Unit A7: Curve Sketching and Optimization单元 A7：曲线描绘与最优化

Find the critical points of $f(x) = x^{3} - 3x^{2} - 9x + 5$.

Differentiate and factor.

Since $f'$ is a polynomial it exists everywhere, so the only critical points come from $f'(x) = 0$, giving $x = 3$ and $x = -1$. These two numbers are the complete candidate list for the local extrema of $f$.

求 $f(x) = x^{3} - 3x^{2} - 9x + 5$ 的临界点。

求导并因式分解。

由于 $f'$ 是多项式，处处存在，所以临界点只能来自 $f'(x) = 0$，即 $x = 3$ 与 $x = -1$。这两个数就是 $f$ 局部极值的完整候选名单。

Find the critical points of $f(x) = x^{2/3}$.

This is never zero, but it fails to exist at $x = 0$, where $f$ is still defined. Hence $x = 0$ is a critical point. Inspecting $f$ shows it has an absolute minimum there: this candidate would be invisible to anyone who only solved $f'(x) = 0$.

求 $f(x) = x^{2/3}$ 的临界点。

它永不为零，但在 $x = 0$ 处不存在，而 $f$ 在该点仍有定义。因此 $x = 0$ 是临界点。考察 $f$ 可知它在此取得绝对最小值：只解 $f'(x) = 0$ 的人会完全漏掉这个候选。

Find the critical points of $f(x) = x - 2\sin x$ on the interval $[0, 2\pi]$.

The function is differentiable everywhere, so every critical point comes from $f'(x) = 0$.

On $[0,2\pi]$ the equation $\cos x = \tfrac12$ has the two solutions $x = \tfrac{\pi}{3}$ and $x = \tfrac{5\pi}{3}$. These are the only interior critical points. When the problem is later treated as an extremum question on the closed interval, the full candidate list will be $\{0,\ \tfrac{\pi}{3},\ \tfrac{5\pi}{3},\ 2\pi\}$: the two critical points plus the two endpoints. The lesson is that solving the trigonometric equation must respect the stated domain, since $\cos x = \tfrac12$ has infinitely many solutions on the whole line.

在区间 $[0, 2\pi]$ 上求 $f(x) = x - 2\sin x$ 的临界点。

该函数处处可微（differentiable），所以每个临界点都来自 $f'(x) = 0$。

在 $[0,2\pi]$ 上，方程 $\cos x = \tfrac12$ 有两个解 $x = \tfrac{\pi}{3}$ 与 $x = \tfrac{5\pi}{3}$，这是仅有的内部临界点。当此题后续作为闭区间上的极值问题处理时，完整的候选名单为 $\{0,\ \tfrac{\pi}{3},\ \tfrac{5\pi}{3},\ 2\pi\}$：两个临界点加两个端点。要点在于解三角方程必须尊重题目给定的定义域，因为 $\cos x = \tfrac12$ 在整条数轴上有无穷多个解。

Show that $f(x) = x$ on the open interval $(0,1)$ has neither an absolute maximum nor an absolute minimum, even though $f$ is continuous and bounded.

Suppose, for contradiction, that $f$ attained a maximum at some $c \in (0,1)$. Then $f(c) = c$, yet the number $\tfrac{c+1}{2}$ also lies in $(0,1)$ and satisfies $\tfrac{c+1}{2} > c$, so $f\big(\tfrac{c+1}{2}\big) > f(c)$, contradicting maximality. The symmetric argument with $\tfrac{c}{2}$ rules out a minimum. The values approach $1$ and $0$ but are never reached. This is exactly why the Extreme Value Theorem insists on a closed interval: the supremum and infimum exist as numbers, but without the endpoints there is no point at which they are attained.

证明 $f(x) = x$ 在开区间 $(0,1)$ 上既无绝对最大值也无绝对最小值，尽管 $f$ 连续且有界。

反证：设 $f$ 在某点 $c \in (0,1)$ 取得最大值。则 $f(c) = c$，但数 $\tfrac{c+1}{2}$ 也属于 $(0,1)$ 且满足 $\tfrac{c+1}{2} > c$，于是 $f\big(\tfrac{c+1}{2}\big) > f(c)$，与最大性矛盾。用 $\tfrac{c}{2}$ 作对称论证可排除最小值。函数值趋于 $1$ 和 $0$，却永远取不到。这正是极值定理坚持要求闭区间的原因：上确界与下确界作为数是存在的，但缺了端点就没有任何一点真正取到它们。

Find every $c$ guaranteed by the Mean Value Theorem for $f(x) = x^{3}$ on $[0,2]$.

