Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 5.1 - 5.12考点 5.1 至 5.12AB
Let $f(x)=x^{2}-3x$ on $[0,4]$. The value $c$ guaranteed by the MVT is设 $f(x)=x^{2}-3x$ 在 $[0,4]$ 上。由均值定理（MVT）保证存在的值 $c$ 为
$f$ is a polynomial, so it is continuous on $[0,4]$ and differentiable on $(0,4)$: the MVT applies. The average rate of change is $\dfrac{f(4)-f(0)}{4-0}=\dfrac{4-0}{4}=1$. (M1)$f$ 为多项式，故在 $[0,4]$ 上连续、在 $(0,4)$ 上可导：均值定理适用。平均变化率为 $\dfrac{f(4)-f(0)}{4-0}=\dfrac{4-0}{4}=1$。(M1)
$f'(x)=2x-3$, so $2c-3=1$ gives $c=2$, which lies in $(0,4)$. (A1)$f'(x)=2x-3$，故 $2c-3=1$ 解得 $c=2$，且 $2\in(0,4)$。(A1)
The critical points of $f(x)=x^{3}-3x$ are$f(x)=x^{3}-3x$ 的临界点为
$f'(x)=3x^{2}-3=3(x-1)(x+1)$, defined for all real $x$. (M1)$f'(x)=3x^{2}-3=3(x-1)(x+1)$，对所有实数 $x$ 均有定义。(M1)
Setting $f'(x)=0$ gives $x=\pm 1$; since $f'$ never fails to exist, these are the only critical points. (A1)令 $f'(x)=0$ 得 $x=\pm 1$；因 $f'$ 处处存在，这就是全部临界点。(A1)
$f(x)=x^{3}-6x^{2}+9x$ is decreasing on$f(x)=x^{3}-6x^{2}+9x$ 在以下区间上递减：
$f'(x)=3x^{2}-12x+9=3(x-1)(x-3)$. (M1)$f'(x)=3x^{2}-12x+9=3(x-1)(x-3)$。(M1)
$f'$ is a positive-leading upward parabola in $(x-1)(x-3)$ form, negative strictly between its roots: $f'(x)<0$ on $(1,3)$, so $f$ is decreasing there. (A1)$f'$ 为开口向上的抛物线，在两根之间为负：$f'(x)<0$ 于 $(1,3)$ 上，故 $f$ 在此区间递减。(A1)
If $f'$ changes from negative to positive at $x=c$, then $f$ has若 $f'$ 在 $x=c$ 处由负变正，则 $f$ 在该点
Negative $f'$ means $f$ is decreasing; positive $f'$ means $f$ is increasing. (M1)$f'$ 为负表示 $f$ 递减；$f'$ 为正表示 $f$ 递增。(M1)
Decreasing then increasing at $x=c$ is exactly the shape of a valley: $f$ has a local minimum at $x=c$. (A1)在 $x=c$ 处先递减后递增，正是"谷"的形状：$f$ 在 $x=c$ 处取得极小值。(A1)
$f(x)=x^{4}-6x^{2}$ is concave up on$f(x)=x^{4}-6x^{2}$ 在以下区间上为凹（concave up）：
$f'(x)=4x^{3}-12x$, so $f''(x)=12x^{2}-12=12(x-1)(x+1)$. (M1)$f'(x)=4x^{3}-12x$，故 $f''(x)=12x^{2}-12=12(x-1)(x+1)$。(M1)
This is a positive-leading upward parabola in $(x-1)(x+1)$ form, positive outside its roots: $f''(x)>0$ on $(-\infty,-1)\cup(1,\infty)$. (A1)该式为开口向上的抛物线，在两根之外为正：$f''(x)>0$ 于 $(-\infty,-1)\cup(1,\infty)$。(A1)
For $f(x)=x^{3}-3x$, at $x=-1$ the function has对于 $f(x)=x^{3}-3x$，函数在 $x=-1$ 处
$f'(x)=3x^{2}-3$, which is $0$ at $x=-1$ (a critical point, as in Q2); $f''(x)=6x$. (M1)$f'(x)=3x^{2}-3$，在 $x=-1$ 处为 $0$（与第 2 题一致，为临界点）；$f''(x)=6x$。(M1)
$f''(-1)=-6<0$, so by the Second Derivative Test, $f$ has a local maximum at $x=-1$. (A1)$f''(-1)=-6<0$，由二阶导数判别法，$f$ 在 $x=-1$ 处取得极大值。(A1)
