Curve Sketching and Optimization · Solutions曲线描绘与最优化 · 解答

(a) Differentiate and factor(a) 求导与因式分解 M1·A1·A1

Differentiate term by term: (M1)逐项求导：(M1)

Since $f'$ is a polynomial it exists everywhere, so critical points come only from $f'(x)=0$: the solutions are $x = 0$, $x = 1$, and $x = 2$. (A1 for factored form, A1 for listing all three.)因为 $f'$ 是多项式，处处存在，所以极值点仅来自 $f'(x)=0$：解为 $x = 0$、$x = 1$、$x = 2$。（A1 给因式分解形式，A1 给列出全部三点。）

(b) Sign chart and monotonicity(b) 符号图与单调性 M1·M1·A1

The three critical points split $\mathbb{R}$ into four intervals. Test one point in each: (M1 for method)三个极值点将 $\mathbb{R}$ 分为四个区间。在每个区间内取一检验点：(M1 给方法)

• $x = -1$: $f'(-1) = 4(-1)(-2)(-3) = -24 < 0$ (decreasing)$x = -1$：$f'(-1) = 4(-1)(-2)(-3) = -24 < 0$（递减）
• $x = \tfrac{1}{2}$: $f'(\tfrac{1}{2}) = 4(\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2}) = \tfrac{3}{2} > 0$ (increasing)$x = \tfrac{1}{2}$：$f'(\tfrac{1}{2}) = 4(\tfrac{1}{2})(-\tfrac{1}{2})(-\tfrac{3}{2}) = \tfrac{3}{2} > 0$（递增）
• $x = \tfrac{3}{2}$: $f'(\tfrac{3}{2}) = 4(\tfrac{3}{2})(\tfrac{1}{2})(-\tfrac{1}{2}) = -\tfrac{3}{2} < 0$ (decreasing)$x = \tfrac{3}{2}$：$f'(\tfrac{3}{2}) = 4(\tfrac{3}{2})(\tfrac{1}{2})(-\tfrac{1}{2}) = -\tfrac{3}{2} < 0$（递减）
• $x = 3$: $f'(3) = 4(3)(2)(1) = 24 > 0$ (increasing)$x = 3$：$f'(3) = 4(3)(2)(1) = 24 > 0$（递增）

Sign pattern: $(-)$ on $(-\infty, 0)$, $(+)$ on $(0,1)$, $(-)$ on $(1,2)$, $(+)$ on $(2,\infty)$. (A1 for each correct interval, up to M1 for using sign chart structure)符号规律：在 $(-\infty, 0)$ 上为 $(-)$，在 $(0,1)$ 上为 $(+)$，在 $(1,2)$ 上为 $(-)$，在 $(2,\infty)$ 上为 $(+)$。（A1 给各正确区间，M1 给使用符号图结构）

$f$ is increasing on $(0,1)$ and $(2, \infty)$; decreasing on $(-\infty,0)$ and $(1,2)$. (A1)$f$ 在 $(0,1)$ 和 $(2, \infty)$ 上递增；在 $(-\infty,0)$ 和 $(1,2)$ 上递减。(A1)

(c) First Derivative Test(c) 一阶导数判别法 M1·A1

Apply the sign changes: (M1)根据符号变化得出：(M1)

• $x = 0$: $f'$ changes from $(-)$ to $(+)$ so $x=0$ is a local minimum. $f(0) = 0$.$x = 0$：$f'$ 从 $(-)$ 变为 $(+)$，故 $x=0$ 为局部极小值点。$f(0) = 0$。
• $x = 1$: $f'$ changes from $(+)$ to $(-)$ so $x=1$ is a local maximum. $f(1) = 1 - 4 + 4 = 1$.$x = 1$：$f'$ 从 $(+)$ 变为 $(-)$，故 $x=1$ 为局部极大值点。$f(1) = 1 - 4 + 4 = 1$。
• $x = 2$: $f'$ changes from $(-)$ to $(+)$ so $x=2$ is a local minimum. $f(2) = 16 - 32 + 16 = 0$.$x = 2$：$f'$ 从 $(-)$ 变为 $(+)$，故 $x=2$ 为局部极小值点。$f(2) = 16 - 32 + 16 = 0$。

