Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: the limit definition, the derivative as a function, slope and rate interpretations, differentiability vs continuity, first-principles computation, higher-order derivatives, where differentiability fails第1至7节：极限定义，导函数，斜率与变化率解释，可微性与连续性，第一性原理计算，高阶导数，可微性失效之处CALC I
(a) Find $f'(2)$ for $f(x)=x^{2}-3x$ via the $h\to 0$ form. (b) Find $f'(1)$ for $f(x)=\frac{1}{x}$ via the $x\to a$ form.(a) 用 $h\to 0$ 形式求 $f(x)=x^{2}-3x$ 在 $f'(2)$。(b) 用 $x\to a$ 形式求 $f(x)=\frac{1}{x}$ 在 $f'(1)$。
Form the difference quotient: (M1)建立差商：(M1)
$$ \frac{f(2+h)-f(2)}{h} = \frac{\bigl[(2+h)^{2}-3(2+h)\bigr]-\bigl[4-6\bigr]}{h}. $$Expand the numerator: $(2+h)^{2}-3(2+h) = 4+4h+h^{2}-6-3h = -2+h+h^{2}$. So:展开分子：$(2+h)^{2}-3(2+h) = 4+4h+h^{2}-6-3h = -2+h+h^{2}$。于是：
$$ \frac{(-2+h+h^{2})-(-2)}{h} = \frac{h+h^{2}}{h} = 1+h. $$(M1) Taking the limit: $f'(2)=\lim_{h\to 0}(1+h)=1$. (A1)(M1) 取极限：$f'(2)=\lim_{h\to 0}(1+h)=1$。(A1)
Form the quotient with $a=1$: (M1)以 $a=1$ 建立商式：(M1)
$$ \frac{f(x)-f(1)}{x-1} = \frac{\frac{1}{x}-1}{x-1} = \frac{\frac{1-x}{x}}{x-1} = \frac{-(x-1)}{x(x-1)} = \frac{-1}{x}. $$(M1) Taking the limit: $f'(1)=\lim_{x\to 1}\frac{-1}{x}=-1$. (A1)(M1) 取极限：$f'(1)=\lim_{x\to 1}\frac{-1}{x}=-1$。(A1)
For $g(x)=|2x-4|$: (a) write as piecewise-linear; (b) compute $g'_{-}(2)$ and $g'_{+}(2)$ from the definition; (c) is $g$ differentiable at $x=2$?对于 $g(x)=|2x-4|$：(a) 写成分段线性形式；(b) 从定义计算 $g'_{-}(2)$ 和 $g'_{+}(2)$；(c) $g$ 在 $x=2$ 处可微吗？
$2x-4<0$ when $x<2$, so $g(x)=-(2x-4)=4-2x$ for $x<2$, and $g(x)=2x-4$ for $x\ge 2$. (B1)当 $x<2$ 时 $2x-4<0$，故 $x<2$ 时 $g(x)=-(2x-4)=4-2x$，$x\ge 2$ 时 $g(x)=2x-4$。(B1)
Left-hand derivative at $x=2$ (using the branch $g(x)=4-2x$ for $h<0$): (M1)$x=2$ 处的左导数（当 $h<0$ 时使用分支 $g(x)=4-2x$）：(M1)
$$ g'_{-}(2)=\lim_{h\to 0^{-}}\frac{g(2+h)-g(2)}{h}=\lim_{h\to 0^{-}}\frac{(4-2(2+h))-0}{h}=\lim_{h\to 0^{-}}\frac{-2h}{h}=-2. $$(A1) Right-hand derivative at $x=2$ (using the branch $g(x)=2x-4$ for $h>0$): (M1)(A1) $x=2$ 处的右导数（当 $h>0$ 时使用分支 $g(x)=2x-4$）：(M1)
$$ g'_{+}(2)=\lim_{h\to 0^{+}}\frac{g(2+h)-g(2)}{h}=\lim_{h\to 0^{+}}\frac{(2(2+h)-4)-0}{h}=\lim_{h\to 0^{+}}\frac{2h}{h}=2. $$(A1)(A1)
Since $g'_{-}(2)=-2\ne 2=g'_{+}(2)$, the two-sided limit $\lim_{h\to 0}\frac{g(2+h)-g(2)}{h}$ does not exist. Therefore $g$ is not differentiable at $x=2$. (R1) The failure is a corner.由于 $g'_{-}(2)=-2\ne 2=g'_{+}(2)$，双侧极限 $\lim_{h\to 0}\frac{g(2+h)-g(2)}{h}$ 不存在。因此 $g$ 在 $x=2$ 处不可微。(R1) 失效类型为角点。