The average rate of change is

Set $f'(c) = 3c^{2}$ equal to $4$: $3c^{2} = 4$, so $c^{2} = 4/3$ and $c = 2/\sqrt{3} \approx 1.155$. The negative root is rejected because it lies outside $(0,2)$.

对 $f(x) = x^{3}$ 在 $[0,2]$ 上，求出中值定理所保证的所有 $c$。

平均变化率为

令 $f'(c) = 3c^{2}$ 等于 $4$：$3c^{2} = 4$，故 $c^{2} = 4/3$，$c = 2/\sqrt{3} \approx 1.155$。负根因落在 $(0,2)$ 之外而舍去。

Given $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, define the auxiliary function that subtracts the secant line.

Then $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ as a difference of such functions. Direct substitution gives $g(a) = 0$ and $g(b) = 0$. By Rolle's Theorem there is a $c \in (a,b)$ with $g'(c) = 0$. Differentiating,

This is precisely the Mean Value Theorem, so the general theorem reduces to the special case $f(a)=f(b)$.

设 $f$ 在 $[a,b]$ 上连续、在 $(a,b)$ 上可微，定义减去割线的辅助函数。

作为这类函数之差，$g$ 在 $[a,b]$ 上连续、在 $(a,b)$ 上可微。直接代入得 $g(a) = 0$ 与 $g(b) = 0$。由罗尔定理，存在 $c \in (a,b)$ 使 $g'(c) = 0$。求导得

这正是中值定理，于是一般情形归结为特例 $f(a)=f(b)$。

Suppose $f$ is differentiable on $\mathbb{R}$ with $f(0) = 3$ and $|f'(x)| \le 2$ for every $x$. How large can $f(5)$ possibly be?

Apply the Mean Value Theorem on $[0,5]$: there is a $c \in (0,5)$ with

Therefore $|f(5) - 3| = 5\,|f'(c)| \le 5 \cdot 2 = 10$, which gives $-7 \le f(5) \le 13$. The largest possible value is $f(5) = 13$, attained when $f'(c) = 2$ throughout, that is when $f(x) = 2x + 3$. This is the model argument for turning a derivative bound into a bound on the function itself, and it underlies error estimates throughout calculus.

设 $f$ 在 $\mathbb{R}$ 上可微，$f(0) = 3$，且对每个 $x$ 有 $|f'(x)| \le 2$。$f(5)$ 最大可能是多少？

在 $[0,5]$ 上用中值定理：存在 $c \in (0,5)$ 使

因此 $|f(5) - 3| = 5\,|f'(c)| \le 5 \cdot 2 = 10$，即 $-7 \le f(5) \le 13$。最大可能值为 $f(5) = 13$，当 $f'(c)$ 全程等于 $2$（即 $f(x) = 2x + 3$）时取得。这是把导数估界转化为函数本身估界的范本论证，也是整个微积分中误差估计的基础。

Prove that $\sin b - \sin a \le b - a$ for all $a \le b$.

If $a = b$ both sides are zero. For $a < b$, apply the Mean Value Theorem to $f(x) = \sin x$ on $[a,b]$: there is a $c \in (a,b)$ with

Since $\cos c \le 1$ and $b - a > 0$, multiplying gives $\sin b - \sin a \le b - a$. The same argument with $|\cos c| \le 1$ yields the sharper statement $|\sin b - \sin a| \le |b - a|$, which says that sine is a nonexpansive (Lipschitz) function with constant $1$.

证明：对一切 $a \le b$ 有 $\sin b - \sin a \le b - a$。

若 $a = b$，两边均为零。若 $a < b$，对 $f(x) = \sin x$ 在 $[a,b]$ 上用中值定理：存在 $c \in (a,b)$ 使

由 $\cos c \le 1$ 且 $b - a > 0$，两边相乘得 $\sin b - \sin a \le b - a$。用 $|\cos c| \le 1$ 作同样论证可得更强的 $|\sin b - \sin a| \le |b - a|$，即正弦是常数为 $1$ 的非扩张（利普希茨，Lipschitz）函数。

Classify the critical points of $f(x) = x^{3} - 3x^{2} - 9x + 5$ from Example 1.1, where $f'(x) = 3(x-3)(x+1)$.

The critical points $x = -1$ and $x = 3$ split the line into three intervals. Test one point in each.