The graph of $y=x^{5}-5x^{4}$ has a point of inflection at$y=x^{5}-5x^{4}$ 的图像的拐点位于
$y'=5x^{4}-20x^{3}$, so $y''=20x^{3}-60x^{2}=20x^{2}(x-3)$, which is zero at $x=0$ and $x=3$. (M1)$y'=5x^{4}-20x^{3}$，故 $y''=20x^{3}-60x^{2}=20x^{2}(x-3)$，在 $x=0$ 和 $x=3$ 处为零。(M1)
Near $x=0$: $x^{2}\ge 0$ never changes sign, and $(x-3)<0$ on both sides of $0$, so $y''<0$ on both sides: no sign change, no inflection at $x=0$. Near $x=3$: $x^{2}>0$ stays positive while $(x-3)$ flips from negative to positive, so $y''$ changes sign there: $x=3$ is the only inflection point. (A1)在 $x=0$ 附近：$x^{2}\ge 0$ 恒不变号，且 $(x-3)<0$ 在 $0$ 的两侧均成立，故 $y''<0$ 两侧皆然：无变号，$x=0$ 处无拐点。在 $x=3$ 附近：$x^{2}>0$ 保持为正，而 $(x-3)$ 由负变正，故 $y''$ 在此变号：$x=3$ 是唯一的拐点。(A1)
The graph of $f'$ is shown. On which interval is $f$ increasing and concave down?如图所示为 $f'$ 的图像。$f$ 在哪个区间上递增且为凸（concave down）？
"Increasing" means $f'>0$ by definition; "concave down" means $f''<0$ by definition. (M1)"递增"根据定义即 $f'>0$；"凸（concave down）"根据定义即 $f''<0$。(M1)
On the shown graph, $f'$ is positive and decreasing over the interval described by option (B)'s condition, which is exactly $f'>0$ and $f''<0$. (A1)在图中，$f'$ 在选项 (B) 所描述的区间上为正且递减，这正对应 $f'>0$ 且 $f''<0$。(A1)
The absolute maximum of $f(x)=x^{3}-3x$ on $[-2,2]$ is$f(x)=x^{3}-3x$ 在 $[-2,2]$ 上的绝对最大值为
From Q2, the critical points are $x=\pm 1$, both inside $[-2,2]$. The full candidate list is $x=-2,-1,1,2$. (M1)由第 2 题，临界点为 $x=\pm 1$，均在 $[-2,2]$ 内。完整候选点列表为 $x=-2,-1,1,2$。(M1)
Evaluate: $f(-2)=-8+6=-2$, $f(-1)=-1+3=2$, $f(1)=1-3=-2$, $f(2)=8-6=2$. The largest value is $2$ (attained at both $x=-1$ and $x=2$). (A1)代入求值：$f(-2)=-8+6=-2$，$f(-1)=-1+3=2$，$f(1)=1-3=-2$，$f(2)=8-6=2$。最大值为 $2$（在 $x=-1$ 与 $x=2$ 处均取得）。(A1)
A rectangle with one side on the $x$-axis has its upper two vertices on the parabola $y=12-x^{2}$. The maximum area of such a rectangle is一个矩形的一边在 $x$ 轴上，上方两个顶点在抛物线 $y=12-x^{2}$ 上。该矩形的最大面积为
By symmetry about the $y$-axis, the vertices are at $(\pm x,12-x^{2})$, giving width $2x$ and height $12-x^{2}$: $A(x)=2x(12-x^{2})=24x-2x^{3}$ for $0
$A''(x)=-12x<0$ at $x=2$, confirming a maximum: $A(2)=2(2)(12-4)=4(8)=32$. (A1)$A''(x)=-12x$ 在 $x=2$ 处为负，确认为最大值：$A(2)=2(2)(12-4)=4(8)=32$。(A1)
The point on $y=\sqrt{x}$ closest to $(3,0)$ has $x$-coordinate$y=\sqrt{x}$ 上距点 $(3,0)$ 最近的点，其 $x$ 坐标为