For $h(x)=ax^{2}+b$ ($x\le 1$), $h(x)=3x-2$ ($x>1$): (a) state the two differentiability conditions; (b) match derivatives to find $a$; (c) use continuity to find $b$.对于 $h(x)=ax^{2}+b$（$x\le 1$），$h(x)=3x-2$（$x>1$）：(a) 陈述两个可微性条件；(b) 匹配导数以求 $a$；(c) 利用连续性求 $b$。
Differentiability at $x=1$ requires first that $h$ be continuous there (necessary, by the theorem of Q6), and second that the one-sided derivatives agree. (R1)在 $x=1$ 处可微首先要求 $h$ 在该点连续（Q6定理的必要条件），其次要求单侧导数相等。(R1)
Continuity: $\lim_{x\to 1^{-}}h(x)=a(1)^{2}+b=a+b$ and $\lim_{x\to 1^{+}}h(x)=3(1)-2=1$, so $a+b=1$. (R1)连续性：$\lim_{x\to 1^{-}}h(x)=a(1)^{2}+b=a+b$，$\lim_{x\to 1^{+}}h(x)=3(1)-2=1$，故 $a+b=1$。(R1)
Left derivative at $x=1$: for $x\le 1$, $h(x)=ax^{2}+b$, so $h'_{-}(1)=2a(1)=2a$. (M1)$x=1$ 处的左导数：当 $x\le 1$ 时，$h(x)=ax^{2}+b$，故 $h'_{-}(1)=2a(1)=2a$。(M1)
Right derivative at $x=1$: for $x>1$, $h(x)=3x-2$, so $h'_{+}(1)=3$. (A1)$x=1$ 处的右导数：当 $x>1$ 时，$h(x)=3x-2$，故 $h'_{+}(1)=3$。(A1)
Setting $h'_{-}(1)=h'_{+}(1)$: $2a=3$, giving $a=\dfrac{3}{2}$. (A1)令 $h'_{-}(1)=h'_{+}(1)$：$2a=3$，得 $a=\dfrac{3}{2}$。(A1)
Substitute $a=\dfrac{3}{2}$ into $a+b=1$: $\dfrac{3}{2}+b=1$, so $b=-\dfrac{1}{2}$. (M1·A1)将 $a=\dfrac{3}{2}$ 代入 $a+b=1$：$\dfrac{3}{2}+b=1$，故 $b=-\dfrac{1}{2}$。(M1·A1)
The complete function is $h(x)=\tfrac{3}{2}x^{2}-\tfrac{1}{2}$ for $x\le 1$ and $h(x)=3x-2$ for $x>1$. One can verify: $h(1)=\tfrac{3}{2}-\tfrac{1}{2}=1$ and $3(1)-2=1$; $h'_{-}(1)=3$ and $h'_{+}(1)=3$. (A1)完整函数为：$x\le 1$ 时 $h(x)=\tfrac{3}{2}x^{2}-\tfrac{1}{2}$，$x>1$ 时 $h(x)=3x-2$。验证：$h(1)=\tfrac{3}{2}-\tfrac{1}{2}=1$ 且 $3(1)-2=1$；$h'_{-}(1)=3$ 且 $h'_{+}(1)=3$。(A1)
For $s(t)=t^{3}-6t^{2}+9t+2$: (a) compute $s'(t)$ from the limit definition; (b) find $s''(t)$; (c) find when the particle is at rest and classify acceleration.对于 $s(t)=t^{3}-6t^{2}+9t+2$：(a) 从极限定义计算 $s'(t)$；(b) 求 $s''(t)$；(c) 求质点静止时刻并判断加速度方向。
Form the difference quotient: (M1)建立差商：(M1)
$$ \frac{s(t+h)-s(t)}{h}. $$Expand $s(t+h)=(t+h)^{3}-6(t+h)^{2}+9(t+h)+2$:展开 $s(t+h)=(t+h)^{3}-6(t+h)^{2}+9(t+h)+2$：
$$ (t+h)^{3}=t^{3}+3t^{2}h+3th^{2}+h^{3}, $$ $$ (t+h)^{2}=t^{2}+2th+h^{2}. $$So $s(t+h)=t^{3}+3t^{2}h+3th^{2}+h^{3}-6t^{2}-12th-6h^{2}+9t+9h+2$.故 $s(t+h)=t^{3}+3t^{2}h+3th^{2}+h^{3}-6t^{2}-12th-6h^{2}+9t+9h+2$。