So $f$ increases on $(-\infty,-1)$, decreases on $(-1,3)$, and increases on $(3,\infty)$. The sign goes $(+)\to(-)$ at $x=-1$, giving a local maximum, and $(-)\to(+)$ at $x=3$, giving a local minimum.

对例题 1.1 中的 $f(x) = x^{3} - 3x^{2} - 9x + 5$（其中 $f'(x) = 3(x-3)(x+1)$）的临界点进行分类。

临界点 $x = -1$ 与 $x = 3$ 把数轴分成三个区间。每个区间各取一点检验。

于是 $f$ 在 $(-\infty,-1)$ 上递增，在 $(-1,3)$ 上递减，在 $(3,\infty)$ 上递增。$x=-1$ 处符号 $(+)\to(-)$，给出局部最大值；$x=3$ 处 $(-)\to(+)$，给出局部最小值。

Classify the critical points of $f(x) = x^{2/3}(x - 5)$ and find the intervals of increase and decrease.

Write $f(x) = x^{5/3} - 5x^{2/3}$ and differentiate.

The derivative is zero at $x = 2$ and undefined at $x = 0$ (which is in the domain), so the critical points are $x = 0$ and $x = 2$. Test the sign of $f'$ on the three intervals, watching the sign of both the numerator $x-2$ and the denominator $x^{1/3}$:

So $f$ increases on $(-\infty,0)$, decreases on $(0,2)$, and increases on $(2,\infty)$. The sign change $(+)\to(-)$ at $x=0$ gives a local maximum (a cusp, since $f'$ is undefined there), and $(-)\to(+)$ at $x=2$ gives a local minimum. This shows why the sign chart must include points where $f'$ is undefined, not only its roots.

对 $f(x) = x^{2/3}(x - 5)$ 的临界点分类，并求出递增、递减区间。

写成 $f(x) = x^{5/3} - 5x^{2/3}$ 再求导。

导数在 $x = 2$ 处为零，在 $x = 0$ 处不存在（而 $0$ 在定义域内），故临界点为 $x = 0$ 与 $x = 2$。在三个区间上检验 $f'$ 的符号，同时关注分子 $x-2$ 与分母 $x^{1/3}$ 的符号：

故 $f$ 在 $(-\infty,0)$ 上递增，在 $(0,2)$ 上递减，在 $(2,\infty)$ 上递增。$x=0$ 处符号变化 $(+)\to(-)$ 给出局部最大值（一个尖点，因为 $f'$ 在此不存在），$x=2$ 处 $(-)\to(+)$ 给出局部最小值。这说明符号表必须包含 $f'$ 不存在的点，而不只是它的根。

Find where $f(x) = x e^{-x}$ is increasing or decreasing, and classify any local extremum.

By the product rule,

Because $e^{-x} > 0$ for every $x$, the sign of $f'$ matches the sign of $1 - x$. Hence $f' > 0$ on $(-\infty,1)$ and $f' < 0$ on $(1,\infty)$: the function increases then decreases. The sign change $(+)\to(-)$ at $x = 1$ gives a local maximum, with value $f(1) = e^{-1} \approx 0.368$. Since this is the only critical point and the function rises to it then falls forever after, it is in fact the absolute maximum on the whole line.

求 $f(x) = x e^{-x}$ 的递增、递减区间，并对任何局部极值进行分类。

由乘积法则（Product Rule），

由于对每个 $x$ 都有 $e^{-x} > 0$，$f'$ 的符号与 $1 - x$ 一致。故 $f'$ 在 $(-\infty,1)$ 上为正、在 $(1,\infty)$ 上为负：函数先增后减。$x = 1$ 处符号变化 $(+)\to(-)$ 给出局部最大值，其值 $f(1) = e^{-1} \approx 0.368$。由于这是唯一的临界点，且函数先升到它、此后永远下降，它实际上是整条数轴上的绝对最大值。

For $f(x) = x^{3} - 3x^{2} - 9x + 5$, analyze concavity. From $f'(x) = 3x^{2}-6x-9$,

Thus $f'' < 0$ for $x < 1$ (concave down) and $f'' > 0$ for $x > 1$ (concave up). The concavity reverses at $x = 1$, where $f(1) = 1 - 3 - 9 + 5 = -6$, so $(1,-6)$ is an inflection point.

Checking the earlier critical points with the Second Derivative Test: $f''(-1) = -12 < 0$ confirms the local maximum, and $f''(3) = 12 > 0$ confirms the local minimum.