$D^{2}=(x-3)^{2}+y^{2}=(x-3)^{2}+x$ since $y^{2}=x$ on the curve. Let $g(x)=(x-3)^{2}+x=x^{2}-5x+9$; then $g'(x)=2x-5=0$ gives $x=\dfrac{5}{2}$. (M1)因曲线上 $y^{2}=x$，故 $D^{2}=(x-3)^{2}+y^{2}=(x-3)^{2}+x$。设 $g(x)=(x-3)^{2}+x=x^{2}-5x+9$，则 $g'(x)=2x-5=0$ 解得 $x=\dfrac{5}{2}$。(M1)
$g''(x)=2>0$, confirming a minimum, so the closest point has $x=\dfrac{5}{2}$. (A1)$g''(x)=2>0$，确认为最小值，故最近点的 $x=\dfrac{5}{2}$。(A1)
If $f'(x)>0$ and $f''(x)<0$ on $(a,b)$, then the graph of $f$ on $(a,b)$ is若在 $(a,b)$ 上 $f'(x)>0$ 且 $f''(x)<0$，则 $f$ 在 $(a,b)$ 上的图像是
$f'>0$ means increasing; $f''<0$ means concave down. (M1)$f'>0$ 表示递增；$f''<0$ 表示为凸（concave down）。(M1)
Combined, the graph is increasing and concave down: option (B). (A1)综合两者，图像递增且为凸：选项 (B)。(A1)
On $[0,2]$, the absolute maximum of $f(x)=x\sqrt{2-x}$ is在 $[0,2]$ 上，$f(x)=x\sqrt{2-x}$ 的绝对最大值为
$f'(x)=\sqrt{2-x}+x\cdot\dfrac{-1}{2\sqrt{2-x}}=\dfrac{2(2-x)-x}{2\sqrt{2-x}}=\dfrac{4-3x}{2\sqrt{2-x}}$. Setting the numerator to $0$ gives $x=\dfrac{4}{3}\in[0,2]$. (M1)$f'(x)=\sqrt{2-x}+x\cdot\dfrac{-1}{2\sqrt{2-x}}=\dfrac{2(2-x)-x}{2\sqrt{2-x}}=\dfrac{4-3x}{2\sqrt{2-x}}$。令分子为 $0$ 得 $x=\dfrac{4}{3}\in[0,2]$。(M1)
Candidates: $f(0)=0$, $f\!\left(\tfrac{4}{3}\right)=\tfrac{4}{3}\sqrt{2-\tfrac{4}{3}}=\tfrac{4}{3}\sqrt{\tfrac{2}{3}}=\dfrac{4\sqrt{6}}{9}$, $f(2)=2\sqrt{0}=0$. The maximum is $\dfrac{4\sqrt{6}}{9}\approx 1.09$. (A1)候选值：$f(0)=0$，$f\!\left(\tfrac{4}{3}\right)=\tfrac{4}{3}\sqrt{2-\tfrac{4}{3}}=\tfrac{4}{3}\sqrt{\tfrac{2}{3}}=\dfrac{4\sqrt{6}}{9}$，$f(2)=2\sqrt{0}=0$。最大值为 $\dfrac{4\sqrt{6}}{9}\approx 1.09$。(A1)
The table gives values of a continuous function $f''$. The graph of $f$ must have an inflection point in which interval(s)?下表给出连续函数 $f''$ 的值。$f$ 的图像在哪个（些）区间上必有拐点？
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|---|
| $f''$ | $-3$ | $-1$ | $2$ | $1$ | $-1$ |
$f''$ is continuous (given), so wherever two consecutive tabulated values of $f''$ have opposite signs, the IVT guarantees a zero, and since the signs differ on either side, that zero is a genuine sign change of $f''$. (M1)$f''$ 连续（已知），故只要相邻两个表格值符号相反，介值定理即保证存在一个零点，且因两侧符号不同，该零点为 $f''$ 的真实变号点。(M1)
$(0,1)$: $-3\to -1$, same sign, no guarantee. $(1,2)$: $-1\to 2$, sign change, inflection guaranteed. $(2,3)$: $2\to 1$, same sign, no guarantee. $(3,4)$: $1\to -1$, sign change, inflection guaranteed. So both (B) $(1,2)$ and (C) $(3,4)$ are guaranteed. (A1)$(0,1)$：$-3\to -1$，同号，无法保证。$(1,2)$：$-1\to 2$，变号，保证有拐点。$(2,3)$：$2\to 1$，同号，无法保证。$(3,4)$：$1\to -1$，变号，保证有拐点。故 (B) $(1,2)$ 与 (C) $(3,4)$ 均可保证。(A1)