Subtract $s(t)=t^{3}-6t^{2}+9t+2$: (M1)减去 $s(t)=t^{3}-6t^{2}+9t+2$：(M1)
$$ s(t+h)-s(t)=3t^{2}h+3th^{2}+h^{3}-12th-6h^{2}+9h. $$Divide by $h$:除以 $h$：
$$ \frac{s(t+h)-s(t)}{h}=3t^{2}+3th+h^{2}-12t-6h+9. $$(A1) Take the limit as $h\to 0$: $s'(t)=3t^{2}-12t+9$. (A1)(A1) 令 $h\to 0$ 取极限：$s'(t)=3t^{2}-12t+9$。(A1)
Differentiate $s'(t)=3t^{2}-12t+9$ using the power rule: (M1)用幂法则对 $s'(t)=3t^{2}-12t+9$ 求导：(M1)
$$ s''(t)=6t-12. $$(A1)(A1)
The particle is at rest when $s'(t)=0$: $3t^{2}-12t+9=3(t-1)(t-3)=0$, giving $t=1$ and $t=3$. (M1)质点静止时 $s'(t)=0$：$3t^{2}-12t+9=3(t-1)(t-3)=0$，得 $t=1$ 和 $t=3$。(M1)
At $t=1$: $s''(1)=6(1)-12=-6<0$, so the particle is decelerating. At $t=3$: $s''(3)=6(3)-12=6>0$, so the particle is accelerating. (A1)当 $t=1$ 时：$s''(1)=6(1)-12=-6<0$，质点在减速。当 $t=3$ 时：$s''(3)=6(3)-12=6>0$，质点在加速。(A1)
For $f(x)=\frac{1}{x^{2}}$: (a) derive $f'(x)$ from the limit definition; (b) verify against the power rule; (c) find tangent and normal lines at $x=-1$.对于 $f(x)=\frac{1}{x^{2}}$：(a) 从极限定义推导 $f'(x)$；(b) 对照幂法则验证；(c) 在 $x=-1$ 处求切线和法线。
Set up the difference quotient: (M1)建立差商：(M1)
$$ \frac{f(x+h)-f(x)}{h}=\frac{\dfrac{1}{(x+h)^{2}}-\dfrac{1}{x^{2}}}{h}. $$Combine the two fractions over the common denominator $x^{2}(x+h)^{2}$: (M1)以公分母 $x^{2}(x+h)^{2}$ 通分：(M1)
$$ =\frac{x^{2}-(x+h)^{2}}{h\cdot x^{2}(x+h)^{2}}. $$Expand the numerator: $x^{2}-(x+h)^{2}=x^{2}-(x^{2}+2xh+h^{2})=-2xh-h^{2}=h(-2x-h)$. (M1)展开分子：$x^{2}-(x+h)^{2}=x^{2}-(x^{2}+2xh+h^{2})=-2xh-h^{2}=h(-2x-h)$。(M1)
$$ \frac{h(-2x-h)}{h\cdot x^{2}(x+h)^{2}}=\frac{-2x-h}{x^{2}(x+h)^{2}}. $$(A1) Take the limit as $h\to 0$: (A1)(A1) 令 $h\to 0$ 取极限：(A1)
$$ f'(x)=\lim_{h\to 0}\frac{-2x-h}{x^{2}(x+h)^{2}}=\frac{-2x}{x^{2}\cdot x^{2}}=\frac{-2}{x^{3}}. $$Writing $f(x)=x^{-2}$, the power rule gives $f'(x)=-2x^{-3}=-\dfrac{2}{x^{3}}$. This matches part (a). (B1)将 $f(x)$ 写为 $x^{-2}$，幂法则给出 $f'(x)=-2x^{-3}=-\dfrac{2}{x^{3}}$，与 (a) 部分一致。(B1)
At $x=-1$: $f(-1)=\dfrac{1}{(-1)^{2}}=1$ and $f'(-1)=-\dfrac{2}{(-1)^{3}}=2$. (M1)在 $x=-1$ 处：$f(-1)=\dfrac{1}{(-1)^{2}}=1$，$f'(-1)=-\dfrac{2}{(-1)^{3}}=2$。(M1)
Tangent line: $y-1=2(x-(-1))$, i.e. $y=2x+3$.切线：$y-1=2(x-(-1))$，即 $y=2x+3$。
The normal line has slope $-\dfrac{1}{2}$ (negative reciprocal): $y-1=-\tfrac{1}{2}(x+1)$, i.e. $y=-\tfrac{1}{2}x+\tfrac{1}{2}$. (A1)法线斜率为 $-\dfrac{1}{2}$（负倒数）：$y-1=-\tfrac{1}{2}(x+1)$，即 $y=-\tfrac{1}{2}x+\tfrac{1}{2}$。(A1)