对 $f(x) = x^{3} - 3x^{2} - 9x + 5$ 分析凹凸性。由 $f'(x) = 3x^{2}-6x-9$，

故 $x < 1$ 时 $f'' < 0$（凹向下），$x > 1$ 时 $f'' > 0$（凹向上）。凹凸性在 $x = 1$ 处反转，此处 $f(1) = 1 - 3 - 9 + 5 = -6$，所以 $(1,-6)$ 是拐点。

用二阶导数判别法核验先前的临界点：$f''(-1) = -12 < 0$ 证实局部最大值，$f''(3) = 12 > 0$ 证实局部最小值。

Suppose $f'' > 0$ on an interval $I$. By the Increasing/Decreasing Test applied to the function $f'$, the fact that $(f')' = f'' > 0$ forces $f'$ to be increasing on $I$. By definition, an increasing first derivative is exactly what concave up means: as $x$ grows, the tangent slope grows, so the curve turns upward and lies above each of its tangent lines on $I$. The concave down case is identical with inequalities reversed.

设在区间 $I$ 上 $f'' > 0$。对函数 $f'$ 应用单调性判别法，$(f')' = f'' > 0$ 迫使 $f'$ 在 $I$ 上递增。按定义，一阶导数递增正是凹向上的含义：随着 $x$ 增大，切线斜率增大，于是曲线向上弯，并在 $I$ 上位于其每条切线之上。凹向下的情形完全相同，只需把不等号反向。

Examine $f(x) = x^4$ at the origin. Compute $f'(x) = 4x^3$ and $f''(x) = 12x^2$. Then $f''(0) = 0$, so one might suspect an inflection point. But $f''(x) = 12x^2 \ge 0$ for all $x$, so the concavity is up on both sides of $0$ and never reverses. There is no inflection point; the origin is in fact the absolute minimum. The lesson is that $f''(c) = 0$ is necessary but not sufficient for an inflection: the second derivative must actually change sign as $x$ crosses $c$.

考察 $f(x) = x^4$ 在原点。算得 $f'(x) = 4x^3$，$f''(x) = 12x^2$。则 $f''(0) = 0$，乍看像是拐点。但对一切 $x$ 都有 $f''(x) = 12x^2 \ge 0$，所以 $0$ 两侧都凹向上，从不反转。这里没有拐点；原点其实是绝对最小值。要点是 $f''(c) = 0$ 对拐点是必要而非充分条件：当 $x$ 跨过 $c$ 时，二阶导数必须真正改变符号。

Classify the critical points of $f(x) = 2\cos x + \cos 2x$ on $(0, 2\pi)$ using the Second Derivative Test, and locate the inflection points.

Differentiate twice.

Setting $f'(x) = 0$ gives $\sin x = 0$ or $\cos x = -\tfrac12$. On $(0,2\pi)$ this yields $x = \pi$ (from $\sin x = 0$) and $x = \tfrac{2\pi}{3},\ \tfrac{4\pi}{3}$ (from $\cos x = -\tfrac12$). Evaluate $f''$:

So the test cleanly classifies all three critical points, illustrating how much faster the Second Derivative Test can be than a full sign chart when $f''$ is easy to evaluate.

用二阶导数判别法对 $f(x) = 2\cos x + \cos 2x$ 在 $(0, 2\pi)$ 上的临界点分类，并找出拐点。

求两次导。

令 $f'(x) = 0$ 得 $\sin x = 0$ 或 $\cos x = -\tfrac12$。在 $(0,2\pi)$ 上由 $\sin x = 0$ 得 $x = \pi$，由 $\cos x = -\tfrac12$ 得 $x = \tfrac{2\pi}{3},\ \tfrac{4\pi}{3}$。代入 $f''$：

于是该判别法干净利落地分类了全部三个临界点，说明当 $f''$ 容易求值时，二阶导数判别法比画完整符号表快得多。

Sketch $f(x) = \dfrac{x^{2}}{x^{2} - 1}$.

Domain and symmetry. Defined for $x \ne \pm 1$. Since $f(-x) = f(x)$, the graph is even and symmetric about the $y$-axis.

Intercepts. $f(0) = 0$, so the only intercept is the origin.

Asymptotes. The denominator vanishes at $x = \pm 1$, giving vertical asymptotes there. For the end behavior,

so $y = 1$ is a horizontal asymptote.

Monotonicity. By the quotient rule,

The denominator is positive, so $f' > 0$ for $x < 0$ and $f' < 0$ for $x > 0$ (excluding the asymptotes). Thus $f$ increases on $(-\infty,-1)$ and $(-1,0)$ and decreases on $(0,1)$ and $(1,\infty)$, with a local maximum at the origin where $f(0)=0$. Combining these facts produces the characteristic three-branch graph hugging $y = 1$.