(A1 for all three correctly classified with values)（A1 给三个极值点均正确分类并附函数值）

(a) First derivative and critical points(a) 一阶导数与极值点 M1·A1·M1·A1

Apply the product rule with $u = x$ and $v = e^{-x^{2}/2}$, noting $v' = -x\,e^{-x^{2}/2}$: (M1)对 $u = x$，$v = e^{-x^{2}/2}$ 应用乘积法则，注意 $v' = -x\,e^{-x^{2}/2}$：(M1)

Since $e^{-x^{2}/2} > 0$ for all $x$, the sign of $g'$ is determined entirely by $(1-x^{2})$. Critical points: $g'(x) = 0$ gives $x = 1$ and $x = -1$. (A1)因为 $e^{-x^{2}/2} > 0$ 对所有 $x$ 成立，$g'$ 的符号完全由 $(1-x^{2})$ 决定。极值点：$g'(x) = 0$ 给出 $x = 1$ 和 $x = -1$。(A1)

Second derivative (product rule on $g' = (1-x^{2})e^{-x^{2}/2}$): (M1)二阶导数（对 $g' = (1-x^{2})e^{-x^{2}/2}$ 应用乘积法则）：(M1)

Evaluate at each critical point:在每个极值点处计算：

• $g''(1) = e^{-1/2}(1)(1-3) = -2e^{-1/2} < 0$ $\Rightarrow$ local maximum. Value: $g(1) = e^{-1/2}$. (A1)$g''(1) = e^{-1/2}(1)(1-3) = -2e^{-1/2} < 0$ $\Rightarrow$ 局部极大值。函数值：$g(1) = e^{-1/2}$。(A1)
• $g''(-1) = e^{-1/2}(-1)(1-3) = 2e^{-1/2} > 0$ $\Rightarrow$ local minimum. Value: $g(-1) = -e^{-1/2}$.$g''(-1) = e^{-1/2}(-1)(1-3) = 2e^{-1/2} > 0$ $\Rightarrow$ 局部极小值。函数值：$g(-1) = -e^{-1/2}$。

(b) Inflection points: zeros of $g''$ with sign change(b) 拐点：$g''$ 的零点处的符号变化 M1·M1·A1·A1

Set $g''(x) = e^{-x^{2}/2}\,x(x^{2}-3) = 0$. Since the exponential is never zero, the candidates are $x = 0$ and $x = \pm\sqrt{3}$. (M1 for finding candidates)令 $g''(x) = e^{-x^{2}/2}\,x(x^{2}-3) = 0$。因指数函数不为零，候选点为 $x = 0$ 和 $x = \pm\sqrt{3}$。(M1 给求候选点)

Verify sign change: factor $x(x^{2}-3)$ changes sign at each of these (all simple roots), confirmed by: (M1 for sign verification)验证符号变化：因子 $x(x^{2}-3)$ 在每处均改变符号（均为单重零点），验证如下：(M1 给符号验证)

• $x = -2$: $(-2)(4-3) = -2 < 0$; $x = -1$: $(-1)(1-3) = 2 > 0$: sign changes at $x = -\sqrt{3}$. (A1)$x = -2$：$(-2)(4-3) = -2 < 0$；$x = -1$：$(-1)(1-3) = 2 > 0$：在 $x = -\sqrt{3}$ 处变号。(A1)
• $x = -0.5$: $(-0.5)(0.25-3) > 0$; $x = 0.5$: $(0.5)(0.25-3) < 0$: sign changes at $x = 0$. (A1)$x = -0.5$：$(-0.5)(0.25-3) > 0$；$x = 0.5$：$(0.5)(0.25-3) < 0$：在 $x = 0$ 处变号。(A1)
• $x = 1.5$: $(1.5)(2.25-3) < 0$; $x = 2$: $(2)(4-3) > 0$: sign changes at $x = \sqrt{3}$.$x = 1.5$：$(1.5)(2.25-3) < 0$；$x = 2$：$(2)(4-3) > 0$：在 $x = \sqrt{3}$ 处变号。

