Chapter 9 — Exercises

These exercises track the chapter's arc: from locating and classifying extrema, through the Mean Value Theorem and L'Hôpital's rule, to assembling everything into a full curve sketch. Work them in order — later problems lean on earlier machinery. Pencil first; the worked solutions in each <details> block are for checking, not reading ahead. Fuller solutions to selected starred problems appear in appendices/answers-to-selected.md.

Difficulty tiers:

Tier Meaning Count
⭐ Mechanical — one tool, one step 10
⭐⭐ Standard — combine two ideas 13
⭐⭐⭐ Challenging — full analysis or applied modeling 11
⭐⭐⭐⭐ Proof / theory — justify, generalize, or construct 5
Total 39

Verify your symbolic work with sympy (§9.11) only after finishing by hand — the point is to build the instinct, not outsource it.

Part A — Critical Points and Absolute Extrema (§9.2–9.3)

A1. ⭐ Find all critical points of $f(x) = x^3 - 12x + 5$.

Answer$f'(x) = 3x^2 - 12 = 3(x-2)(x+2)$, so the critical points are $x = -2$ and $x = 2$ (both from $f' = 0$; $f'$ exists everywhere).

A2. ⭐ Find all critical points of $f(x) = x^4 - 4x^3$.

Answer$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$, giving critical points $x = 0$ and $x = 3$.

A3. ⭐ Find the critical points of $f(x) = |x - 3|$, being careful about where $f'$ fails to exist.

AnswerFor $x \ne 3$, $f'(x) = \pm 1 \ne 0$. At $x = 3$ there is a corner, so $f'(3)$ does not exist. The single critical point is $x = 3$ — an absolute minimum (§9.3 pitfall: critical points include where $f'$ is undefined).

A4. ⭐⭐ Use the Closed Interval Method to find the absolute maximum and minimum of $f(x) = x^3 - 3x^2 - 9x + 1$ on $[-2, 4]$.

Answer$f'(x) = 3x^2 - 6x - 9 = 3(x-3)(x+1)$; interior critical points $x = -1$ and $x = 3$. Candidates: $f(-2) = -8 - 12 + 18 + 1 = -1$, $f(-1) = -1 - 3 + 9 + 1 = 6$, $f(3) = 27 - 27 - 27 + 1 = -26$, $f(4) = 64 - 48 - 36 + 1 = -19$. Absolute max $= 6$ at $x = -1$; absolute min $= -26$ at $x = 3$.

A5. ⭐⭐ Find the absolute extrema of $f(x) = x + \dfrac{1}{x}$ on $\left[\tfrac12, 3\right]$.

Answer$f'(x) = 1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1 \Rightarrow x = 1$ (only $x = 1$ lies in the interval). Candidates: $f(\tfrac12) = \tfrac12 + 2 = 2.5$, $f(1) = 2$, $f(3) = 3 + \tfrac13 \approx 3.33$. Absolute min $= 2$ at $x = 1$; absolute max $= \tfrac{10}{3}$ at $x = 3$.

A6. ⭐⭐ Find the absolute extrema of $f(x) = x\sqrt{4 - x^2}$ on $[-2, 2]$.

Answer$f'(x) = \sqrt{4-x^2} + x \cdot \frac{-x}{\sqrt{4-x^2}} = \frac{(4 - x^2) - x^2}{\sqrt{4-x^2}} = \frac{4 - 2x^2}{\sqrt{4-x^2}}$. This is zero when $x^2 = 2$, i.e. $x = \pm\sqrt{2}$; $f'$ is undefined at $x = \pm 2$ (endpoints). Values: $f(\sqrt2) = \sqrt2\cdot\sqrt2 = 2$, $f(-\sqrt2) = -2$, $f(\pm 2) = 0$. Absolute max $= 2$ at $x = \sqrt2$; absolute min $= -2$ at $x = -\sqrt2$.

A7. ⭐⭐⭐ A continuous function $g$ on $[0, 6]$ has exactly two interior critical points, at $x = 2$ and $x = 5$, with values $g(0) = 4$, $g(2) = 9$, $g(5) = 1$, $g(6) = 7$. State the absolute maximum and minimum and explain why no other information is needed.

AnswerBy the Extreme Value Theorem the extrema exist, and by Fermat's theorem they occur only among critical points and endpoints. The candidate values are exactly $\{4, 9, 1, 7\}$. Absolute max $= 9$ at $x = 2$; absolute min $= 1$ at $x = 5$. No values *between* the listed points can exceed or undercut these, because any such extremum would itself be a critical point, and we are told there are no others.