描绘 $f(x) = \dfrac{x^{2}}{x^{2} - 1}$。

定义域与对称性。在 $x \ne \pm 1$ 时有定义。由 $f(-x) = f(x)$ 知图像为偶函数，关于 $y$ 轴对称。

截距。$f(0) = 0$，故唯一的截距是原点。

渐近线。分母在 $x = \pm 1$ 处为零，给出该处的竖直渐近线。对于端点行为，

所以 $y = 1$ 是水平渐近线。

单调性。由商的法则（Quotient Rule），

分母为正，故 $x < 0$ 时 $f' > 0$，$x > 0$ 时 $f' < 0$（不含渐近线处）。于是 $f$ 在 $(-\infty,-1)$ 与 $(-1,0)$ 上递增，在 $(0,1)$ 与 $(1,\infty)$ 上递减，在原点处 $f(0)=0$ 取局部最大值。综合这些事实即得贴着 $y = 1$ 的特征性三支图像。

Sketch $f(x) = \dfrac{x^{2} + 1}{x}$ and identify all asymptotes.

Domain and symmetry. Defined for $x \ne 0$. Since $f(-x) = -f(x)$, the graph is odd and symmetric about the origin.

Asymptotes. Long division gives $f(x) = x + \dfrac{1}{x}$. As $x \to 0$ the term $1/x$ dominates, so $x = 0$ is a vertical asymptote. As $x \to \pm\infty$ the term $1/x \to 0$, so the graph approaches the line $y = x$. This line is a slant (oblique) asymptote, which appears whenever the degree of the numerator is exactly one more than that of the denominator.

Monotonicity and shape. From $f'(x) = 1 - \dfrac{1}{x^{2}} = \dfrac{x^{2}-1}{x^{2}}$, the critical points are $x = \pm 1$. The derivative is positive for $|x| > 1$ and negative for $0 < |x| < 1$, so $x = -1$ is a local maximum with $f(-1) = -2$ and $x = 1$ is a local minimum with $f(1) = 2$. Notice the local minimum value $2$ exceeds the local maximum value $-2$; this is possible only because a vertical asymptote separates the two branches.

描绘 $f(x) = \dfrac{x^{2} + 1}{x}$ 并找出所有渐近线。

定义域与对称性。在 $x \ne 0$ 时有定义。由 $f(-x) = -f(x)$ 知图像为奇函数，关于原点对称。

渐近线。多项式除法给出 $f(x) = x + \dfrac{1}{x}$。当 $x \to 0$ 时 $1/x$ 项占主导，所以 $x = 0$ 是竖直渐近线。当 $x \to \pm\infty$ 时 $1/x \to 0$，所以图像趋近直线 $y = x$。这条直线是斜渐近线（slant asymptote），只要分子次数恰好比分母次数高一次就会出现。

单调性与形状。由 $f'(x) = 1 - \dfrac{1}{x^{2}} = \dfrac{x^{2}-1}{x^{2}}$，临界点为 $x = \pm 1$。导数在 $|x| > 1$ 时为正、在 $0 < |x| < 1$ 时为负，故 $x = -1$ 是局部最大值 $f(-1) = -2$，$x = 1$ 是局部最小值 $f(1) = 2$。注意局部最小值 $2$ 竟大于局部最大值 $-2$；这只有在竖直渐近线把两支分开时才可能发生。

Assemble the graph of $f(x) = x^{3} - 3x^{2} - 9x + 5$, the function studied throughout this unit.

Collecting the earlier results: $f$ increases on $(-\infty,-1)$, decreases on $(-1,3)$, and increases on $(3,\infty)$, with a local maximum at $(-1, 10)$ and a local minimum at $(3, -22)$. The concavity is down on $(-\infty,1)$ and up on $(1,\infty)$, with an inflection point at $(1,-6)$. The end behavior is $f \to -\infty$ as $x \to -\infty$ and $f \to +\infty$ as $x \to +\infty$, with no asymptotes since $f$ is a polynomial.

Plotting these features in order, rising to the crest at $(-1,10)$ while concave down, bending through the inflection at $(1,-6)$, sinking to the trough at $(3,-22)$ while concave up, then rising again, reproduces the standard cubic with one hump and one valley. The independent pieces, monotonicity from $f'$ and bending from $f''$, lock together into a single unambiguous shape.

