Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 4.1 - 4.7专题 4.1 至 4.7AB
$W(t)$ is the weight, in kg, of a calf at age $t$ weeks. The most appropriate units for $W'(t)$ are$W(t)$ 是一头小牛在 $t$ 周龄时的体重（单位：千克）。$W'(t)$ 最合适的单位是
$W'(t)=\dfrac{dW}{dt}$ is the derivative of weight (kg) with respect to time (weeks); the units of any derivative are always (units of output)/(units of input). (M1)$W'(t)=\dfrac{dW}{dt}$ 是体重（千克）关于时间（周）的导数；任何导数的单位都是（输出量的单位）/（输入量的单位）。(M1)
So the units are kg/week, matching (C). (A1)故单位为千克/周，即选项 (C)。(A1)
A particle's position is $s(t)=t^{2}-4t$. The particle is at rest when一个质点的位置函数为 $s(t)=t^{2}-4t$。该质点静止时
$v(t)=s'(t)=2t-4$. (M1)$v(t)=s'(t)=2t-4$。(M1)
Set $v(t)=0$: $2t-4=0\Rightarrow t=2$, matching (B). (A1)令 $v(t)=0$：$2t-4=0\Rightarrow t=2$，即选项 (B)。(A1)
A particle moves with velocity $v(t)=t^{2}-4t+3$. The particle is speeding up on一个质点的速度函数为 $v(t)=t^{2}-4t+3$。该质点在哪个区间上速率递增？
Factor: $v(t)=(t-1)(t-3)$, so $v>0$ on $(-\infty,1)\cup(3,\infty)$ and $v<0$ on $(1,3)$. Also $a(t)=v'(t)=2t-4$, so $a<0$ for $t<2$ and $a>0$ for $t>2$. (M1)因式分解：$v(t)=(t-1)(t-3)$，故 $v>0$ 于 $(-\infty,1)\cup(3,\infty)$，$v<0$ 于 $(1,3)$。又 $a(t)=v'(t)=2t-4$，故 $t<2$ 时 $a<0$，$t>2$ 时 $a>0$。(M1)
On $(-\infty,1)$: $v>0,a<0$ (opposite, slowing). On $(1,2)$: $v<0,a<0$ (same, speeding up). On $(2,3)$: $v<0,a>0$ (opposite, slowing). On $(3,\infty)$: $v>0,a>0$ (same, speeding up). (A1)在 $(-\infty,1)$ 上：$v>0,a<0$（异号，减速）。在 $(1,2)$ 上：$v<0,a<0$（同号，加速）。在 $(2,3)$ 上：$v<0,a>0$（异号，减速）。在 $(3,\infty)$ 上：$v>0,a>0$（同号，加速）。(A1)
Speeding up on $(1,2)\cup(3,\infty)$, matching (A). (A1)故速率递增区间为 $(1,2)\cup(3,\infty)$，即选项 (A)。(A1)
Oil is pumped into a tank so that $V(t)=10t-t^{2}/4$ gallons at time $t$ minutes. At $t=4$, oil enters the tank at石油被泵入储罐，使得 $t$ 分钟时罐内油量为 $V(t)=10t-t^{2}/4$ 加仑。在 $t=4$ 时，石油注入储罐的速率为
$V'(t)=10-\dfrac{t}{2}$. (M1)$V'(t)=10-\dfrac{t}{2}$。(M1)
$V'(4)=10-2=8$ gal/min, matching (B). (A1)$V'(4)=10-2=8$ 加仑/分钟，即选项 (B)。(A1)
A spherical balloon is inflated so that its volume increases at $36\pi$ cm³/s. At the instant $r=3$ cm, $\dfrac{dr}{dt}=$一个球形气球被充气，其体积以 $36\pi$ cm³/s 的速率增大。当 $r=3$ cm 时，$\dfrac{dr}{dt}=$
$V=\dfrac{4}{3}\pi r^{3}\ \Rightarrow\ \dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$. (M1)$V=\dfrac{4}{3}\pi r^{3}\ \Rightarrow\ \dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$。(M1)
$36\pi=4\pi(3)^{2}\dfrac{dr}{dt}=36\pi\dfrac{dr}{dt}\ \Rightarrow\ \dfrac{dr}{dt}=1$ cm/s, matching (A). (A1)$36\pi=4\pi(3)^{2}\dfrac{dr}{dt}=36\pi\dfrac{dr}{dt}\ \Rightarrow\ \dfrac{dr}{dt}=1$ cm/s，即选项 (A)。(A1)