Part B — Monotonicity and the First Derivative Test (§9.4–9.5)

B1. ⭐ State the intervals where $f(x) = x^2 - 6x + 8$ is increasing and decreasing.

Answer$f'(x) = 2x - 6 = 2(x - 3)$. Decreasing on $(-\infty, 3)$, increasing on $(3, \infty)$.

B2. ⭐ For $f(x) = x^3 - 3x^2 + 4$, build the sign chart of $f'$ and read off the intervals of increase and decrease.

Answer$f'(x) = 3x^2 - 6x = 3x(x - 2)$. Increasing on $(-\infty, 0)$, decreasing on $(0, 2)$, increasing on $(2, \infty)$.

B3. ⭐⭐ Classify every critical point of $f(x) = 3x^4 - 8x^3 + 6$ using the first derivative test.

Answer$f'(x) = 12x^3 - 24x^2 = 12x^2(x - 2)$; critical points $x = 0$ and $x = 2$. Sign of $f'$: negative on $(-\infty,0)$ ($12x^2 > 0$, $x-2<0$), still negative on $(0,2)$ (no sign change at $0$), positive on $(2,\infty)$. So $x = 0$ is **neither** (no sign change — a flat spot like $x^3$), and $x = 2$ is a **local minimum** ($-\to+$).

B4. ⭐⭐ Find and classify the local extrema of $f(x) = x e^{-x}$ on $\mathbb{R}$ using the first derivative test.

Answer$f'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x)$. Since $e^{-x} > 0$ always, the sign follows $1 - x$: positive for $x < 1$, negative for $x > 1$. So $f'$ goes $+\to-$ at $x = 1$: a **local (and absolute) maximum**, value $f(1) = e^{-1} \approx 0.368$.

B5. ⭐⭐ Find the local extrema of $f(x) = x^{2/3}(x - 5)$.

AnswerWrite $f = x^{5/3} - 5x^{2/3}$. Then $f'(x) = \tfrac53 x^{2/3} - \tfrac{10}{3}x^{-1/3} = \tfrac{5}{3}x^{-1/3}(x - 2) = \frac{5(x-2)}{3x^{1/3}}$. Critical points: $x = 2$ (from $f'=0$) and $x = 0$ (where $f'$ is undefined). Sign chart: for $x<0$, $x^{1/3}<0$ and $x-2<0$ so $f'>0$; on $(0,2)$, $x^{1/3}>0$, $x-2<0$ so $f'<0$; on $(2,\infty)$, $f'>0$. Thus $x=0$ is a **local maximum** ($+\to-$, value $0$) and $x=2$ is a **local minimum** ($-\to+$, value $2^{2/3}(-3) = -3\sqrt[3]{4}\approx -4.76$).

B6. ⭐⭐⭐ A function satisfies $f'(x) = (x+1)^2(x-2)(x-4)$. Without finding $f$, determine where $f$ is increasing and decreasing and classify each critical point.

AnswerCritical points $x = -1, 2, 4$. The factor $(x+1)^2$ is $\ge 0$, so it never changes the sign — it only contributes a zero. Sign of $f'$ is governed by $(x-2)(x-4)$ except at $x=-1$: on $(-\infty,-1)$ positive, on $(-1,2)$ positive (no sign change at $-1$), on $(2,4)$ negative, on $(4,\infty)$ positive. Increasing on $(-\infty,2)$ and $(4,\infty)$; decreasing on $(2,4)$. At $x=-1$: **neither** (even factor, no sign change). At $x=2$: **local max** ($+\to-$). At $x=4$: **local min** ($-\to+$).

Part C — Concavity, the Second Derivative Test, and Inflection Points (§9.6)

C1. ⭐ Find the intervals of concavity for $f(x) = x^3 - 6x^2 + 5$.

Answer$f''(x) = 6x - 12 = 6(x - 2)$. Concave down on $(-\infty, 2)$, concave up on $(2, \infty)$.

C2. ⭐ Locate the inflection point(s) of $f(x) = x^3 - 3x$.

Answer$f''(x) = 6x$, which changes sign at $x = 0$. Inflection point at $(0, 0)$ since $f(0) = 0$.