A page must have a printed area of $96$ in², with $1$ in. margins on each side and $1.5$ in. margins top and bottom. The dimensions (width × height) that minimize total page area are closest to一页纸的印刷区域面积为 $96$ 平方英寸，左右各留 $1$ 英寸页边距，上下各留 $1.5$ 英寸页边距。使整页面积最小的尺寸（宽 × 高）最接近
Let $x$ be the printed width, so the printed height is $\dfrac{96}{x}$ (from $xy=96$). The page width adds $1$ in. on each side, and the page height adds $1.5$ in. top and bottom: page width $=x+2$, page height $=\dfrac{96}{x}+3$. (M1)设 $x$ 为印刷宽度，则印刷高度为 $\dfrac{96}{x}$（由 $xy=96$）。页面宽度左右各加 $1$ 英寸，页面高度上下各加 $1.5$ 英寸：页面宽度 $=x+2$，页面高度 $=\dfrac{96}{x}+3$。(M1)
Total page area: $A(x)=(x+2)\!\left(\dfrac{96}{x}+3\right)=96+3x+\dfrac{192}{x}+6=3x+\dfrac{192}{x}+102$. (M1)整页面积：$A(x)=(x+2)\!\left(\dfrac{96}{x}+3\right)=96+3x+\dfrac{192}{x}+6=3x+\dfrac{192}{x}+102$。(M1)
$A'(x)=3-\dfrac{192}{x^{2}}=0$ gives $x^{2}=64$, so $x=8$ (rejecting $x=-8$). Then the printed height is $\dfrac{96}{8}=12$, so page dimensions are $(8+2)\times(12+3)=10\times 15$. Since $A''(x)=\dfrac{384}{x^{3}}>0$ for $x>0$, this is a minimum, giving minimum total area $10\times 15=150$ in². (A1)$A'(x)=3-\dfrac{192}{x^{2}}=0$ 得 $x^{2}=64$，故 $x=8$（舍去 $x=-8$）。此时印刷高度为 $\dfrac{96}{8}=12$，页面尺寸为 $(8+2)\times(12+3)=10\times 15$。因 $A''(x)=\dfrac{384}{x^{3}}>0$（$x>0$），确认为最小值，最小整页面积为 $10\times 15=150$ 平方英寸。(A1)
If the graph of $f'$ crosses the $x$-axis at $x=1$ (from + to -) and has a local min at $x=3$, then $f$ has若 $f'$ 的图像在 $x=1$ 处穿越 $x$ 轴（由正变负），并在 $x=3$ 处有局部极小值，则 $f$ 在
At $x=1$, $f'$ itself changes sign (+ to -): by the First Derivative Test, $f$ has a local maximum at $x=1$. (M1)在 $x=1$ 处，$f'$ 本身变号（正变负）：由一阶导数判别法，$f$ 在 $x=1$ 处取得极大值。(M1)
At $x=3$, $f'$ has a local minimum, meaning $f'$ turns from decreasing to increasing there, so $f''$ changes from negative to positive: this is a change in concavity, i.e. an inflection point of $f$, not necessarily a zero of $f'$ itself. (A1)在 $x=3$ 处，$f'$ 取得局部极小值，意味着 $f'$ 由递减转为递增，故 $f''$ 由负变正：这是凹凸性的改变，即 $f$ 的拐点，而非 $f'$ 本身的零点。(A1)
Which hypothesis of Rolle's Theorem is violated for $f(x)=|x|$ on $[-1,1]$?对于 $f(x)=|x|$ 在 $[-1,1]$ 上，罗尔定理的哪个条件不满足？
$f(x)=|x|$ is continuous everywhere, so continuity on $[-1,1]$ holds. $f(-1)=1=f(1)$, so the endpoint condition holds. (M1)$f(x)=|x|$ 处处连续，故在 $[-1,1]$ 上的连续性成立。$f(-1)=1=f(1)$，故端点条件成立。(M1)