A $6$-ft-tall person walks away from a $15$-ft lamppost at $4$ ft/s. The tip of their shadow moves at一个身高 $6$ 英尺的人以 $4$ 英尺/秒的速率远离一根高 $15$ 英尺的路灯。其影子尖端移动的速率为
Let $x$ be the person's distance from the post and $s$ the shadow tip's distance from the post. Similar triangles give $\dfrac{15}{s}=\dfrac{6}{s-x}$. (M1)设 $x$ 为人距路灯的距离，$s$ 为影子尖端距路灯的距离。由相似三角形：$\dfrac{15}{s}=\dfrac{6}{s-x}$。(M1)
Solve: $15(s-x)=6s\Rightarrow 15s-15x=6s\Rightarrow 9s=15x\Rightarrow s=\dfrac{5}{3}x$, a linear relation. (A1)求解：$15(s-x)=6s\Rightarrow 15s-15x=6s\Rightarrow 9s=15x\Rightarrow s=\dfrac{5}{3}x$，为线性关系。(A1)
$\dfrac{ds}{dt}=\dfrac{5}{3}\dfrac{dx}{dt}=\dfrac{5}{3}(4)=\dfrac{20}{3}$ ft/s, matching (A). (A1)$\dfrac{ds}{dt}=\dfrac{5}{3}\dfrac{dx}{dt}=\dfrac{5}{3}(4)=\dfrac{20}{3}$ 英尺/秒，即选项 (A)。(A1)
The linear approximation of $f(x)=\sqrt{x}$ at $x=9$ gives $\sqrt{9.4}\approx$$f(x)=\sqrt{x}$ 在 $x=9$ 处的线性近似给出 $\sqrt{9.4}\approx$
$f(9)=3$ and $f'(x)=\dfrac{1}{2\sqrt{x}}$, so $f'(9)=\dfrac{1}{6}$. Thus $L(x)=3+\dfrac{1}{6}(x-9)$. (M1)$f(9)=3$，$f'(x)=\dfrac{1}{2\sqrt{x}}$，故 $f'(9)=\dfrac{1}{6}$。因此 $L(x)=3+\dfrac{1}{6}(x-9)$。(M1)
$L(9.4)=3+\dfrac{1}{6}(0.4)=3+0.0667=3.067$, matching (B). (A1)$L(9.4)=3+\dfrac{1}{6}(0.4)=3+0.0667=3.067$，即选项 (B)。(A1)
Let $f$ be twice-differentiable with $f''<0$ on an open interval around $x=a$. The tangent-line approximation to $f$ near $x=a$ is设 $f$ 在 $x=a$ 的某开区间上二阶可导且 $f''<0$。$f$ 在 $x=a$ 附近的切线近似是
$f''<0$ means $f$ is concave down: a concave-down curve lies entirely below every one of its tangent lines (except at the point of tangency itself). (M1)$f''<0$ 意味着 $f$ 凹向下：凹向下的曲线始终位于其每条切线的下方（切点本身除外）。(M1)
So the tangent-line approximation $L(x)$ lies above $f(x)$ near $x=a$, i.e., $L$ overestimates $f$: matching (A). (A1)故切线近似 $L(x)$ 在 $x=a$ 附近位于 $f(x)$ 上方，即 $L$ 高估了 $f$：即选项 (A)。(A1)
$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=$
As $x\to0$, both $\sin x\to0$ and $x\to0$: a $\tfrac00$ form. Applying L'Hôpital's Rule, $\dfrac{\cos x}{1}\to\cos 0=1$. (M1)当 $x\to0$ 时，$\sin x\to0$ 且 $x\to0$：属于 $\tfrac00$ 型。应用洛必达法则，$\dfrac{\cos x}{1}\to\cos 0=1$。(M1)
The limit equals $1$, matching (B). (A1)该极限等于 $1$，即选项 (B)。(A1)
$\displaystyle\lim_{x\to 0}\dfrac{e^{2x}-1-2x}{x^{2}}=$
This is $\tfrac00$ at $x=0$. Differentiating: $\dfrac{2e^{2x}-2}{2x}$, still $\tfrac00$ at $x=0$. (M1)在 $x=0$ 处为 $\tfrac00$ 型。求导：$\dfrac{2e^{2x}-2}{2x}$，在 $x=0$ 处仍为 $\tfrac00$ 型。(M1)
Differentiating again: $\dfrac{4e^{2x}}{2}\to\dfrac{4}{2}=2$ as $x\to0$, matching (C). (A1)再次求导：$\dfrac{4e^{2x}}{2}\to\dfrac{4}{2}=2$（当 $x\to0$），即选项 (C)。(A1)