拼出本单元始终研究的函数 $f(x) = x^{3} - 3x^{2} - 9x + 5$ 的图像。

汇总前面的结果：$f$ 在 $(-\infty,-1)$ 上递增，在 $(-1,3)$ 上递减，在 $(3,\infty)$ 上递增，在 $(-1, 10)$ 取局部最大值、在 $(3, -22)$ 取局部最小值。凹凸性在 $(-\infty,1)$ 上为凹向下、在 $(1,\infty)$ 上为凹向上，拐点在 $(1,-6)$。端点行为是 $x \to -\infty$ 时 $f \to -\infty$、$x \to +\infty$ 时 $f \to +\infty$，因 $f$ 是多项式而无渐近线。

按顺序画出这些特征：凹向下地升到 $(-1,10)$ 的峰顶，经 $(1,-6)$ 处拐点弯转，凹向上地降到 $(3,-22)$ 的谷底，再次上升，就重现了一峰一谷的标准三次曲线。两块独立的信息——来自 $f'$ 的单调性与来自 $f''$ 的弯曲——咬合成唯一确定的形状。

Find the absolute maximum and minimum of $f(x) = x^{3} - 3x + 1$ on $[0, 3]$.

Differentiate: $f'(x) = 3x^{2} - 3 = 3(x-1)(x+1)$. The critical points are $x = \pm 1$, but only $x = 1$ lies in $(0,3)$, so the candidate list is $\{0, 1, 3\}$.

The absolute maximum is $19$, attained at $x = 3$ (an endpoint), and the absolute minimum is $-1$, attained at the interior critical point $x = 1$.

求 $f(x) = x^{3} - 3x + 1$ 在 $[0, 3]$ 上的绝对最大值与最小值。

求导：$f'(x) = 3x^{2} - 3 = 3(x-1)(x+1)$。临界点为 $x = \pm 1$，但只有 $x = 1$ 落在 $(0,3)$ 内，故候选名单为 $\{0, 1, 3\}$。

绝对最大值为 $19$，在端点 $x = 3$ 处取得；绝对最小值为 $-1$，在内部临界点 $x = 1$ 处取得。

Find the absolute extrema of $f(x) = x^{2/3}(5 - x)$ on $[-1, 4]$.

Write $f(x) = 5x^{2/3} - x^{5/3}$, so $f'(x) = \tfrac{10}{3}x^{-1/3} - \tfrac{5}{3}x^{2/3} = \dfrac{5}{3}\cdot\dfrac{2 - x}{x^{1/3}}$, which is zero at $x = 2$ and undefined at $x = 0$. Both lie in $(-1,4)$, so the candidate list is $\{-1,\ 0,\ 2,\ 4\}$: two endpoints and two critical points.

The absolute maximum is $6$ at the endpoint $x = -1$, and the absolute minimum is $0$ at the interior critical point $x = 0$ where the derivative is undefined. Had we tested only the zeros of $f'$, the true minimum would have been missed.

求 $f(x) = x^{2/3}(5 - x)$ 在 $[-1, 4]$ 上的绝对极值。

写成 $f(x) = 5x^{2/3} - x^{5/3}$，则 $f'(x) = \tfrac{10}{3}x^{-1/3} - \tfrac{5}{3}x^{2/3} = \dfrac{5}{3}\cdot\dfrac{2 - x}{x^{1/3}}$，它在 $x = 2$ 处为零、在 $x = 0$ 处不存在。两者都落在 $(-1,4)$ 内，故候选名单为 $\{-1,\ 0,\ 2,\ 4\}$：两个端点加两个临界点。

绝对最大值为 $6$，在端点 $x = -1$ 处；绝对最小值为 $0$，在导数不存在的内部临界点 $x = 0$ 处。若只检验 $f'$ 的零点，就会漏掉真正的最小值。

Find the absolute extrema of $f(x) = x - 2\sin x$ on $[0, 2\pi]$, continuing Example 1.3.

The interior critical points are $x = \tfrac{\pi}{3}$ and $x = \tfrac{5\pi}{3}$, so the candidate list is $\{0,\ \tfrac{\pi}{3},\ \tfrac{5\pi}{3},\ 2\pi\}$. Evaluate, using $\sin\tfrac{\pi}{3} = \tfrac{\sqrt3}{2}$ and $\sin\tfrac{5\pi}{3} = -\tfrac{\sqrt3}{2}$:

The absolute maximum is $\tfrac{5\pi}{3} + \sqrt3$ at the interior critical point $x = \tfrac{5\pi}{3}$, and the absolute minimum is $\tfrac{\pi}{3} - \sqrt3$ at $x = \tfrac{\pi}{3}$. Note the maximum is interior and exceeds both endpoint values, which is why endpoints alone never suffice.