However, $|x|$ has a corner at $x=0\in(-1,1)$ where the left and right derivatives ($-1$ and $1$) disagree, so $f$ fails to be differentiable on the open interval. (A1)但 $|x|$ 在 $x=0\in(-1,1)$ 处存在拐角，左、右导数（$-1$ 与 $1$）不相等，故 $f$ 在开区间上不可导。(A1)
Which of the following must be true if $f''(x)>0$ for all $x$?若对所有 $x$ 均有 $f''(x)>0$，以下哪项必定成立？
$f''$ is by definition the derivative of $f'$, so $f''>0$ everywhere means $f'$ is increasing everywhere: this is exactly option (B), true by definition alone. (M1)$f''$ 按定义即为 $f'$ 的导数，故 $f''>0$ 处处成立意味着 $f'$ 处处递增：这正是选项 (B)，仅凭定义即可成立。(M1)
The other options fail by counterexample: $f(x)=x^{2}$ has $f''>0$ everywhere but is decreasing for $x<0$, ruling out (A); $f(x)=e^{x}$ has $f''>0$ everywhere but no minimum, ruling out (C); a linear function has $f''=0$, not $f''>0$, ruling out (D). (A1)其余选项均可用反例排除：$f(x)=x^{2}$ 处处 $f''>0$，但当 $x<0$ 时递减，排除 (A)；$f(x)=e^{x}$ 处处 $f''>0$，但无最小值，排除 (C)；线性函数有 $f''=0$（非 $f''>0$），排除 (D)。(A1)
Let $f(x)=x^{3}-6x^{2}+9x+2$.设 $f(x)=x^{3}-6x^{2}+9x+2$。
$f'(x)=3x^{2}-12x+9=3(x-1)(x-3)$, so the critical points are $x=1$ and $x=3$. (M1)$f'(x)=3x^{2}-12x+9=3(x-1)(x-3)$，故临界点为 $x=1$ 与 $x=3$。(M1)
$f(1)=1-6+9+2=6$ and $f(3)=27-54+27+2=2$: critical points $(1,6)$ and $(3,2)$. (A1)$f(1)=1-6+9+2=6$，$f(3)=27-54+27+2=2$：临界点为 $(1,6)$ 与 $(3,2)$。(A1)
$f'(x)=3(x-1)(x-3)$: positive on $(-\infty,1)$, negative on $(1,3)$, positive on $(3,\infty)$. (M1)$f'(x)=3(x-1)(x-3)$：在 $(-\infty,1)$ 上为正，在 $(1,3)$ 上为负，在 $(3,\infty)$ 上为正。(M1)
At $x=1$, $f'$ changes $+$ to $-$: local maximum, $f(1)=6$. (A1)在 $x=1$ 处，$f'$ 由正变负：极大值，$f(1)=6$。(A1)
At $x=3$, $f'$ changes $-$ to $+$: local minimum, $f(3)=2$. (A1)在 $x=3$ 处，$f'$ 由负变正：极小值，$f(3)=2$。(A1)
$f''(x)=6x-12=6(x-2)$: negative for $x<2$ (concave down), positive for $x>2$ (concave up). (M1)$f''(x)=6x-12=6(x-2)$：当 $x<2$ 时为负（凸），当 $x>2$ 时为正（凹）。(M1)
The sign change at $x=2$ confirms an inflection point; $f(2)=8-24+18+2=4$, so the inflection point is $(2,4)$. (A1)$x=2$ 处的变号确认存在拐点；$f(2)=8-24+18+2=4$，故拐点为 $(2,4)$。(A1)
$f$ is concave down on $(-\infty,2)$ and concave up on $(2,\infty)$. (A1)$f$ 在 $(-\infty,2)$ 上为凸，在 $(2,\infty)$ 上为凹。(A1)
Let $f(x)=\dfrac{x}{x^{2}+1}$ on $[0,3]$.设 $f(x)=\dfrac{x}{x^{2}+1}$ 在 $[0,3]$ 上。
$f'(x)=\dfrac{(x^{2}+1)(1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$. (M1)$f'(x)=\dfrac{(x^{2}+1)(1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$。(M1)
The denominator is never $0$, so set the numerator to $0$: $1-x^{2}=0$ gives $x=\pm 1$; only $x=1$ lies in $[0,3]$. (A1)分母恒不为 $0$，故令分子为 $0$：$1-x^{2}=0$ 得 $x=\pm 1$；只有 $x=1$ 落在 $[0,3]$ 内。(A1)