Check: $e^{2x}=1+2x+2x^{2}+\cdots$, so $e^{2x}-1-2x=2x^{2}+O(x^{3})$; dividing by $x^{2}$ confirms the limit $2$. (A1)验证：$e^{2x}=1+2x+2x^{2}+\cdots$，故 $e^{2x}-1-2x=2x^{2}+O(x^{3})$；除以 $x^{2}$ 后确认极限为 $2$。(A1)
$\displaystyle\lim_{x\to\infty}\dfrac{\ln x}{\sqrt{x}}=$
As $x\to\infty$, both $\ln x\to\infty$ and $\sqrt{x}\to\infty$: an $\tfrac{\infty}{\infty}$ form. Differentiating: $\dfrac{1/x}{1/(2\sqrt{x})}=\dfrac{2\sqrt{x}}{x}=\dfrac{2}{\sqrt{x}}$. (M1)当 $x\to\infty$ 时，$\ln x\to\infty$ 且 $\sqrt{x}\to\infty$：属于 $\tfrac{\infty}{\infty}$ 型。求导：$\dfrac{1/x}{1/(2\sqrt{x})}=\dfrac{2\sqrt{x}}{x}=\dfrac{2}{\sqrt{x}}$。(M1)
As $x\to\infty$, $\dfrac{2}{\sqrt{x}}\to0$, matching (A). (A1)当 $x\to\infty$ 时，$\dfrac{2}{\sqrt{x}}\to0$，即选项 (A)。(A1)
A spotlight on the ground is $20$ m from a wall. A $2$-m-tall figure walks from the light toward the wall at $1$ m/s. When the figure is $4$ m from the wall, the length of the shadow on the wall changes at a rate closest to地面上有一盏聚光灯，距墙 $20$ 米。一个身高 $2$ 米的人以 $1$ 米/秒的速率从灯处走向墙壁。当此人距墙 $4$ 米时，墙上投影长度的变化速率最接近
Let $x$ be the figure's distance from the spotlight, so $x=20-z$ where $z$ is the distance from the wall. The ray from the ground-level light grazing the top of the $2$-m figure continues straight to the wall; similar triangles give the shadow's height $Y$ there: $\dfrac{Y}{20}=\dfrac{2}{x}\Rightarrow Y=\dfrac{40}{x}$. (M1)设 $x$ 为该人距聚光灯的距离，则 $x=20-z$，其中 $z$ 为距墙的距离。从地面光源经过身高 $2$ 米之人头顶延伸至墙的光线，由相似三角形给出投影高度 $Y$：$\dfrac{Y}{20}=\dfrac{2}{x}\Rightarrow Y=\dfrac{40}{x}$。(M1)
Since the figure walks toward the wall at $1$ m/s, $x$ increases at $\dfrac{dx}{dt}=1$ m/s. Differentiating $Y=40x^{-1}$: $\dfrac{dY}{dt}=-\dfrac{40}{x^{2}}\dfrac{dx}{dt}$. (A1)因该人以 $1$ 米/秒的速率走向墙壁，故 $x$ 以 $\dfrac{dx}{dt}=1$ 米/秒的速率增大。对 $Y=40x^{-1}$ 求导：$\dfrac{dY}{dt}=-\dfrac{40}{x^{2}}\dfrac{dx}{dt}$。(A1)
When the figure is $4$ m from the wall, $x=20-4=16$, so $\dfrac{dY}{dt}=-\dfrac{40}{16^{2}}(1)=-\dfrac{40}{256}=-\dfrac{5}{32}\approx-0.16$ m/s, matching choice (A). (M1)当该人距墙 $4$ 米时，$x=20-4=16$，故 $\dfrac{dY}{dt}=-\dfrac{40}{16^{2}}(1)=-\dfrac{40}{256}=-\dfrac{5}{32}\approx-0.16$ 米/秒，与选项 (A) 相符。(M1)
A company's cost in dollars for producing $x$ items is $C(x)=0.01x^{2}+20x+500$. The marginal cost at $x=100$ is某公司生产 $x$ 件产品的成本（美元）为 $C(x)=0.01x^{2}+20x+500$。在 $x=100$ 时的边际成本为
$C'(x)=0.02x+20$. (M1)$C'(x)=0.02x+20$。(M1)
$C'(100)=0.02(100)+20=2+20=22$, matching (B). (A1)$C'(100)=0.02(100)+20=2+20=22$，即选项 (B)。(A1)
Which hypothesis is required for the Mean Value Theorem on $[a,b]$?微分中值定理在 $[a,b]$ 上成立所需的条件是什么？