C3. ⭐⭐ Classify the critical points of $f(x) = x^4 - 8x^2 + 3$ with the second derivative test, and report the inflection points.

Answer$f'(x) = 4x^3 - 16x = 4x(x^2 - 4)$; critical points $x = 0, \pm 2$. $f''(x) = 12x^2 - 16$. At $x=0$: $f''=-16<0$, **local max** ($f(0)=3$). At $x=\pm2$: $f''=48-16=32>0$, **local minima** ($f(\pm2)=16-32+3=-13$). Inflection: $f''=0$ at $x = \pm\sqrt{4/3} = \pm\frac{2}{\sqrt3}$, and $f''$ changes sign at each, so both are inflection points.

C4. ⭐⭐ Show that $f(x) = x^4$ has a local minimum at $x = 0$ even though the second derivative test is inconclusive there.

Answer$f'(x) = 4x^3$ and $f''(x) = 12x^2$, so $f''(0) = 0$ — the second derivative test says nothing. But $f'$ changes from negative ($x<0$) to positive ($x>0$), so the **first derivative test** gives a local minimum at $x = 0$. (Indeed $x^4 \ge 0$ with equality only at $0$, an absolute minimum.)

C5. ⭐⭐ Find all inflection points of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$.

Answer$f''(x) = -\sin x - \cos x = -(\sin x + \cos x)$. Zero when $\tan x = -1$, i.e. $x = \tfrac{3\pi}{4}$ and $x = \tfrac{7\pi}{4}$. Each is a genuine sign change of $f''$, so both are inflection points; $f(\tfrac{3\pi}{4}) = \tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2} = 0$ and $f(\tfrac{7\pi}{4}) = -\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2} = 0$.

C6. ⭐⭐⭐ For $f(x) = \dfrac{x}{x^2 + 1}$ (the bounded rational function of §9.10), confirm by hand that the inflection points occur at $x = 0$ and $x = \pm\sqrt{3}$.

AnswerFrom §9.10, $f'(x) = \frac{1 - x^2}{(x^2+1)^2}$. Differentiate again with the quotient rule. Numerator derivative: $\frac{d}{dx}[(1-x^2)] \cdot (x^2+1)^2 - (1-x^2)\cdot \frac{d}{dx}[(x^2+1)^2]$, all over $(x^2+1)^4$. Compute: $(-2x)(x^2+1)^2 - (1-x^2)\cdot 2(x^2+1)(2x)$. Factor out $2x(x^2+1)$: $2x(x^2+1)\big[-(x^2+1) - 2(1-x^2)\big] = 2x(x^2+1)(x^2 - 3)$. Divide by $(x^2+1)^4$: $f''(x) = \frac{2x(x^2 - 3)}{(x^2+1)^3}$. This is zero at $x = 0$ and $x = \pm\sqrt3$, and each is a sign change, so all three are inflection points.

C7. ⭐⭐⭐ Give a function with a critical point at $x = 1$ where $f''(1) = 0$ and $x = 1$ is not an extremum. Then explain why such a point must instead be an inflection point — i.e. why "critical, $f''=0$, no extremum" forces a concavity change.

AnswerTake $f(x) = (x-1)^3$. Then $f'(x) = 3(x-1)^2$, so $f'(1) = 0$ (a horizontal-tangent critical point), and $f''(x) = 6(x-1)$, so $f''(1) = 0$. The first derivative test: $f' = 3(x-1)^2 \ge 0$ on both sides, no sign change, so $x=1$ is **not** an extremum. And $f''$ goes from negative ($x<1$) to positive ($x>1$), so concavity flips — $x=1$ is an **inflection point** with a horizontal tangent (exactly like $x^3$ at the origin in §9.6). The general lesson: at a critical point where $f''(c)=0$, the second-derivative test is silent, so the first derivative test decides. If $f'$ does **not** change sign, there is no extremum — and for a smooth function passing through with nonzero higher behavior, the curvature must change, giving an inflection. The point is never simply "nothing happens."

Part D — The Mean Value Theorem and Rolle's Theorem (§9.7)

D1. ⭐ Verify the Mean Value Theorem for $f(x) = x^2 - 2x$ on $[0, 4]$: find every $c$ guaranteed by the theorem.