(a) Prove: differentiability at $a$ implies continuity at $a$. (b) Give a specific counterexample showing continuity does not imply differentiability.(a) 证明：在 $a$ 处可微蕴含在 $a$ 处连续。(b) 给出一个具体反例，说明连续不蕴含可微。
Assume $f$ is differentiable at $x=a$, so $f'(a)=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists as a finite number. (M1)设 $f$ 在 $x=a$ 处可微，即 $f'(a)=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ 作为有限数存在。(M1)
We want to show $\lim_{h\to 0}f(a+h)=f(a)$, i.e. $\lim_{h\to 0}[f(a+h)-f(a)]=0$.我们需要证明 $\lim_{h\to 0}f(a+h)=f(a)$，即 $\lim_{h\to 0}[f(a+h)-f(a)]=0$。
Write the identity: (M1)写出恒等式：(M1)
$$ f(a+h)-f(a)=\frac{f(a+h)-f(a)}{h}\cdot h. $$(This requires $h\ne 0$, which is fine since we are taking a limit.) (M1)（此式要求 $h\ne 0$，这在取极限时是允许的。）(M1)
Take the limit of both sides as $h\to 0$. The product rule for limits applies because both factors have limits: (R1)对两边令 $h\to 0$ 取极限。由于两个因子均有极限，可用极限的乘积法则：(R1)
$$ \lim_{h\to 0}[f(a+h)-f(a)]=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot\lim_{h\to 0}h=f'(a)\cdot 0=0. $$Therefore $\lim_{h\to 0}f(a+h)=f(a)$, which is exactly the definition of continuity of $f$ at $a$. (A1)因此 $\lim_{h\to 0}f(a+h)=f(a)$，这正是 $f$ 在 $a$ 处连续的定义。(A1)
Let $f(x)=|x|$. Then $f(0)=0$ and $\lim_{x\to 0}|x|=0=f(0)$, so $f$ is continuous at $x=0$. (M1)设 $f(x)=|x|$。则 $f(0)=0$ 且 $\lim_{x\to 0}|x|=0=f(0)$，故 $f$ 在 $x=0$ 处连续。(M1)
Check differentiability: (A1)检验可微性：(A1)
$$ \lim_{h\to 0^{+}}\frac{|h|-0}{h}=\lim_{h\to 0^{+}}\frac{h}{h}=1, \qquad \lim_{h\to 0^{-}}\frac{|h|-0}{h}=\lim_{h\to 0^{-}}\frac{-h}{h}=-1. $$Since the one-sided limits of the difference quotient are $+1\ne -1$, the limit does not exist: $f$ is not differentiable at $x=0$. (R1)由于差商的两个单侧极限为 $+1\ne -1$，极限不存在，故 $f$ 在 $x=0$ 处不可微。(R1)
For $f(x)=x^{3}$: (a) prove $f'(a)=3a^{2}$ from the limit definition; (b) interpret the secant slope and its limit; (c) where is the tangent horizontal?对于 $f(x)=x^{3}$：(a) 从极限定义证明 $f'(a)=3a^{2}$；(b) 解释割线斜率及其极限；(c) 切线在哪里水平？
Form the difference quotient: (M1)建立差商：(M1)
$$ \frac{(a+h)^{3}-a^{3}}{h}. $$Expand $(a+h)^{3}=a^{3}+3a^{2}h+3ah^{2}+h^{3}$. (M1) Subtract $a^{3}$:展开 $(a+h)^{3}=a^{3}+3a^{2}h+3ah^{2}+h^{3}$。(M1) 减去 $a^{3}$：
$$ \frac{3a^{2}h+3ah^{2}+h^{3}}{h}=3a^{2}+3ah+h^{2}. $$(M1·A1) Take the limit: (A1)(M1·A1) 取极限：(A1)