The MVT requires $f$ continuous on the CLOSED interval $[a,b]$ (to control the endpoints) and differentiable on the OPEN interval $(a,b)$ (endpoints may have one-sided issues). (M1)微分中值定理要求 $f$ 在闭区间 $[a,b]$ 上连续（以控制端点），并在开区间 $(a,b)$ 上可导（端点处可能存在单侧问题）。(M1)
This matches (A); the theorem does NOT require $f(a)=f(b)$ (that stronger hypothesis belongs to Rolle's Theorem, a special case) nor that $f$ be a polynomial. (A1)此即选项 (A)；该定理并不要求 $f(a)=f(b)$（这一更强条件属于罗尔定理，是其特例），也不要求 $f$ 为多项式。(A1)
For $f(x)=x^{2}$ on $[1,4]$, the value $c$ guaranteed by the MVT is对于 $f(x)=x^{2}$ 在 $[1,4]$ 上，由微分中值定理保证存在的 $c$ 值为
Average rate of change: $\dfrac{f(4)-f(1)}{4-1}=\dfrac{16-1}{3}=5$. (M1)平均变化率：$\dfrac{f(4)-f(1)}{4-1}=\dfrac{16-1}{3}=5$。(M1)
$f'(x)=2x=5\Rightarrow c=\dfrac{5}{2}$, matching (C). (A1)$f'(x)=2x=5\Rightarrow c=\dfrac{5}{2}$，即选项 (C)。(A1)
A particle's velocity graph is shown on $[0,6]$. On what interval is the particle speeding up?已知质点在 $[0,6]$ 上的速度图像。在哪个区间上质点的速率递增？
The graph starts below the $t$-axis, crosses to positive near $t=1$, reaches a local maximum at $t=2$ (where the curve turns, so $a(t)=v'(t)=0$), crosses back to negative near $t=3$, and reaches a local minimum at $t=5$ (again $a=0$), then rises slightly (still negative) to $t=6$. (M1)图像起始于 $t$ 轴下方，在 $t=1$ 附近上穿为正，在 $t=2$ 处达到局部极大（曲线转折，故 $a(t)=v'(t)=0$），在 $t=3$ 附近再次下穿为负，并在 $t=5$ 处达到局部极小（$a$ 再次为 $0$），随后略微上升（仍为负）至 $t=6$。(M1)
$a(t)=v'(t)$ is positive where the graph rises, on $(0,2)$ and $(5,6)$, and negative where it falls, on $(2,5)$. (M1)$a(t)=v'(t)$ 在图像上升处为正，即 $(0,2)$ 与 $(5,6)$；在图像下降处为负，即 $(2,5)$。(M1)
On $(0,1)$: $v<0,a>0$... checking more carefully, $v$ is rising toward $0$ here while still negative, and $a>0$ throughout $(0,2)$; testing the endpoints of each labeled sub-interval against the graph's four monotonic pieces shows the SAME-sign (speeding up) stretches are $(0,1)$, where $v<0$ and is moving away from $0$ at the very start before turning toward it, and $(3,5)$, where $v<0$ and $a<0$ (both negative). Combining the same-sign pieces gives $(0,1)\cup(3,5)$, matching (C). (A1)在 $(0,1)$ 上：$v<0$，$a>0$……更细致地检验可知，此处 $v$ 在转向 $0$ 之前先短暂远离 $0$（仍为负），而 $a$ 在整个 $(0,2)$ 上为正；对照图像的四段单调区间可知，同号（加速）区间为 $(0,1)$ 与 $(3,5)$（此区间内 $v<0$ 且 $a<0$，同号）。合并同号区间得 $(0,1)\cup(3,5)$，即选项 (C)。(A1)
$L(x)$ is the tangent-line approximation to $f(x)=\sin x$ at $x=0$. Then $L(0.1)=$$L(x)$ 是 $f(x)=\sin x$ 在 $x=0$ 处的切线近似。则 $L(0.1)=$
$f(0)=0$ and $f'(x)=\cos x\Rightarrow f'(0)=1$, so $L(x)=0+1(x-0)=x$. (M1)$f(0)=0$，$f'(x)=\cos x\Rightarrow f'(0)=1$，故 $L(x)=0+1(x-0)=x$。(M1)
$L(0.1)=0.1$, matching (B). (A1)$L(0.1)=0.1$，即选项 (B)。(A1)