AnswerAverage rate: $\frac{f(4)-f(0)}{4-0} = \frac{8 - 0}{4} = 2$. Set $f'(c) = 2c - 2 = 2 \Rightarrow c = 2 \in (0,4)$. ✓

D2. ⭐⭐ Verify Rolle's theorem for $f(x) = x^3 - x$ on $[-1, 1]$ and find all valid $c$.

Answer$f(-1) = 0 = f(1)$, and $f$ is continuous and differentiable, so Rolle applies. $f'(x) = 3x^2 - 1 = 0 \Rightarrow x = \pm\frac{1}{\sqrt3}$. Both lie in $(-1,1)$, so $c = \pm\frac{1}{\sqrt3}$.

D3. ⭐⭐ The function $f(x) = 1 - x^{2/3}$ satisfies $f(-1) = f(1) = 0$, yet there is no $c \in (-1,1)$ with $f'(c) = 0$. Which hypothesis of Rolle's theorem fails, and where?

Answer$f'(x) = -\tfrac23 x^{-1/3}$, which is never zero and is **undefined at $x = 0$** — a cusp. Rolle's theorem requires differentiability on the *entire* open interval $(-1,1)$; that fails at $x=0$, so the conclusion is not guaranteed. (Compare the $|x|$ counterexample in §9.7.1.)

D4. ⭐⭐⭐ A car passes milepost 0 at 1:00 PM and milepost 80 at 2:15 PM. The speed limit is 60 mph. Use the Mean Value Theorem to prove the driver broke the speed limit at some instant.

AnswerLet $s(t)$ be position (miles) at time $t$ (hours after 1:00). Elapsed time is $1.25$ h, distance $80$ mi, so average speed $= \frac{80 - 0}{1.25} = 64$ mph. Position is continuous (the car does not teleport) and differentiable (it has a velocity at every instant), so the MVT applies: there exists an instant $c \in (0, 1.25)$ with $s'(c) = 64$ mph $> 60$. The driver exceeded the limit at that instant. This is exactly the average-speed-camera principle of §9.7.

D5. ⭐⭐⭐⭐ Use the Mean Value Theorem to prove that $|\sin a - \sin b| \le |a - b|$ for all real $a, b$.

AnswerIf $a = b$ both sides are $0$. Otherwise apply the MVT to $f(x) = \sin x$ on the interval with endpoints $a, b$: there is a $c$ strictly between them with $\frac{\sin a - \sin b}{a - b} = f'(c) = \cos c$. Taking absolute values, $\frac{|\sin a - \sin b|}{|a - b|} = |\cos c| \le 1$. Multiplying through gives $|\sin a - \sin b| \le |a - b|$. (A function with bounded derivative is *Lipschitz* with that bound.)

D6. ⭐⭐⭐⭐ Suppose $f$ is differentiable on $\mathbb{R}$ and $f'(x) = 0$ for every $x$. Prove from the Mean Value Theorem alone that $f$ is constant.

AnswerTake any two points $x_1 < x_2$. The MVT on $[x_1, x_2]$ gives a $c$ with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0 \cdot (x_2 - x_1) = 0$. Hence $f(x_2) = f(x_1)$. Since $x_1, x_2$ were arbitrary, $f$ takes the same value everywhere — it is constant. (This is the Constant Function Theorem of §9.7.3.)

D7. ⭐⭐⭐⭐ Prove that a polynomial of degree $n$ has at most $n - 1$ critical points, and at most $n$ real roots, and connect the two counts via Rolle's theorem.

AnswerA degree-$n$ polynomial $p$ has derivative $p'$ of degree $n-1$, which has at most $n-1$ real roots (Fundamental Theorem of Algebra bounds roots by degree); those roots are exactly the critical points, so there are at most $n-1$ of them. For the root count: suppose, for contradiction, $p$ had more than $n$ real roots. Between any two consecutive roots $r_i < r_{i+1}$, $p(r_i)=p(r_{i+1})=0$, so Rolle's theorem gives a critical point in $(r_i, r_{i+1})$. With $n+1$ roots there would be $n$ such disjoint intervals, forcing $\ge n$ critical points — but $p'$ has degree $n-1$ and at most $n-1$ roots. Contradiction. Hence at most $n$ real roots.

Part E — L'Hôpital's Rule (§9.8)

E1. ⭐ Evaluate $\displaystyle\lim_{x \to 0} \frac{\tan x}{x}$.

AnswerForm $\tfrac00$. $\lim_{x\to0}\frac{\sec^2 x}{1} = \sec^2 0 = 1$.