$$ f'(a)=\lim_{h\to 0}(3a^{2}+3ah+h^{2})=3a^{2}. $$The slope of the secant joining $(a, a^{3})$ to $(a+h, (a+h)^{3})$ is (M1)连接 $(a, a^{3})$ 与 $(a+h, (a+h)^{3})$ 的割线斜率为 (M1)
$$ m_{\text{sec}}=\frac{(a+h)^{3}-a^{3}}{(a+h)-a}=3a^{2}+3ah+h^{2}. $$(A1) As $h\to 0$, the two points approach each other and the secant becomes the tangent. The secant slope $3a^{2}+3ah+h^{2}$ approaches $3a^{2}$, which is the instantaneous slope $f'(a)$. (R1) The two extra terms $3ah+h^{2}$ represent the error in approximating the tangent by a secant: they vanish when $h\to 0$.(A1) 当 $h\to 0$ 时，两点相互趋近，割线趋于切线。割线斜率 $3a^{2}+3ah+h^{2}$ 趋于 $3a^{2}$，即瞬时斜率 $f'(a)$。(R1) 额外两项 $3ah+h^{2}$ 代表用割线近似切线的误差，当 $h\to 0$ 时消失。
The tangent is horizontal when $f'(a)=0$: $3a^{2}=0$, so $a=0$. (M1) The only point with a horizontal tangent is $(0, 0^{3})=(0,0)$. (A1)切线水平时 $f'(a)=0$：$3a^{2}=0$，故 $a=0$。(M1) 唯一具有水平切线的点为 $(0, 0^{3})=(0,0)$。(A1)
For $y=\sqrt{2x+1}$: (a) derive $\frac{d}{dx}[\sqrt{2x+1}]=\frac{1}{\sqrt{2x+1}}$ from the definition; (b) find the tangent at $x=4$; (c) find the normal at $x=4$ and confirm perpendicularity.对于 $y=\sqrt{2x+1}$：(a) 从定义推导 $\frac{d}{dx}[\sqrt{2x+1}]=\frac{1}{\sqrt{2x+1}}$；(b) 求 $x=4$ 处的切线；(c) 求 $x=4$ 处的法线并验证垂直性。
Let $f(x)=\sqrt{2x+1}$. Form the difference quotient: (M1)设 $f(x)=\sqrt{2x+1}$，建立差商：(M1)
$$ \frac{f(x+h)-f(x)}{h}=\frac{\sqrt{2(x+h)+1}-\sqrt{2x+1}}{h}=\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}. $$Rationalise by multiplying numerator and denominator by the conjugate $\sqrt{2x+2h+1}+\sqrt{2x+1}$: (M1)乘以共轭因子 $\sqrt{2x+2h+1}+\sqrt{2x+1}$ 进行有理化：(M1)
$$ =\frac{(2x+2h+1)-(2x+1)}{h\left(\sqrt{2x+2h+1}+\sqrt{2x+1}\right)}=\frac{2h}{h\left(\sqrt{2x+2h+1}+\sqrt{2x+1}\right)}=\frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}. $$(M1) Take the limit as $h\to 0$: (A1)(M1) 令 $h\to 0$ 取极限：(A1)
$$ \frac{d}{dx}\!\left[\sqrt{2x+1}\right]=\lim_{h\to 0}\frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}. $$At $x=4$: $f(4)=\sqrt{9}=3$ and $f'(4)=\dfrac{1}{\sqrt{9}}=\dfrac{1}{3}$. (M1)在 $x=4$ 处：$f(4)=\sqrt{9}=3$，$f'(4)=\dfrac{1}{\sqrt{9}}=\dfrac{1}{3}$。(M1)
Point-slope form: $y-3=\dfrac{1}{3}(x-4)$. (A1) Simplify: $y=\dfrac{1}{3}x-\dfrac{4}{3}+3=\dfrac{1}{3}x+\dfrac{5}{3}$. (A1)点斜式：$y-3=\dfrac{1}{3}(x-4)$。(A1) 化简：$y=\dfrac{1}{3}x-\dfrac{4}{3}+3=\dfrac{1}{3}x+\dfrac{5}{3}$。(A1)