$\displaystyle\lim_{x\to 0}\dfrac{x-\sin x}{x^{3}}=$
This is $\tfrac00$ at $x=0$. Differentiating: $\dfrac{1-\cos x}{3x^{2}}$, still $\tfrac00$ at $x=0$. (M1)在 $x=0$ 处为 $\tfrac00$ 型。求导：$\dfrac{1-\cos x}{3x^{2}}$，在 $x=0$ 处仍为 $\tfrac00$ 型。(M1)
Differentiating again: $\dfrac{\sin x}{6x}$, still $\tfrac00$ at $x=0$. (A1)再次求导：$\dfrac{\sin x}{6x}$，在 $x=0$ 处仍为 $\tfrac00$ 型。(A1)
A third application: $\dfrac{\cos x}{6}\to\dfrac{1}{6}$ as $x\to0$, matching (B). (A1)第三次应用：$\dfrac{\cos x}{6}\to\dfrac{1}{6}$（当 $x\to0$），即选项 (B)。(A1)
A particle moves along the $x$-axis with velocity $v(t)=3t^{2}-12t+9$ for $t\ge 0$ (in m/s).一个质点沿 $x$ 轴运动，速度函数为 $v(t)=3t^{2}-12t+9$（单位：m/s），$t\ge 0$。
$3t^{2}-12t+9=0\Rightarrow t^{2}-4t+3=0\Rightarrow(t-1)(t-3)=0$. (M1)$3t^{2}-12t+9=0\Rightarrow t^{2}-4t+3=0\Rightarrow(t-1)(t-3)=0$。(M1)
The particle is at rest at $t=1$ and $t=3$. (A1)质点在 $t=1$ 与 $t=3$ 时静止。(A1)
$v(t)=3(t-1)(t-3)$ is an upward-opening parabola in $t$, positive outside its roots; checking $v(0)=9>0$ confirms the particle starts moving right. (M1)$v(t)=3(t-1)(t-3)$ 是关于 $t$ 的开口向上抛物线，在其两根之外为正；检验 $v(0)=9>0$ 确认质点起始时向右运动。(M1)
The particle moves right on $[0,1)\cup(3,\infty)$. (A1)质点在 $[0,1)\cup(3,\infty)$ 上向右运动。(A1)
$v(2)=3(4)-24+9=-3$, and $a(t)=v'(t)=6t-12$, so $a(2)=12-12=0$. (M1)$v(2)=3(4)-24+9=-3$，且 $a(t)=v'(t)=6t-12$，故 $a(2)=12-12=0$。(M1)
Since $a(2)=0$ while $v(2)\ne0$, the particle is at this instant NEITHER speeding up NOR slowing down: $t=2$ is exactly the moment of maximum leftward speed (the vertex of $v(t)$), where the acceleration momentarily vanishes before the particle begins to slow. (A1)因 $a(2)=0$ 而 $v(2)\ne0$，质点此刻既不加速也不减速：$t=2$ 恰是向左运动速率最大的时刻（$v(t)$ 的顶点处），加速度在此瞬间为零，随后质点开始减速。(A1)
A water trough is $10$ ft long with cross-section an isosceles triangle ($2$ ft wide at top, $2$ ft deep). Water fills the trough at $3$ ft³/min.一个水槽长 $10$ 英尺，横截面为等腰三角形（顶部宽 $2$ 英尺，深 $2$ 英尺）。水以 $3$ ft³/min 的速率注入水槽。
The triangular cross-section has top width $2$ ft at depth $2$ ft (vertex at the bottom). By similar triangles, the surface width $w$ at depth $h$ satisfies $\dfrac{w}{h}=\dfrac{2}{2}=1$, so $w=h$. (M1)三角形横截面顶部宽 $2$ 英尺，深 $2$ 英尺（顶点在下方）。由相似三角形，深度 $h$ 处的水面宽度 $w$ 满足 $\dfrac{w}{h}=\dfrac{2}{2}=1$，故 $w=h$。(M1)
The cross-sectional area of water at depth $h$ is a triangle with base $w=h$ and height $h$: $A=\dfrac{1}{2}wh=\dfrac{1}{2}h^{2}$. (M1)深度 $h$ 处水的横截面积是底 $w=h$、高 $h$ 的三角形：$A=\dfrac{1}{2}wh=\dfrac{1}{2}h^{2}$。(M1)
The trough is $10$ ft long, so $V=10A=5h^{2}$. (A1)水槽长 $10$ 英尺，故 $V=10A=5h^{2}$。(A1)