E2. ⭐ Evaluate $\displaystyle\lim_{x \to 0} \frac{e^{2x} - 1}{x}$.

AnswerForm $\tfrac00$. $\lim_{x\to0}\frac{2e^{2x}}{1} = 2$.

E3. ⭐⭐ Evaluate $\displaystyle\lim_{x \to \infty} \frac{\ln x}{x}$ and state what it says about growth.

AnswerForm $\tfrac{\infty}{\infty}$. $\lim_{x\to\infty}\frac{1/x}{1} = 0$. So $\ln x$ grows *slower* than any positive power of $x$ — logarithmic growth is the slowest unbounded growth (mirror image of "$e^x$ beats every polynomial," §9.8.1).

E4. ⭐⭐ Evaluate $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

AnswerForm $\tfrac00$. Apply once: $\lim_{x\to0}\frac{\sin x}{2x}$, still $\tfrac00$. Apply again: $\lim_{x\to0}\frac{\cos x}{2} = \tfrac12$.

E5. ⭐⭐ Evaluate $\displaystyle\lim_{x \to \infty} x \sin\!\left(\frac{1}{x}\right)$.

AnswerForm $\infty \cdot 0$. Let $u = 1/x \to 0^+$: $\lim_{u\to0^+}\frac{\sin u}{u} = 1$. (Or rewrite as $\frac{\sin(1/x)}{1/x}$ and apply L'Hôpital — same answer, $1$.)

E6. ⭐⭐⭐ Evaluate $\displaystyle\lim_{x \to 0^+} x^{\sin x}$ (form $0^0$).

AnswerLet $y = x^{\sin x}$, so $\ln y = \sin x \cdot \ln x$. As $x\to0^+$ this is form $0\cdot(-\infty)$. Write $\frac{\ln x}{1/\sin x} = \frac{\ln x}{\csc x}$ (form $\tfrac{-\infty}{\infty}$): L'Hôpital gives $\frac{1/x}{-\csc x\cot x} = \frac{-\sin x \tan x}{x} = -\frac{\sin x}{x}\cdot\tan x \to -(1)(0) = 0$. So $\ln y \to 0$ and $y \to e^0 = 1$.

E7. ⭐⭐⭐ Evaluate $\displaystyle\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{x}$ (form $1^\infty$), and connect to continuous compounding (§9.8.2).

AnswerLet $y = (1+3/x)^x$, so $\ln y = x\ln(1+3/x)$, form $\infty\cdot 0$. Rewrite as $\frac{\ln(1+3/x)}{1/x}$ (form $\tfrac00$); L'Hôpital: numerator derivative $\frac{-3/x^2}{1+3/x}$, denominator derivative $-1/x^2$, ratio $= \frac{3}{1+3/x} \to 3$. So $\ln y \to 3$ and $y \to e^3$. This is the $r=3$ case of continuous compounding $\lim(1+r/n)^n = e^r$.

E8. ⭐⭐⭐ Show that L'Hôpital's rule fails to settle $\displaystyle\lim_{x \to \infty} \frac{x + \cos x}{x}$, then find the true value.

AnswerThe form is $\tfrac{\infty}{\infty}$, so L'Hôpital is *attemptable*: it gives $\lim_{x\to\infty}\frac{1 - \sin x}{1}$, which oscillates and **does not exist**. The hypothesis "the limit of the ratio of derivatives exists" is violated, so the rule yields nothing. Elementary method: $\frac{x+\cos x}{x} = 1 + \frac{\cos x}{x} \to 1 + 0 = 1$. The original limit exists and equals $1$ (compare §9.8.3).

Part F — Full Curve Sketching (§9.9–9.10)

F1. ⭐⭐ Carry out the full procedure for $f(x) = x^3 - 3x^2 - 9x + 5$.

AnswerDomain $\mathbb{R}$; $y$-intercept $f(0)=5$; no symmetry; no asymptotes. $f'(x)=3x^2-6x-9=3(x-3)(x+1)$: increasing on $(-\infty,-1)$ and $(3,\infty)$, decreasing on $(-1,3)$. Local max at $(-1, 10)$ ($f(-1)=-1-3+9+5=10$); local min at $(3, -22)$ ($f(3)=27-27-27+5=-22$). $f''(x)=6x-6=6(x-1)$: concave down on $(-\infty,1)$, up on $(1,\infty)$; inflection at $(1, -6)$ ($f(1)=1-3-9+5=-6$). Standard cubic silhouette: rise to $(-1,10)$, fall through the inflection to $(3,-22)$, rise again.