The normal has slope $-\dfrac{1}{1/3}=-3$. (M1) Normal line through $(4, 3)$: $y-3=-3(x-4)$, i.e. $y=-3x+15$. (A1)法线斜率为 $-\dfrac{1}{1/3}=-3$。(M1) 过 $(4, 3)$ 的法线：$y-3=-3(x-4)$，即 $y=-3x+15$。(A1)
Product of slopes: $\dfrac{1}{3}\cdot(-3)=-1$. Since the product of slopes equals $-1$, the tangent and normal are perpendicular. (R1)斜率之积：$\dfrac{1}{3}\cdot(-3)=-1$。由于斜率之积等于 $-1$，切线与法线相互垂直。(R1)
For $V(r)=\frac{4}{3}\pi r^{3}$: (a) average rate over $[2, 2+h]$; (b) instantaneous rate $V'(r)$ and its geometric meaning; (c) rate of volume increase at $r=3$ when $\dot{r}=0.5$ cm/s; (d) geometric meaning of the average rate on the graph.对于 $V(r)=\frac{4}{3}\pi r^{3}$：(a) $[2, 2+h]$ 上的平均变化率；(b) 瞬时变化率 $V'(r)$ 及其几何意义；(c) 当 $\dot{r}=0.5$ cm/s 时 $r=3$ 处体积的变化率；(d) 平均变化率在图像上的几何意义。
Compute $V(2+h)-V(2)$: (M1)计算 $V(2+h)-V(2)$：(M1)
$$ V(2+h)=\tfrac{4}{3}\pi(2+h)^{3}=\tfrac{4}{3}\pi(8+12h+6h^{2}+h^{3}), \quad V(2)=\tfrac{4}{3}\pi(8). $$ $$ V(2+h)-V(2)=\tfrac{4}{3}\pi(12h+6h^{2}+h^{3}). $$Divide by $h$: (A1)除以 $h$：(A1)
$$ \frac{V(2+h)-V(2)}{h}=\tfrac{4}{3}\pi(12+6h+h^{2}). $$(A1)(A1)
At general $r$, the difference quotient for $V(r)=\tfrac{4}{3}\pi r^{3}$ is (M1)对一般的 $r$，$V(r)=\tfrac{4}{3}\pi r^{3}$ 的差商为 (M1)
$$ \frac{V(r+h)-V(r)}{h}=\tfrac{4}{3}\pi\frac{(r+h)^{3}-r^{3}}{h}=\tfrac{4}{3}\pi(3r^{2}+3rh+h^{2}). $$Taking the limit: $V'(r)=\tfrac{4}{3}\pi\cdot 3r^{2}=4\pi r^{2}$. (A1)取极限：$V'(r)=\tfrac{4}{3}\pi\cdot 3r^{2}=4\pi r^{2}$。(A1)
The formula $4\pi r^{2}$ is the surface area of a sphere of radius $r$. Geometrically: a thin shell of thickness $\Delta r$ has volume $\approx 4\pi r^{2}\Delta r$, so the rate of change of volume with respect to radius equals the surface area. (R1)公式 $4\pi r^{2}$ 是半径为 $r$ 的球的表面积。几何上：厚度为 $\Delta r$ 的薄球壳体积约为 $4\pi r^{2}\Delta r$，因此体积关于半径的变化率等于表面积。(R1)
$V'(3)=4\pi(3)^{2}=36\pi$ cm$^{3}$/cm. If the radius increases at $0.5$ cm/s, the volume increases at: (M1)$V'(3)=4\pi(3)^{2}=36\pi$ cm$^{3}$/cm。若半径以 $0.5$ cm/s 增大，则体积的增加率为：(M1)
$$ \dot{V}\approx V'(3)\cdot\dot{r}=36\pi\times 0.5=18\pi\approx 56.5\text{ cm}^{3}\text{/s}. $$(A1) Units: cm$^{3}$/s.(A1) 单位：cm$^{3}$/s。
The average rate of change $\frac{V(r_{0}+h)-V(r_{0})}{h}$ is the slope of the secant line on the graph of $V$ versus $r$ joining the points $(r_{0}, V(r_{0}))$ and $(r_{0}+h, V(r_{0}+h))$. (B1)平均变化率 $\frac{V(r_{0}+h)-V(r_{0})}{h}$ 是 $V$ 关于 $r$ 的图像上连接点 $(r_{0}, V(r_{0}))$ 与 $(r_{0}+h, V(r_{0}+h))$ 的割线斜率。(B1)