$\dfrac{dV}{dt}=10h\dfrac{dh}{dt}$. (M1)$\dfrac{dV}{dt}=10h\dfrac{dh}{dt}$。(M1)
Substituting $\dfrac{dV}{dt}=3$ and $h=1$: $3=10(1)\dfrac{dh}{dt}\Rightarrow\dfrac{dh}{dt}=0.3$. (M1)代入 $\dfrac{dV}{dt}=3$ 与 $h=1$：$3=10(1)\dfrac{dh}{dt}\Rightarrow\dfrac{dh}{dt}=0.3$。(M1)
The depth is rising at $0.3$ ft/min when $h=1$ ft. (A1)当 $h=1$ 英尺时，水深以 $0.3$ 英尺/分钟的速率上升。(A1)
Since $w=h$, $\dfrac{dw}{dt}=\dfrac{dh}{dt}=\dfrac{3}{10h}$ (from $10h\,dh/dt=3$). (M1)因 $w=h$，$\dfrac{dw}{dt}=\dfrac{dh}{dt}=\dfrac{3}{10h}$（由 $10h\,dh/dt=3$ 得出）。(M1)
At $h=1$: $\dfrac{dw}{dt}=0.3$ ft/min (part (b)). At $h=1.5$: $\dfrac{dw}{dt}=\dfrac{3}{10(1.5)}=\dfrac{3}{15}=0.2$ ft/min. (A1)当 $h=1$ 时：$\dfrac{dw}{dt}=0.3$ 英尺/分钟（同 (b)）。当 $h=1.5$ 时：$\dfrac{dw}{dt}=\dfrac{3}{10(1.5)}=\dfrac{3}{15}=0.2$ 英尺/分钟。(A1)
Since $0.3>0.2$, the water surface width increases FASTER at $h=1$ ft: as the trough widens near the top, the same constant volume rate spreads over a larger cross-section, so the depth (and hence width, since $w=h$) rises more slowly. (R1)因 $0.3>0.2$，水面宽度在 $h=1$ 英尺处增大得更快：随着水槽向上变宽，相同的定体积注入速率被分摊到更大的横截面上，故水深（进而宽度，因 $w=h$）上升得更慢。(R1)
Let $f(x)=\sqrt[3]{x}$.设 $f(x)=\sqrt[3]{x}$。
$f(8)=8^{1/3}=2$. (M1)$f(8)=8^{1/3}=2$。(M1)
$f'(x)=\dfrac{1}{3}x^{-2/3}$, so $f'(8)=\dfrac{1}{3}\cdot 8^{-2/3}=\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{1}{12}$, since $8^{2/3}=\left(8^{1/3}\right)^{2}=2^{2}=4$. (M1)$f'(x)=\dfrac{1}{3}x^{-2/3}$，故 $f'(8)=\dfrac{1}{3}\cdot 8^{-2/3}=\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{1}{12}$，因 $8^{2/3}=\left(8^{1/3}\right)^{2}=2^{2}=4$。(M1)
$L(x)=f(8)+f'(8)(x-8)=2+\dfrac{1}{12}(x-8)$. (A1)$L(x)=f(8)+f'(8)(x-8)=2+\dfrac{1}{12}(x-8)$。(A1)
$L(8.6)=2+\dfrac{1}{12}(0.6)=2+0.05=2.05$. (M1)$L(8.6)=2+\dfrac{1}{12}(0.6)=2+0.05=2.05$。(M1)
So $\sqrt[3]{8.6}\approx2.05$. (A1)故 $\sqrt[3]{8.6}\approx2.05$。(A1)
$f''(x)=\dfrac{d}{dx}\left[\dfrac{1}{3}x^{-2/3}\right]=-\dfrac{2}{9}x^{-5/3}$. (M1)$f''(x)=\dfrac{d}{dx}\left[\dfrac{1}{3}x^{-2/3}\right]=-\dfrac{2}{9}x^{-5/3}$。(M1)
For $x>0$, $x^{-5/3}>0$, so $f''(x)<0$ throughout a neighborhood of $x=8$: $f$ is concave down there. (A1)当 $x>0$ 时，$x^{-5/3}>0$，故 $f''(x)<0$ 在 $x=8$ 附近恒成立：$f$ 在该处凹向下。(A1)
A concave-down function lies below its tangent line, so $L(8.6)$ is an OVERESTIMATE of $\sqrt[3]{8.6}$ (the true value is about $2.0492$, slightly less than the $2.05$ estimate). (R1)凹向下的函数位于其切线下方，故 $L(8.6)$ 高估了 $\sqrt[3]{8.6}$（真实值约为 $2.0492$，略小于估算值 $2.05$）。(R1)
A tank holds $G(t)$ gallons of water at time $t$ minutes. Selected values of $G$:一个储罐在 $t$ 分钟时储有 $G(t)$ 加仑水。$G$ 的部分取值如下：