Evaluate $f$ at the critical point and both endpoints: $f(0)=0$, $f(1)=\dfrac{1}{2}$, $f(3)=\dfrac{3}{10}$. (M1)在临界点及两端点处求值：$f(0)=0$，$f(1)=\dfrac{1}{2}$，$f(3)=\dfrac{3}{10}$。(M1)
Comparing $0,\ \tfrac{1}{2}=0.5,\ \tfrac{3}{10}=0.3$: the absolute maximum is $f(1)=\dfrac{1}{2}$. (A1)比较 $0,\ \tfrac{1}{2}=0.5,\ \tfrac{3}{10}=0.3$：绝对最大值为 $f(1)=\dfrac{1}{2}$。(A1)
The absolute minimum is $f(0)=0$. (A1)绝对最小值为 $f(0)=0$。(A1)
Quotient rule on $f'$ gives $f''(x)=\dfrac{2x(x^{2}-3)}{(x^{2}+1)^{3}}$. (M1)对 $f'$ 再次应用商法则，得 $f''(x)=\dfrac{2x(x^{2}-3)}{(x^{2}+1)^{3}}$。(M1)
On $[0,3]$ the denominator is always positive and $x\ge 0$, so the sign of $f''$ matches the sign of $x^{2}-3$: negative on $(0,\sqrt{3})$, positive on $(\sqrt{3},3)$. (A1)在 $[0,3]$ 上分母恒为正，且 $x\ge 0$，故 $f''$ 的符号与 $x^{2}-3$ 一致：在 $(0,\sqrt{3})$ 上为负，在 $(\sqrt{3},3)$ 上为正。(A1)
So $f$ is concave up on $(\sqrt{3},3)$, with an inflection point at $x=\sqrt{3}$ where $f(\sqrt{3})=\dfrac{\sqrt{3}}{4}$. (A1)故 $f$ 在 $(\sqrt{3},3)$ 上为凹，拐点在 $x=\sqrt{3}$ 处，$f(\sqrt{3})=\dfrac{\sqrt{3}}{4}$。(A1)
A rectangular box with a square base and open top has volume $256$ in³. Material for the base costs $\$6$/in² and for the sides costs $\$2$/in².一个正方形底面、开口顶部的长方体，体积为 $256$ 立方英寸。底面材料每平方英寸 $\$6$，侧面材料每平方英寸 $\$2$。
Let $x$ be the base edge and $h$ the height: $x^{2}h=256$, so $h=\dfrac{256}{x^{2}}$. (M1)设 $x$ 为底边长，$h$ 为高：$x^{2}h=256$，故 $h=\dfrac{256}{x^{2}}$。(M1)
The base has area $x^{2}$ (cost $6x^{2}$); the four open-top sides each have area $xh$, total side area $4xh$ (cost $8xh$): $C=6x^{2}+8xh$. (M1)底面面积为 $x^{2}$（费用 $6x^{2}$）；开口顶部意味着四个侧面，每面面积 $xh$，总侧面面积 $4xh$（费用 $8xh$）：$C=6x^{2}+8xh$。(M1)
Substituting $h$: $C(x)=6x^{2}+8x\!\left(\dfrac{256}{x^{2}}\right)=6x^{2}+\dfrac{2048}{x}$, for $x>0$. (A1)代入 $h$：$C(x)=6x^{2}+8x\!\left(\dfrac{256}{x^{2}}\right)=6x^{2}+\dfrac{2048}{x}$，其中 $x>0$。(A1)
$C'(x)=12x-\dfrac{2048}{x^{2}}$. (M1)$C'(x)=12x-\dfrac{2048}{x^{2}}$。(M1)
Setting $C'(x)=0$: $12x^{3}=2048$, so $x^{3}=\dfrac{512}{3}$, giving $x=\dfrac{8}{\sqrt[3]{3}}=\dfrac{8\sqrt[3]{9}}{3}\approx 5.55$ in. (A1)令 $C'(x)=0$：$12x^{3}=2048$，故 $x^{3}=\dfrac{512}{3}$，解得 $x=\dfrac{8}{\sqrt[3]{3}}=\dfrac{8\sqrt[3]{9}}{3}\approx 5.55$ 英寸。(A1)