承接例题 1.3，求 $f(x) = x - 2\sin x$ 在 $[0, 2\pi]$ 上的绝对极值。

内部临界点为 $x = \tfrac{\pi}{3}$ 与 $x = \tfrac{5\pi}{3}$，故候选名单为 $\{0,\ \tfrac{\pi}{3},\ \tfrac{5\pi}{3},\ 2\pi\}$。利用 $\sin\tfrac{\pi}{3} = \tfrac{\sqrt3}{2}$ 与 $\sin\tfrac{5\pi}{3} = -\tfrac{\sqrt3}{2}$ 求值：

绝对最大值为 $\tfrac{5\pi}{3} + \sqrt3$，在内部临界点 $x = \tfrac{5\pi}{3}$ 处；绝对最小值为 $\tfrac{\pi}{3} - \sqrt3$，在 $x = \tfrac{\pi}{3}$ 处。注意最大值在内部，且超过两个端点的值，这正是为何仅靠端点永远不够。

A farmer has $100$ meters of fencing to enclose a rectangular field. What dimensions maximize the area?

Let the rectangle have width $x$ and height $y$. The perimeter constraint is $2x + 2y = 100$, so $y = 50 - x$. The area to maximize is

Then $A'(x) = 50 - 2x$, which vanishes at $x = 25$. Comparing candidates, $A(0) = 0$, $A(50) = 0$, and $A(25) = 25 \cdot 25 = 625$. The maximum area is $625$ square meters, achieved by the square with $x = y = 25$.

一位农夫有 $100$ 米篱笆，要围出一块矩形田地。什么尺寸能使面积最大？

设矩形宽 $x$、高 $y$。周长约束为 $2x + 2y = 100$，故 $y = 50 - x$。要最大化的面积为

则 $A'(x) = 50 - 2x$，在 $x = 25$ 处为零。比较各候选：$A(0) = 0$、$A(50) = 0$、$A(25) = 25 \cdot 25 = 625$。最大面积为 $625$ 平方米，由 $x = y = 25$ 的正方形实现。

Minimize the surface area of a closed cylindrical can of fixed volume $V$. With radius $r$ and height $h$, the volume constraint is $V = \pi r^{2} h$, so $h = V/(\pi r^{2})$. The surface area is

Differentiate and set to zero.

Since $S''(r) = 4\pi + 4V/r^{3} > 0$ for all $r > 0$, this critical point is the absolute minimum. Substituting back, $h = V/(\pi r^{2}) = 2r$: the most economical can has height equal to its diameter.

使固定体积 $V$ 的密闭圆柱形罐头表面积最小。设半径 $r$、高 $h$，体积约束为 $V = \pi r^{2} h$，故 $h = V/(\pi r^{2})$。表面积为

求导并令其为零。

由于对一切 $r > 0$ 有 $S''(r) = 4\pi + 4V/r^{3} > 0$，这个临界点是绝对最小值。回代得 $h = V/(\pi r^{2}) = 2r$：最省料的罐头其高等于其直径。

Find the point on the parabola $y = x^{2}$ closest to the point $(0, \tfrac{3}{2})$.

A point on the curve is $(x, x^{2})$. Rather than minimize the distance, minimize the squared distance, which has the same minimizer and avoids the square root.

Differentiate and simplify.

The critical points are $x = 0$ and $x = \pm 1$. Since $D''(x) = 12x^2 - 4$, we have $D''(0) = -4 < 0$ (a local max of squared distance) and $D''(\pm 1) = 8 > 0$ (local minima). Comparing $D(0) = \tfrac{9}{4}$ with $D(\pm 1) = 1 + \tfrac14 = \tfrac54$, the minimum distance occurs at $x = \pm 1$. The two closest points are $(1,1)$ and $(-1,1)$, each at distance $\sqrt{5}/2$. Minimizing the square is a standard device worth remembering.

求抛物线 $y = x^{2}$ 上离点 $(0, \tfrac{3}{2})$ 最近的点。

曲线上的点为 $(x, x^{2})$。与其最小化距离，不如最小化距离的平方，二者最小值点相同，且可避开平方根。

求导并化简。

临界点为 $x = 0$ 与 $x = \pm 1$。由 $D''(x) = 12x^2 - 4$，有 $D''(0) = -4 < 0$（平方距离的局部最大值）与 $D''(\pm 1) = 8 > 0$（局部最小值）。比较 $D(0) = \tfrac{9}{4}$ 与 $D(\pm 1) = 1 + \tfrac14 = \tfrac54$，最短距离在 $x = \pm 1$ 处。两个最近点是 $(1,1)$ 与 $(-1,1)$，各距 $\sqrt{5}/2$。最小化平方是值得记住的标准手法。

An open-top box is made from a $12 \times 12$ square sheet by cutting a square of side $x$ from each corner and folding up the sides. Find $x$ that maximizes the volume.