F2. ⭐⭐ Sketch $f(x) = \dfrac{1}{x^2 - 4}$. Identify the asymptotes and symmetry.

AnswerDomain $x \ne \pm 2$. Even function ($f(-x)=f(x)$), symmetric about the $y$-axis. Vertical asymptotes at $x = \pm 2$; horizontal asymptote $y = 0$ (degree of denominator exceeds numerator). $f'(x) = \frac{-2x}{(x^2-4)^2}$: $f'>0$ for $x<0$, $f'<0$ for $x>0$, so a local max at $(0, -\tfrac14)$. On $(-2,2)$ the curve is a downward hump peaking at $(0,-\tfrac14)$ and diving to $-\infty$ near $\pm2$; outside, two branches in the upper half-plane approaching $0$ from above.

F3. ⭐⭐⭐ Sketch $f(x) = \dfrac{x^2}{x - 2}$, including the oblique asymptote.

AnswerLong division: $x^2 = (x-2)(x+2) + 4$, so $f(x) = x + 2 + \frac{4}{x-2}$. Oblique asymptote $y = x+2$; vertical asymptote $x = 2$. $f'(x) = 1 - \frac{4}{(x-2)^2} = \frac{(x-2)^2 - 4}{(x-2)^2}$, zero when $(x-2)^2 = 4 \Rightarrow x = 0$ or $x = 4$. $f(0) = 0$ (local max, since $f$ decreasing into it from the left branch) and $f(4) = \frac{16}{2} = 8$ (local min). Check: $f''(x) = \frac{8}{(x-2)^3}$, negative for $x<2$ (so $x=0$ is a max), positive for $x>2$ (so $x=4$ is a min) — consistent.

F4. ⭐⭐⭐ Sketch $f(x) = x e^{-x^2}$: extrema, inflection points, end behavior, symmetry.

AnswerOdd function ($f(-x) = -f(x)$). As $x\to\pm\infty$, $f\to 0$ (exponential dominates), so $y=0$ is a horizontal asymptote both ways. $f'(x) = e^{-x^2}(1 - 2x^2)$, zero at $x = \pm\frac{1}{\sqrt2}$: local max at $x=\frac1{\sqrt2}$ (value $\frac{1}{\sqrt2}e^{-1/2}\approx 0.43$), local min at $x=-\frac1{\sqrt2}$. $f''(x) = e^{-x^2}(4x^3 - 6x) = 2x e^{-x^2}(2x^2 - 3)$, zero at $x = 0$ and $x = \pm\sqrt{3/2}$ — three inflection points. An odd S-through-origin with a hump on each side decaying to the axis.

F5. ⭐⭐⭐ Sketch $f(x) = \dfrac{\ln x}{x}$ on $(0, \infty)$.

AnswerDomain $x>0$; $x$-intercept at $x=1$ ($\ln 1 = 0$). As $x\to0^+$, $\ln x\to-\infty$ and $x\to0^+$ so $f\to-\infty$ (vertical asymptote at $x=0$). As $x\to\infty$, $f\to 0$ (Exercise E3), so $y=0$ is a horizontal asymptote. $f'(x) = \frac{1 - \ln x}{x^2}$, zero when $\ln x = 1 \Rightarrow x = e$: local (and absolute) max at $(e, 1/e)$. $f''(x) = \frac{2\ln x - 3}{x^3}$, zero when $\ln x = \tfrac32 \Rightarrow x = e^{3/2}$: inflection there. Curve rises from $-\infty$, crosses zero at $x=1$, peaks at $(e, 1/e)$, then descends toward the axis.

Part G — Applications Across Fields (§9.12)

G1. ⭐⭐ (Biology — pharmacokinetics.) A drug's blood concentration is $C(t) = 50\,e^{-0.2 t}$ (mg/L). Show $C$ is decreasing and concave up for $t \ge 0$, and compute the half-life.

Answer$C'(t) = -10 e^{-0.2t} < 0$ (always decreasing); $C''(t) = 2 e^{-0.2t} > 0$ (always concave up) — a steep drop that flattens. Half-life: $e^{-0.2 t_{1/2}} = \tfrac12 \Rightarrow -0.2 t_{1/2} = \ln\tfrac12 = -\ln 2 \Rightarrow t_{1/2} = \frac{\ln 2}{0.2} \approx 3.47$ hours.