$C''(x)=12+\dfrac{4096}{x^{3}}>0$ for every $x>0$, so this critical point is a minimum; since $C(x)\to\infty$ as $x\to 0^{+}$ or $x\to\infty$, it is the global minimum on $x>0$. (R1)对所有 $x>0$，$C''(x)=12+\dfrac{4096}{x^{3}}>0$，故该临界点为最小值；又因当 $x\to 0^{+}$ 或 $x\to\infty$ 时 $C(x)\to\infty$，它是 $x>0$ 上的全局最小值。(R1)
$h=\dfrac{256}{x^{2}}=4\sqrt[3]{9}\approx 8.32$ in. (A1)$h=\dfrac{256}{x^{2}}=4\sqrt[3]{9}\approx 8.32$ 英寸。(A1)
$C_{\min}=6x^{2}+\dfrac{2048}{x}=384\sqrt[3]{3}\approx\$553.82$ (nearest cent). (A1)$C_{\min}=6x^{2}+\dfrac{2048}{x}=384\sqrt[3]{3}\approx\$553.82$（精确到分）。(A1)
The graph of $f'$, the derivative of a function $f$, is shown on $[-2,6]$. The graph consists of two line segments and a parabolic arc.函数 $f$ 的导数 $f'$ 的图像在 $[-2,6]$ 上如图所示，由两段线段和一段抛物线弧组成。
On $[-2,0]$, $f'$ is the line through $(-2,-1.5)$ and $(0,1.5)$: $f'(x)=1.5x+1.5$, crossing zero at $x=-1$ ($-$ to $+$, a local min, not max). On $[0,2]$, $f'\equiv 1.5>0$ (flat). On $[2,6]$, fitting a parabola through the three plotted points $(2,1.5)$, $(4,-1.5)$, $(6,0.5)$ gives $f'(x)=\tfrac{5}{8}x^{2}-\tfrac{21}{4}x+\tfrac{19}{2}$, with zeros at $x=\dfrac{21\pm\sqrt{61}}{5}\approx 2.64,\,5.76$. (M1)在 $[-2,0]$ 上，$f'$ 为过 $(-2,-1.5)$ 与 $(0,1.5)$ 的直线：$f'(x)=1.5x+1.5$，在 $x=-1$ 处穿越零点（负变正，为极小值而非极大值）。在 $[0,2]$ 上，$f'\equiv 1.5>0$（平坦）。在 $[2,6]$ 上，用图上三个标出点 $(2,1.5)$、$(4,-1.5)$、$(6,0.5)$ 拟合抛物线，得 $f'(x)=\tfrac{5}{8}x^{2}-\tfrac{21}{4}x+\tfrac{19}{2}$，零点为 $x=\dfrac{21\pm\sqrt{61}}{5}\approx 2.64,\,5.76$。(M1)
$f'\ge 0$ throughout $(-1,\,2.64)$, touching $0$ only at the endpoints, so the only place $f'$ changes from $+$ to $-$ in $(-2,6)$ is $x=\dfrac{21-\sqrt{61}}{5}\approx 2.64$. (A1)在整个 $(-1,\,2.64)$ 上 $f'\ge 0$，仅在端点处触及 $0$，故在 $(-2,6)$ 上 $f'$ 由正变负的唯一位置为 $x=\dfrac{21-\sqrt{61}}{5}\approx 2.64$。(A1)
By the First Derivative Test, this is the unique local maximum of $f$ on $(-2,6)$. (R1)由一阶导数判别法，这是 $f$ 在 $(-2,6)$ 上唯一的局部极大值。(R1)
On the two line segments, $f'$ is monotonic (increasing, then constant), so $f''\ge 0$ throughout $(-2,2)$ with no interior sign change: the corner at $x=0$ is a jump in slope, not a change in concavity. (M1)在两段线段上，$f'$ 单调（先递增后不变），故在 $(-2,2)$ 上 $f''\ge 0$ 处处成立，无内部变号：$x=0$ 处的拐角是斜率的跳变，而非凹凸性的改变。(M1)
On the parabolic arc, $f'(x)=\tfrac{5}{8}x^{2}-\tfrac{21}{4}x+\tfrac{19}{2}$ opens upward, with vertex (minimum) at $x=-\dfrac{b}{2a}=\dfrac{21}{5}=4.2$, $f'(4.2)=-\dfrac{61}{40}=-1.525$. (A1)在抛物线弧上，$f'(x)=\tfrac{5}{8}x^{2}-\tfrac{21}{4}x+\tfrac{19}{2}$ 开口向上，顶点（最小值）在 $x=-\dfrac{b}{2a}=\dfrac{21}{5}=4.2$，$f'(4.2)=-\dfrac{61}{40}=-1.525$。(A1)