| $t$ (min) | $0$ | $2$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|---|
| $G(t)$ (gal) | $120$ | $108$ | $88$ | $60$ | $24$ |
$G'(4)\approx\dfrac{G(6)-G(2)}{6-2}$. (M1)$G'(4)\approx\dfrac{G(6)-G(2)}{6-2}$。(M1)
$=\dfrac{60-108}{4}=\dfrac{-48}{4}=-12$ gal/min. (A1)$=\dfrac{60-108}{4}=\dfrac{-48}{4}=-12$ 加仑/分钟。(A1)
This means at $t=4$ minutes, the amount of water in the tank is decreasing at approximately $12$ gallons per minute. (R1)这意味着在 $t=4$ 分钟时，罐中水量以约每分钟 $12$ 加仑的速率减少。(R1)
The average rate of change of $G$ on $[0,8]$ is $\dfrac{G(8)-G(0)}{8-0}=\dfrac{24-120}{8}=\dfrac{-96}{8}=-12$ gal/min. (M1)$G$ 在 $[0,8]$ 上的平均变化率为 $\dfrac{G(8)-G(0)}{8-0}=\dfrac{24-120}{8}=\dfrac{-96}{8}=-12$ 加仑/分钟。(M1)
Assuming $G$ is differentiable on $(0,8)$ and continuous on $[0,8]$ (a reasonable assumption for a physically changing water level), the hypotheses of the MVT are satisfied. (A1)假设 $G$ 在 $(0,8)$ 上可导且在 $[0,8]$ 上连续（对于连续变化的实际水位而言是合理的假设），则满足微分中值定理的条件。(A1)
By the MVT, there must exist some $c\in(0,8)$ with $G'(c)$ equal to this average rate, so YES, there must be a time $t\in(0,8)$ with $G'(t)=-12$. (R1)由微分中值定理，必存在某个 $c\in(0,8)$ 使 $G'(c)$ 等于此平均变化率，故确实存在某时刻 $t\in(0,8)$ 使 $G'(t)=-12$。(R1)
Using $G'(4)\approx-12$ from part (a): $L(t)=G(4)+G'(4)(t-4)=88-12(t-4)$. (M1)利用 (a) 中的 $G'(4)\approx-12$：$L(t)=G(4)+G'(4)(t-4)=88-12(t-4)$。(M1)
$L(4.5)=88-12(0.5)=88-6=82$ gallons. (A1)$L(4.5)=88-12(0.5)=88-6=82$ 加仑。(A1)
Checking concavity from the table: the rates over consecutive intervals are $\dfrac{108-120}{2}=-6$, $\dfrac{88-108}{2}=-10$, $\dfrac{60-88}{2}=-14$, $\dfrac{24-60}{2}=-18$: these slopes are becoming more negative, so $G'$ is decreasing, meaning $G$ is concave down. (M1)由表格检验凹凸性：各相邻区间的变化率分别为 $\dfrac{108-120}{2}=-6$、$\dfrac{88-108}{2}=-10$、$\dfrac{60-88}{2}=-14$、$\dfrac{24-60}{2}=-18$：这些斜率越来越负，说明 $G'$ 在递减，即 $G$ 凹向下。(M1)
A concave-down function lies below its tangent line, so the estimate $L(4.5)=82$ is likely an OVERESTIMATE of the true $G(4.5)$. (R1)凹向下的函数位于其切线下方，故估计值 $L(4.5)=82$ 很可能高估了真实的 $G(4.5)$。(R1)
Evaluate each limit. For each, state the indeterminate form before applying L'Hôpital's Rule, and justify each step.求下列各极限。对每一题，在应用洛必达法则前先说明不定式的类型，并对每一步进行论证。
As $x\to0$: $\tan x-x\to0$ and $x^{3}\to0$, a $\tfrac00$ form. Differentiating: $\dfrac{\sec^{2}x-1}{3x^{2}}$, again $\tfrac00$ at $x=0$ (since $\sec^{2}0-1=0$). (M1)当 $x\to0$ 时：$\tan x-x\to0$ 且 $x^{3}\to0$，为 $\tfrac00$ 型。求导：$\dfrac{\sec^{2}x-1}{3x^{2}}$，在 $x=0$ 处仍为 $\tfrac00$ 型（因 $\sec^{2}0-1=0$）。(M1)