After folding, the base measures $(12 - 2x)$ by $(12 - 2x)$ and the height is $x$, so

The physical domain is $[0,6]$ because the cut cannot exceed half the side. Differentiate using the product rule.

Setting $V'(x) = 0$ gives $x = 6$ or $x = 2$. The endpoints $x = 0$ and $x = 6$ both yield $V = 0$ (a flat sheet or a sealed-up box of zero base), so compare the interior candidate: $V(2) = 2(8)^{2} = 128$. The maximum volume is $128$ cubic units at $x = 2$. The boundary check is essential, since the algebra alone offers $x = 6$ as a spurious critical value sitting at the useless endpoint.

用一张 $12 \times 12$ 的方形纸板，在四角各剪去边长为 $x$ 的小方块再折起四边，做成一个无盖盒子。求使体积最大的 $x$。

折起后，底面为 $(12 - 2x)$ 乘 $(12 - 2x)$，高为 $x$，故

实际定义域是 $[0,6]$，因为剪口不能超过边长的一半。用乘积法则求导。

令 $V'(x) = 0$ 得 $x = 6$ 或 $x = 2$。端点 $x = 0$ 与 $x = 6$ 都给出 $V = 0$（一张平板，或底面为零的封死盒子），故比较内部候选：$V(2) = 2(8)^{2} = 128$。最大体积为 $128$ 立方单位，在 $x = 2$ 处取得。边界检查至关重要，因为单凭代数会给出 $x = 6$ 这个落在无用端点上的虚假临界值。

A classic application of optimization is to derive the law that governs how light bends. Suppose light travels from a point $A$ in one medium to a point $B$ in another, crossing a flat boundary, and that its speed is $v_1$ above the boundary and $v_2$ below. Fermat's principle of least time says the actual path is the one that minimizes travel time.

Place the boundary on the $x$-axis. Let the horizontal distance between $A$ and $B$ be $d$, with the crossing point at horizontal position $x$ measured from the foot of $A$. If $A$ is at height $a$ and $B$ at depth $b$, the total time is

Differentiate and set $T'(x) = 0$:

Now recognize the two fractions as sines of the angles each ray makes with the vertical (the normal to the boundary): if $\theta_1$ and $\theta_2$ are those angles of incidence and refraction, then $\sin\theta_1 = \dfrac{x}{\sqrt{a^{2}+x^{2}}}$ and $\sin\theta_2 = \dfrac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}$. The stationarity condition becomes

which is exactly Snell's law. Because $T(x) \to \infty$ as $x \to \pm\infty$ and $T$ is smooth with a single critical point, that critical point is the global minimum. A purely geometric law of physics falls out of a one-variable minimization, a striking illustration of why optimization is a centerpiece of first-year calculus.

最优化的一个经典应用是推导支配光线弯折的定律。设光线从一种介质中的点 $A$ 出发，穿过一条平直界面，到达另一种介质中的点 $B$，且其速率在界面上方为 $v_1$、下方为 $v_2$。费马的最短时间原理指出，真实路径是使传播时间最短的那条。

把界面放在 $x$ 轴上。设 $A$ 与 $B$ 的水平距离为 $d$，过界点的水平位置为 $x$（从 $A$ 的垂足起算）。若 $A$ 在高度 $a$、$B$ 在深度 $b$，则总时间为

求导并令 $T'(x) = 0$：

现把这两个分式认作每条光线与竖直方向（界面法线）夹角的正弦：若入射角与折射角分别为 $\theta_1$ 与 $\theta_2$，则 $\sin\theta_1 = \dfrac{x}{\sqrt{a^{2}+x^{2}}}$，$\sin\theta_2 = \dfrac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}$。驻点条件化为

这正是斯涅尔定律（Snell's law）。由于 $x \to \pm\infty$ 时 $T(x) \to \infty$，且 $T$ 光滑、只有一个临界点，该临界点就是全局最小值。一条纯几何的物理定律竟从一元最小化中落出，鲜明地说明了为何最优化是大学一年级微积分的核心。