$f'$ decreases into this vertex, then increases out of it, so $f''$ changes from negative to positive there: $x=\dfrac{21}{5}=4.2$ is the unique point of inflection of $f$ on $(-2,6)$. (R1)$f'$ 在该顶点前递减、之后递增，故 $f''$ 在此处由负变正：$x=\dfrac{21}{5}=4.2$ 是 $f$ 在 $(-2,6)$ 上唯一的拐点。(R1)
$f(0)-f(-2)=\displaystyle\int_{-2}^{0}f'(x)\,dx$, the net signed area between the graph of $f'$ and the $x$-axis. (M1)$f(0)-f(-2)=\displaystyle\int_{-2}^{0}f'(x)\,dx$，即 $f'$ 的图像与 $x$ 轴之间的净有向面积。(M1)
The line from $(-2,-1.5)$ to $(0,1.5)$ crosses zero at $x=-1$, forming two congruent triangles: one below the axis on $[-2,-1]$ (area $\tfrac{1}{2}(1)(1.5)=0.75$, counted negative) and one above the axis on $[-1,0]$ (area $0.75$, counted positive). (A1)从 $(-2,-1.5)$ 到 $(0,1.5)$ 的直线在 $x=-1$ 处穿越零点，形成两个全等三角形：一个在 $[-2,-1]$ 上位于轴下方（面积 $\tfrac{1}{2}(1)(1.5)=0.75$，计为负），一个在 $[-1,0]$ 上位于轴上方（面积 $0.75$，计为正）。(A1)
The two areas cancel: $\displaystyle\int_{-2}^{0}f'(x)\,dx=-0.75+0.75=0$, so $f(0)-f(-2)=0$, giving $f(0)=f(-2)=3$. (R1)两块面积恰好抵消：$\displaystyle\int_{-2}^{0}f'(x)\,dx=-0.75+0.75=0$，故 $f(0)-f(-2)=0$，即 $f(0)=f(-2)=3$。(R1)
A farmer has $800$ meters of fencing and wants to enclose a rectangular field along a straight river (no fencing is needed along the river) and then divide it with one fence parallel to the river.一位农民有 $800$ 米的围栏，要沿一条直河围一块矩形农田（沿河一侧不需要围栏），并用一道平行于河流的围栏将其分为两部分。
The far side opposite the river (length $x$), the two ends perpendicular to the river (each length $y$), and the interior divider parallel to the river (also length $x$) all need fencing; the river side needs none. (M1)与河流相对的远边（长 $x$）、垂直于河流的两条边（每条长 $y$）、以及平行于河流的内部分隔围栏（同样长 $x$），均需围栏；沿河一侧无需围栏。(M1)
Total fencing: $x+2y+x=2x+2y=800$. (A1)总围栏长度：$x+2y+x=2x+2y=800$。(A1)
Solve the constraint for $y$: $y=400-x$ (from $x+y=400$). (M1)由约束条件解出 $y$：$y=400-x$（由 $x+y=400$）。(M1)
$A=xy=x(400-x)=400x-x^{2}$, with domain $0
$A'(x)=400-2x$; setting $A'(x)=0$ gives $x=200$. (M1)$A'(x)=400-2x$；令 $A'(x)=0$ 得 $x=200$。(M1)
$A''(x)=-2$, a negative constant. (A1)$A''(x)=-2$，为负常数。(A1)
Since $A''(x)<0$ everywhere, $x=200$ gives a maximum; because $A$ is a downward parabola in $x$, it is the unique global maximum on the domain. (R1)因 $A''(x)<0$ 处处成立，$x=200$ 处取得最大值；又因 $A$ 是关于 $x$ 的开口向下抛物线，这是定义域上唯一的全局最大值。(R1)
$x=200$ m. (A1)$x=200$ 米。(A1)
$y=400-200=200$ m. (A1)$y=400-200=200$ 米。(A1)
Maximum area $A(200)=200\times 200=40{,}000\text{ m}^2$. (A1)最大面积 $A(200)=200\times 200=40{,}000$ 平方米。(A1)