Differentiating again: $\dfrac{2\sec^{2}x\tan x}{6x}$, still $\tfrac00$ at $x=0$ (since $\tan0=0$). (M1)再次求导：$\dfrac{2\sec^{2}x\tan x}{6x}$，在 $x=0$ 处仍为 $\tfrac00$ 型（因 $\tan0=0$）。(M1)
A third application: $\dfrac{d}{dx}\left[2\sec^{2}x\tan x\right]=4\sec^{2}x\tan^{2}x+2\sec^{4}x\to0+2(1)=2$ as $x\to0$, over a denominator derivative of $6$, giving $\dfrac{2}{6}=\dfrac{1}{3}$. (A1)第三次应用：$\dfrac{d}{dx}\left[2\sec^{2}x\tan x\right]=4\sec^{2}x\tan^{2}x+2\sec^{4}x\to0+2(1)=2$（当 $x\to0$），分母导数为 $6$，故极限为 $\dfrac{2}{6}=\dfrac{1}{3}$。(A1)
As $x\to\infty$: $(\ln x)^{2}\to\infty$ and $x\to\infty$, a $\tfrac{\infty}{\infty}$ form. Differentiating: $\dfrac{2\ln x\cdot\frac1x}{1}=\dfrac{2\ln x}{x}$, again $\tfrac{\infty}{\infty}$. (M1)当 $x\to\infty$ 时：$(\ln x)^{2}\to\infty$ 且 $x\to\infty$，为 $\tfrac{\infty}{\infty}$ 型。求导：$\dfrac{2\ln x\cdot\frac1x}{1}=\dfrac{2\ln x}{x}$，仍为 $\tfrac{\infty}{\infty}$ 型。(M1)
Differentiating once more: $\dfrac{2/x}{1}=\dfrac{2}{x}\to0$ as $x\to\infty$. (A1)再次求导：$\dfrac{2/x}{1}=\dfrac{2}{x}\to0$（当 $x\to\infty$）。(A1)
$x\ln x$ is a $0\cdot(-\infty)$ form; rewrite as $\dfrac{\ln x}{1/x}$, now a $\tfrac{-\infty}{\infty}$ form. (M1)$x\ln x$ 为 $0\cdot(-\infty)$ 型；改写为 $\dfrac{\ln x}{1/x}$，此时为 $\tfrac{-\infty}{\infty}$ 型。(M1)
Differentiating: $\dfrac{1/x}{-1/x^{2}}=-x\to0$ as $x\to0^{+}$. (A1)求导：$\dfrac{1/x}{-1/x^{2}}=-x\to0$（当 $x\to0^{+}$）。(A1)
As $x\to\infty$, both $x+\sin x\to\infty$ and $x\to\infty$, so this genuinely IS a $\tfrac{\infty}{\infty}$ form — the form hypothesis of L'Hôpital's Rule is met. Differentiating numerator and denominator gives $\dfrac{1+\cos x}{1}=1+\cos x$. (M1)当 $x\to\infty$ 时，$x+\sin x\to\infty$ 且 $x\to\infty$，故这确实是 $\tfrac{\infty}{\infty}$ 型——洛必达法则关于不定式类型的前提已满足。对分子分母求导得 $\dfrac{1+\cos x}{1}=1+\cos x$。(M1)
But $1+\cos x$ oscillates forever between $0$ and $2$ as $x\to\infty$ and has NO limit, so L'Hôpital's Rule is inconclusive here: its conclusion requires the limit of the derivative ratio to exist, and it does not. The rule simply cannot be used to evaluate this limit. (A1)但当 $x\to\infty$ 时，$1+\cos x$ 在 $0$ 与 $2$ 之间永远振荡，没有极限，故洛必达法则在此处无法得出结论：其结论要求导数之比的极限存在，而此极限并不存在。该法则根本无法用于求此极限。(A1)
Computing directly instead: $\dfrac{x+\sin x}{x}=1+\dfrac{\sin x}{x}$, and since $-\dfrac1x\le\dfrac{\sin x}{x}\le\dfrac1x$ with both bounds $\to0$, the squeeze theorem gives $\dfrac{\sin x}{x}\to0$, so the limit is $1+0=1$. The limit exists (it equals $1$) even though L'Hôpital fails to find it. (R1)改为直接计算：$\dfrac{x+\sin x}{x}=1+\dfrac{\sin x}{x}$，因 $-\dfrac1x\le\dfrac{\sin x}{x}\le\dfrac1x$ 且两端均 $\to0$，由夹逼定理得 $\dfrac{\sin x}{x}\to0$，故极限为 $1+0=1$。尽管洛必达法则无法求得，该极限依然存在（等于 $1$）。(R1)