$(-3,-1)$、$(-1,7)$、$(1,3)$、$(3,-5)$、$(5,-5)$，依次连线：
| Volunteers $n$ | $2$ | $3$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|---|
| Time $T$ (minutes) | $120$ | $81$ | $59$ | $41$ | $30$ |
A park manager records the time $T$, in minutes, that a group of $n$ volunteers needs to collect the litter in a park. A calculator is allowed.
(a) Use the data to explain why a model of the form $T(n)=\dfrac{k}{n}$, for a constant $k$, is appropriate.
(b) The manager uses the model $T(n)=\dfrac{240}{n}$. Find the residual at $n=6$. Does the model overestimate or underestimate the actual time?
(c) Find the average rate of change of $T$ from $n=4$ to $n=8$. Interpret your answer in context, with units.
(d) What is the least number of volunteers needed to finish in $25$ minutes or less, according to the model? Then give one reason why the model would not be reasonable for very large values of $n$.
(a) $nT$ 的值：$240,\ 243,\ 236,\ 246,\ 240$，都约等于 240——人数乘时间（总工作量）大致不变，这正是反比关系 $T=\dfrac kn$ 的特征。（线性模型不行：人数每多 1，时间减少的量越来越小，一阶差商不是常数。）
(b) 残差 = 实际值 − 预测值 $=41-\dfrac{240}{6}=41-40=1$ 分钟。残差为正，模型低估了实际时间。
(c) $\dfrac{T(8)-T(4)}{8-4}=\dfrac{30-60}{4}=-7.5$ 分钟/人：志愿者从 4 人增加到 8 人，平均每多 1 人，所需时间减少 7.5 分钟。
(d) $\dfrac{240}{n}\le25\Rightarrow n\ge9.6$，人数必须是整数，所以至少 10 人。
模型的局限（答出一条即可）：$n$ 很大时模型预测时间趋近于 0（如 $n=240$ 只要 1 分钟），但走遍公园本身就需要一定时间，人太多也会互相妨碍；所以模型只适用于一个合理的人数范围。
A rideshare company charges $C(d)=4+1.5d$ dollars for a trip of $d$ miles, where $d>0$. For a fare of $c$ dollars, the driver earns $E(c)=0.75c-1$ dollars.
(a) Find $C^{-1}(25)$, and interpret its meaning in context.
(b) Write an expression for $E\big(C(d)\big)$ in terms of $d$. Find $E\big(C(10)\big)$ and interpret it in context.
(c) What is the shortest trip, in miles, for which the driver earns at least $\$20$?
(d) Suppose instead the company charges a flat fare of $\$10$ for any trip of $4$ miles or less, and $C(d)$ dollars for longer trips. Explain why this new fare function does not have an inverse function.
(a) 解 $4+1.5d=25$：$d=14$，所以 $C^{-1}(25)=14$：车费为 25 美元的行程是 14 英里。（$C^{-1}$ 的输入是车费，输出是英里数。）
(b) $E\big(C(d)\big)=0.75(4+1.5d)-1=2+1.125d$。$E\big(C(10)\big)=2+11.25=13.25$：一趟 10 英里的行程，司机挣 13.25 美元。
(c) $2+1.125d\ge20\Rightarrow d\ge16$：至少 16 英里。
(d) 新的车费函数在 $0<d\le4$ 上恒为 10，例如 1 英里和 3 英里的车费都是 10 美元——两个不同的输入对应同一个输出，函数不是一一对应的（水平线 $y=10$ 与图像交于无数个点），所以没有反函数：知道车费是 10 美元，无法确定行程有多长。