G2. ⭐⭐⭐ (Economics — marginal analysis.) $R(q) = 60q - q^2$, $C(q) = q^3 - 9q^2 + 30q + 20$. Maximize profit $\pi(q) = R(q) - C(q)$.

Answer$\pi(q) = 60q - q^2 - q^3 + 9q^2 - 30q - 20 = -q^3 + 8q^2 + 30q - 20$. $\pi'(q) = -3q^2 + 16q + 30$. Set to zero: $3q^2 - 16q - 30 = 0 \Rightarrow q = \frac{16 \pm\sqrt{256 + 360}}{6} = \frac{16\pm\sqrt{616}}{6}$. $\sqrt{616}\approx 24.82$, so $q \approx \frac{40.82}{6} \approx 6.80$ (reject the negative root). $\pi''(q) = -6q + 16$; at $q\approx6.80$, $\pi'' \approx -24.8 < 0$, confirming a **maximum**. The firm produces about $6{,}800$ units, where marginal revenue equals marginal cost (§9.12).

G3. ⭐⭐⭐ (Physics — potential energy.) $U(x) = x^4 - 4x^2$. Find and classify the equilibria.

Answer$U'(x) = 4x^3 - 8x = 4x(x^2 - 2)$; equilibria at $x = 0$ and $x = \pm\sqrt2$. $U''(x) = 12x^2 - 8$. At $x=0$: $U''=-8<0$ — **unstable** (local max of $U$). At $x=\pm\sqrt2$: $U''=24-8=16>0$ — **stable** (minima, the bottom of two wells). A classic double-well potential.

G4. ⭐⭐⭐ (Data science — loss surface.) $L(w) = (w - 3)^2 + \tfrac12 w^2$. Find $w^*$ and verify convexity.

Answer$L'(w) = 2(w-3) + w = 3w - 6$, zero at $w^* = 2$. $L''(w) = 3 > 0$ for all $w$, so $L$ is convex (concave up everywhere) and $w^*=2$ is the **global** minimum — gradient descent converges to it from any start (the Chapter 6 anchor).

Part H — Synthesis and Proof (§9.3–9.7)

H1. ⭐⭐⭐ Sketch a single continuous $f$ on $[-3, 5]$ with: $f'>0$ on $(-3,0)\cup(2,5)$; $f'<0$ on $(0,2)$; $f''>0$ on $(1,5)$; $f''<0$ on $(-3,1)$; $f(0)=3$, $f(2)=-1$.

AnswerFrom $f'$: increasing up to $x=0$ (local **max** at $(0,3)$), decreasing to $x=2$ (local **min** at $(2,-1)$), increasing after. From $f''$: concave down left of $x=1$, concave up right — **inflection** at $x=1$. A valid sketch rises concave-down to the peak $(0,3)$, falls (still concave down) to the inflection at $x=1$, continues falling now concave up to the trough $(2,-1)$, then rises concave up to the right end. The two pieces of information agree: the max sits in a concave-down region, the min in a concave-up region.

H2. ⭐⭐⭐⭐ Prove the first derivative test: if $f'$ changes from positive to negative at a critical point $c$, then $f$ has a local maximum at $c$.

AnswerSuppose $f$ is continuous at $c$ and $f' > 0$ on an interval $(a, c)$ and $f' < 0$ on $(c, b)$. By the Increasing/Decreasing Test (§9.4) — itself a corollary of the Mean Value Theorem (§9.7.3) — $f$ is increasing on $(a, c]$, so $f(x) \le f(c)$ for $x \in (a, c]$; and $f$ is decreasing on $[c, b)$, so $f(x) \le f(c)$ for $x \in [c, b)$. Therefore $f(x) \le f(c)$ for all $x$ in $(a,b)$, which is precisely a local maximum at $c$. The result we invoke is the Increasing/Decreasing Test, and the theorem underneath it is the MVT.

A note on checking your work. For any sketching or extremum problem, a 20-second sympy check (§9.11) catches arithmetic slips: sp.solve(sp.diff(f, x), x) for critical points, sp.diff(f, x, 2) evaluated at each, and sp.limit(...) for the L'Hôpital problems. But do the hand analysis first — Section 9.11's whole point is that the structure you find by hand is what the machine merely confirms.