All three are genuine inflection points. Coordinates: $g(0) = 0$; $g(\sqrt{3}) = \sqrt{3}\,e^{-3/2}$; $g(-\sqrt{3}) = -\sqrt{3}\,e^{-3/2}$.三点均为真实拐点。坐标：$g(0) = 0$；$g(\sqrt{3}) = \sqrt{3}\,e^{-3/2}$；$g(-\sqrt{3}) = -\sqrt{3}\,e^{-3/2}$。

(a) EVT guarantee R1

$h$ is a polynomial, hence continuous everywhere. In particular it is continuous on the closed bounded interval $[-\tfrac{1}{2}, 4]$. The Extreme Value Theorem therefore guarantees that $h$ attains both an absolute maximum value and an absolute minimum value on this interval. (R1)

(b) Interior critical points M1·A1

Differentiate: (M1)

Setting $h'(x) = 0$ gives $x = 1$ and $x = 2$, both of which lie in the open interval $(-\tfrac{1}{2}, 4)$. (A1)

(c) Evaluate candidates and read off extrema M1·A1·A1

The full candidate list is $\{-\tfrac{1}{2},\, 1,\, 2,\, 4\}$. Evaluate $h$ at each: (M1)

The largest value is $28$, attained at the endpoint $x = 4$: this is the absolute maximum. (A1) The smallest value is $-\tfrac{25}{2}$, attained at the endpoint $x = -\tfrac{1}{2}$: this is the absolute minimum. (A1)

(a) Domain, intercepts, symmetry, and asymptotes M1·A1

Domain: The denominator vanishes at $x = \pm 1$, so the domain is $\{x \in \mathbb{R} : x \ne \pm 1\}$. (M1)

Symmetry: $f(-x) = \dfrac{(-x)^{2}-4}{(-x)^{2}-1} = \dfrac{x^{2}-4}{x^{2}-1} = f(x)$, so $f$ is an even function and its graph is symmetric about the $y$-axis.

Intercepts: $f(0) = \dfrac{-4}{-1} = 4$, giving the $y$-intercept $(0,4)$. For $x$-intercepts, set $x^{2} - 4 = 0$: $x = \pm 2$, giving intercepts $(\pm 2, 0)$.

Asymptotes: Since $x^{2}-1 \to 0$ as $x \to \pm 1$ while the numerator is non-zero there, $x = 1$ and $x = -1$ are vertical asymptotes. For the horizontal asymptote,

so $y = 1$ is a horizontal asymptote. (A1 for asymptotes with supporting limits)

(b) $f'$, monotonicity, and local extrema M1·A1

By the quotient rule: (M1)

The denominator $(x^{2}-1)^{2}$ is strictly positive on the domain. The sign of $f'$ is therefore the sign of $6x$. Sign chart (including $x=\pm 1$ as partition points due to asymptotes):

• $x < -1$: $6x < 0$, $f$ is decreasing.
• $-1 < x < 0$: $6x < 0$, $f$ is decreasing.
• $0 < x < 1$: $6x > 0$, $f$ is increasing.
• $x > 1$: $6x > 0$, $f$ is increasing.

The only zero of $f'$ in the domain is $x = 0$, where $f'$ changes from $(-)$ to $(+)$: by the First Derivative Test, $x = 0$ is a local minimum with $f(0) = 4$. (A1)

(c) $f''$, concavity, and inflection points M1·A1

Differentiate $f'(x) = 6x(x^{2}-1)^{-2}$ using the product rule: (M1)

Since $3x^{2}+1 > 0$ always, the sign of $f''$ is determined by $-1/(x^{2}-1)^{3}$:

• $|x| > 1$: $(x^{2}-1)^{3} > 0$, so $f'' < 0$: concave down on each of $(-\infty,-1)$ and $(1,\infty)$.
• $|x| < 1$: $(x^{2}-1)^{3} < 0$, so $f'' > 0$: concave up on $(-1, 1)$.