(a) $f(x)=|x^{2}-1|$ at $x=1$: classify the failure. (b) $g(x)=x^{2/3}$ at $x=0$: classify the failure. (c) Sketch qualitative behaviour near the respective points.(a) $f(x)=|x^{2}-1|$ 在 $x=1$ 处：分类失效类型。(b) $g(x)=x^{2/3}$ 在 $x=0$ 处：分类失效类型。(c) 在相应点附近画出定性图形。
Near $x=1$: $x^{2}-1=(x-1)(x+1)$, which is negative for $x\in(-1,1)$ and positive for $|x|>1$. So just left of $x=1$, $f(x)=-(x^{2}-1)=1-x^{2}$, and just right of $x=1$, $f(x)=x^{2}-1$. Compute the one-sided derivatives: (M1)在 $x=1$ 附近：$x^{2}-1=(x-1)(x+1)$，当 $x\in(-1,1)$ 时为负，当 $|x|>1$ 时为正。因此在 $x=1$ 稍左处 $f(x)=-(x^{2}-1)=1-x^{2}$，在 $x=1$ 稍右处 $f(x)=x^{2}-1$。计算单侧导数：(M1)
$$ f'_{-}(1)=\lim_{h\to 0^{-}}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^{-}}\frac{(1-(1+h)^{2})-0}{h}=\lim_{h\to 0^{-}}\frac{-2h-h^{2}}{h}=-2. $$ $$ f'_{+}(1)=\lim_{h\to 0^{+}}\frac{(1+h)^{2}-1-0}{h}=\lim_{h\to 0^{+}}\frac{2h+h^{2}}{h}=2. $$(A1) Since $f'_{-}(1)=-2\ne 2=f'_{+}(1)$, the derivative does not exist at $x=1$. The failure is a corner: both one-sided derivatives exist as finite numbers but are unequal. (R1)(A1) 由于 $f'_{-}(1)=-2\ne 2=f'_{+}(1)$，导数在 $x=1$ 处不存在。失效类型为角点：两个单侧导数均作为有限数存在，但不相等。(R1)
Compute the one-sided limits of the difference quotient at $x=0$: (M1)计算 $x=0$ 处差商的单侧极限：(M1)
$$ \lim_{h\to 0^{+}}\frac{h^{2/3}-0}{h}=\lim_{h\to 0^{+}}h^{-1/3}=+\infty. $$ $$ \lim_{h\to 0^{-}}\frac{h^{2/3}-0}{h}=\lim_{h\to 0^{-}}h^{-1/3}=-\infty. $$(A1) Both one-sided limits of the difference quotient are infinite (and have opposite signs). This is a cusp: the tangent line approaches vertical from both sides, but on opposite sides. (R1) The function is continuous at $x=0$ (since $0^{2/3}=0$) but not differentiable there.(A1) 差商的两个单侧极限均为无穷大（且符号相反）。这是一个尖点：切线从两侧趋向竖直，但在相反方向。(R1) 函数在 $x=0$ 处连续（因为 $0^{2/3}=0$），但在该处不可微。
For $f(x)=|x^{2}-1|$ near $x=1$: the graph comes from above-left with slope $-2$ and leaves to the right with slope $+2$, creating a sharp corner at $(1,0)$. Label: "corner". (B1)对于 $x=1$ 附近的 $f(x)=|x^{2}-1|$：图像从左上方以斜率 $-2$ 到达，从右侧以斜率 $+2$ 离去，在 $(1,0)$ 处形成尖锐的角点。标注："角点"。(B1)
For $g(x)=x^{2/3}$ near $x=0$: the graph is symmetric about the $y$-axis, concave down near $0$, and comes to a sharp point (like a rounded vee) at the origin with the two arms nearly vertical. Label: "cusp". (B1)对于 $x=0$ 附近的 $g(x)=x^{2/3}$：图像关于 $y$ 轴对称，在 $0$ 附近下凹，在原点处形成尖锐的点（像一个圆润的V形），两条臂几乎竖直。标注："尖点"。(B1)