There are no points in the domain where $f''=0$ (the equation $3x^{2}+1=0$ has no real solutions), so there are no inflection points. The concavity changes at the asymptotes $x=\pm 1$, not at any point on the graph. (A1)

Summary of key features: $y$-intercept $(0,4)$; $x$-intercepts $(\pm 2, 0)$; local minimum $(0,4)$; vertical asymptotes $x=\pm 1$; horizontal asymptote $y=1$; concave down on $(-\infty,-1)$ and $(1,\infty)$; concave up on $(-1,1)$; no inflection points.

(a) Statement of Rolle's Theorem R1·R1

Rolle's Theorem. Suppose $f$ satisfies the following three hypotheses: (i) $f$ is continuous on the closed interval $[a,b]$; (ii) $f$ is differentiable on the open interval $(a,b)$; (iii) $f(a) = f(b)$. Then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. (R1 for all three hypotheses, R1 for the conclusion)

(b) Application to $p(x)= x^{3} - x$ on $[-1,1]$ R1·M1·A1

Verify the three hypotheses on $[a,b] = [-1,1]$: (R1)

• (i) $p$ is a polynomial, hence continuous everywhere, so continuous on $[-1,1]$.
• (ii) $p$ is a polynomial, hence differentiable everywhere, so differentiable on $(-1,1)$.
• (iii) $p(-1) = -1 + 1 = 0$ and $p(1) = 1 - 1 = 0$, so $p(-1) = p(1)$.

All three hold. By Rolle's Theorem, there exists $c \in (-1,1)$ with $p'(c) = 0$. Set $p'(x) = 3x^{2} - 1 = 0$: (M1)

Both values lie in $(-1,1)$, so the theorem guarantees two values of $c$ in this instance. (A1)

(c) $q(x) = x^{5} + 3x + 1$ has at most one real root R1·M1·A1

Suppose, for contradiction, that $q$ has two distinct real roots $a < b$, so $q(a) = q(b) = 0$. (R1 for setting up contradiction)

Since $q$ is a polynomial it is continuous on $[a,b]$ and differentiable on $(a,b)$, and $q(a) = q(b) = 0$. By Rolle's Theorem, there exists $c \in (a,b)$ such that $q'(c) = 0$. (M1)

But $q'(x) = 5x^{4} + 3$. Since $5x^{4} \ge 0$ for all $x$, we have $q'(x) \ge 3 > 0$ for all $x \in \mathbb{R}$. In particular $q'(c) \ge 3 > 0$, contradicting $q'(c) = 0$. (A1)

The assumption that $q$ has two distinct real roots is false. Therefore $q$ has at most one real root.

(a) Statement of the Mean Value Theorem R1·R1

Mean Value Theorem. Suppose $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists at least one $c \in (a,b)$ such that

(R1 for both hypotheses, R1 for the conclusion)

(b) MVT applied to $\ln x$ on $[1, e]$ R1·M1·A1

Verify hypotheses: $f(x) = \ln x$ is continuous on $[1,e]$ (it is continuous for all $x>0$) and differentiable on $(1,e)$, with $f'(x) = 1/x$. Both hypotheses hold. (R1)

The MVT guarantees $c \in (1,e)$ with (M1)

Setting $1/c = 1/(e-1)$ gives $c = e - 1 \approx 1.718$. Since $1 < e-1 < e$, this $c$ lies in $(1,e)$ as required. (A1)

(c) Proving $|\sin b - \sin a| \le |b-a|$ M1·R1·A1

If $a = b$ then both sides are $0$ and the inequality holds trivially. Suppose $a \ne b$; without loss of generality assume $a < b$. Apply the MVT to $f(x) = \sin x$ on $[a,b]$: (M1) since $\sin$ is continuous and differentiable everywhere, there exists $c \in (a,b)$ with

The key property is that $|\cos c| \le 1$ for every $c \in \mathbb{R}$. (R1 for naming this) Therefore

(A1) This holds for all $a,b \in \mathbb{R}$.

(d) Upper bound for $f(7)$ M1·A1

Apply the MVT to $f$ on the interval $[2, 7]$. The hypothesis that $f$ is differentiable on $\mathbb{R}$ implies it is continuous on $[2,7]$ and differentiable on $(2,7)$, so the MVT applies. There exists $c \in (2,7)$ with: (M1)

Since $f'(c) \le 3$, we get $\dfrac{f(7)-5}{5} \le 3$, so $f(7) - 5 \le 15$, giving $f(7) \le 20$. (A1) The bound $f(7) = 20$ is attained when $f(x) = 3x - 1$.

(a) Constant function theorem R1·M1·A1·A1

Let $x_{1}, x_{2} \in I$ with $x_{1} < x_{2}$. We must show $f(x_{1}) = f(x_{2})$.

Because $f'(x) = 0$ on all of $I$, the function $f$ is differentiable on $(x_{1}, x_{2}) \subset I$ and hence continuous on $[x_{1}, x_{2}]$. The MVT therefore applies on $[x_{1}, x_{2}]$: there exists $c \in (x_{1}, x_{2})$ such that (M1)

But $f'(c) = 0$ by hypothesis (since $c \in I$), so $f(x_{2}) - f(x_{1}) = 0$, i.e.\ $f(x_{2}) = f(x_{1})$. (A1 for applying MVT, A1 for conclusion)

Since $x_{1}$ and $x_{2}$ were arbitrary points of $I$, $f$ takes the same value at every pair of points: $f$ is constant on $I$. (R1 for the quantifier argument)

(b) Antiderivative uniqueness M1·A1

Define $F = f - g$. Then $F' = f' - g' = 0$ on $I$ (since $f'=g'$ throughout $I$). (M1) By part (a), $F$ is constant on $I$: there is a constant $C$ such that $F(x) = C$ for all $x \in I$. That is, $f(x) - g(x) = C$, equivalently $f(x) = g(x) + C$ for all $x \in I$. (A1)

(c) The inequality $e^{x} \ge 1 + x$ for $x \ge 0$ M1·A1

For $x = 0$: $e^{0} = 1 = 1 + 0$, so equality holds and the inequality is satisfied.

For $x > 0$: apply the MVT to $F(t) = e^{t}$ on $[0, x]$. Since $e^{t}$ is differentiable everywhere, the hypotheses are met. There exists $c \in (0, x)$ such that: (M1)

Since $c > 0$ we have $e^{c} > e^{0} = 1$, so $xe^{c} > x$ (using $x > 0$). Therefore $e^{x} - 1 > x$, i.e.\ $e^{x} > 1 + x$. (A1)

Combining both cases, $e^{x} \ge 1 + x$ for all $x \ge 0$, with equality only at $x = 0$.

(a) Constraint, objective function, and feasible domain M1·M1·A1

Name the variables: let $x$ be the length of the sides perpendicular to the internal dividers and $y$ be the length of the sides parallel to them (i.e.\ the sides the dividers run along). The outer perimeter uses two sides of length $x$ and two sides of length $y$. The two internal dividers each have length $x$. Total fencing: (M1)

Objective: total area $A = xy$. Substituting: (M1)

Feasible domain: $x > 0$ and $y = 60-2x > 0$ require $0 < x < 30$. (A1)

(b) Critical point and Second Derivative Test M1·A1·A1

Differentiate: (M1)

$A''(x) = -4 < 0$ for all $x$, confirming $x=15$ is a maximum by the Second Derivative Test. (A1 for critical point, A1 for SDT confirmation)

(c) Optimal dimensions and maximum area M1·A1

(M1) $y = 60 - 2(15) = 30\,\text{m}$. Maximum area: $A(15) = 15 \times 30 = 450\,\text{m}^{2}$. (A1)

The entire enclosure measures $15\,\text{m}$ (perpendicular to the dividers) by $30\,\text{m}$ (parallel to the dividers).

(a) Constraint and expressing $h$ in terms of $x$ M1·A1

Volume constraint: $x^{2}h = 32$, so $h = \dfrac{32}{x^{2}}$. (M1) Feasible domain: $x > 0$ (no upper bound; as $x \to \infty$, $h \to 0$, which is physically fine). (A1)

(b) Total cost function M1·A1

The base has area $x^{2}$ at $\$3/\text{cm}^{2}$: base cost $= 3x^{2}$. Each of the four sides has area $xh$: total side area $= 4xh$ at $\$1/\text{cm}^{2}$, so side cost $= 4xh$. Substituting $h = 32/x^{2}$: (M1)

(A1)

(c) Critical point and confirmation of minimum M1·M1·A1·A1

Differentiate: (M1)

Set $C'(x) = 0$: $6x = \dfrac{128}{x^{2}}$, so $6x^{3} = 128$, giving $x^{3} = \dfrac{64}{3}$, hence (M1)

(A1) Confirm minimum: $C''(x) = 6 + \dfrac{256}{x^{3}} > 0$ for all $x > 0$. (A1 for SDT) The cost function is convex everywhere on $(0,\infty)$, so the single critical point is the global minimum. (As $x \to 0^{+}$, the term $128/x \to \infty$, and as $x \to \infty$, $3x^{2} \to \infty$, confirming no finite minimum elsewhere.)

(d) Dimensions and minimum cost M1·A1

(M1) With $x = 4 \cdot 3^{-1/3}$, we have $x^{2} = 16 \cdot 3^{-2/3}$, so

Minimum cost:

Numerically: $48/3^{2/3} \approx 48/2.080 \approx 23.08$ and $32 \times 3^{1/3} \approx 32 \times 1.442 \approx 46.15$. So minimum cost $\approx \$69.23$. (A1)

(a) Squared distance and equivalence R1·A1

A point on $y = \sqrt{x}$ has coordinates $(x, \sqrt{x})$ for $x \ge 0$. The squared distance from this point to $(2,0)$ is: (R1 for setting up $D(x)$)

Minimising $D(x)$ is equivalent to minimising the actual distance $\sqrt{D(x)}$, since the square root is an increasing function: the minimiser of $D$ is also the minimiser of $\sqrt{D}$. Feasible domain: $x \ge 0$. (A1)

(b) Critical points of $D$ M1·M1·A1·A1

Differentiate: (M1)

Set $D'(x) = 0$: $2x - 3 = 0 \implies x = \dfrac{3}{2}$. (M1) The value $x = \tfrac{3}{2}$ lies in the interior of the feasible domain $(x \ge 0)$. (A1 for finding $x=3/2$)

Note: the factored-polynomial step mentioned in the question is needed for more complex squared-distance expressions (e.g.\ $y = x^{2}$). Here the derivative is already linear in $x$. (A1 for recognising the domain location)

(c) Confirm minimum and compare with boundary M1·A1

$D''(x) = 2 > 0$ for all $x$, so $x = \tfrac{3}{2}$ is a local minimum by the Second Derivative Test. (M1) Compare with the boundary value at $x = 0$:

Since $\tfrac{7}{4} < 4$, the global minimum of $D$ on $[0,\infty)$ occurs at $x = \tfrac{3}{2}$. (A1) As $x \to \infty$, $(x-2)^{2} + x \to \infty$, so the minimum is not at infinity.

(d) Closest point and minimum distance M1·A1

(M1) At $x = \tfrac{3}{2}$: $y = \sqrt{\tfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$. The closest point on $y = \sqrt{x}$ to $(2,0)$ is $\left(\dfrac{3}{2},\, \dfrac{\sqrt{6}}{2}\right)$.

Minimum distance: $\sqrt{D\!\left(\tfrac{3}{2}\right)} = \sqrt{\tfrac{7}{4}} = \dfrac{\sqrt{7}}{2}$. (A1)