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1. 1.Computation · 7 min

   Find the absolute maximum and minimum values of each function on the given interval by the Closed Interval Method. Begin each part by saying why the Extreme Value Theorem guarantees that both values exist.

   1. (a)$f(x)=2x^3-9x^2+12x$ on $[0,3]$.
   2. (b)$g(x)=|x^2-4|$ on $[-1,3]$.
   3. (c)In (b), one of the two absolute extrema sits at a point where $g^{\prime }$ does not exist. Explain why solving $g^{\prime }(x)=0$ alone would have missed it, and which clause of the definition of a critical number catches it.
   Show the solution to question 1

   (a) Check the hypotheses, then list the candidates. $f$ is a polynomial, so it is continuous on the closed interval $[0,3]$ and the Extreme Value Theorem promises an absolute maximum and minimum. Differentiate and factor:

   $$f^{\prime }(x)=6x^2-18x+12=6\left(x^2-3x+2\right)=6(x-1)(x-2).$$

   $f^{\prime }$ exists everywhere and vanishes at $x=1$ and $x=2$, both inside $(0,3)$.

   Evaluate at the critical numbers and the endpoints. $f(1)=2-9+12=5$, $f(2)=16-36+24=4$, $f(0)=0$, $f(3)=54-81+36=9$.

   Comparing $5,4,0,9$: the absolute maximum is $f(3)=9$ and the absolute minimum is $f(0)=0$. Both sit at endpoints; the interior critical numbers give only a local maximum ($5$) and a local minimum ($4$). A method that checked critical numbers alone would have got both answers wrong.

   (b) Remove the absolute value first. $g$ is continuous, being the composition of the continuous function $|\cdot |$ with a polynomial, so the theorem applies on $[-1,3]$. Since $x^2-4=(x-2)(x+2)$ and $x+2>0$ throughout $[-1,3]$, the sign of $x^2-4$ is the sign of $x-2$:

   $$g(x)=\{\begin{array}{ll}4-x^2,& -1\le x\le 2,\\ x^2-4,& 2\le x\le 3.\end{array}$$

   On $(-1,2)$, $g^{\prime }(x)=-2x$, which is zero at $x=0$. On $(2,3)$, $g^{\prime }(x)=2x$, never zero. At $x=2$ the left-hand derivative is $-4$ and the right-hand derivative is $+4$, so $g^{\prime }(2)$ does not exist. The critical numbers in $(-1,3)$ are therefore $x=0$ and $x=2$.

   Values: $g(-1)=3$, $g(0)=4$, $g(2)=0$, $g(3)=5$. The absolute maximum is $g(3)=5$ and the absolute minimum is $g(2)=0$.

   (c) The equation $g^{\prime }(x)=0$ has the single solution $x=0$, and that is the local maximum, not the minimum. The minimum lives at the corner $x=2$, where the graph is a V: the tangent slope jumps from $-4$ to $+4$ without ever being $0$. Fermat's Theorem only says $f^{\prime }(c)=0$ when $f^{\prime }(c)$ exists, so it is silent at a corner. The definition of a critical number has a second clause precisely for this: a number $c$ in the domain where $f^{\prime }(c)$ does not exist is also a critical number, and the Closed Interval Method must test it.

2. 2.Proof · 8 min

   Quote the Mean Value Theorem or Rolle's Theorem by name wherever you use one, and check its hypotheses each time.

   1. (a)Find every number $c$ that satisfies the conclusion of the Mean Value Theorem for $f(x)=\sqrt{x}$ on $[1,4]$.
   2. (b)Prove that $\ln (1+x)<x$ for every $x>0$.
   3. (c)Let $a$ and $b$ be any real constants. Prove that the equation $x^4+ax+b=0$ has at most two real solutions.
   Show the solution to question 2

   (a) Hypotheses, secant slope, then solve. $\sqrt{x}$ is continuous on $[1,4]$ and differentiable on $(1,4)$, with $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ defined for every $x>0$. The Mean Value Theorem applies. The average rate of change is

   $$\frac{f(4)-f(1)}{4-1}=\frac{2-1}{3}=\frac{1}{3}.$$

   Solve $f^{\prime }(c)=\frac{1}{3}$: $\frac{1}{2\sqrt{c}}=\frac{1}{3}$ gives $\sqrt{c}=\frac{3}{2}$, so $c=\frac{9}{4}$. It lies in $(1,4)$, and it is the only solution because $f^{\prime }$ is one-to-one on $(0,\infty )$. So $c=\frac{9}{4}$.

   Sanity check: the tangent at $x=2.25$ has slope $\frac{1}{2\cdot 1.5}=\frac{1}{3}$, matching the secant.

   (b) Proof. Fix $x>0$ and apply the Mean Value Theorem to $f(t)=\ln t$ on $[1,1+x]$. The hypotheses hold: $\ln$ is continuous on $[1,1+x]$ and differentiable on $(1,1+x)$, with $f^{\prime }(t)=\frac{1}{t}$. So there is a $c$ in $(1,1+x)$ with

   $$\ln (1+x)-\ln 1=\frac{1}{c}((1+x)-1),\text{that is}\ln (1+x)=\frac{x}{c}.$$

   Since $c>1$ we have $\frac{1}{c}<1$, and since $x>0$, multiplying gives $\frac{x}{c}<x$. Hence $\ln (1+x)<x$. $■$

   The theorem did the whole job: it replaced the unknown value $\ln (1+x)$ by a slope we could bound.

   (c) Proof by contradiction, via Rolle. Let $p(x)=x^4+ax+b$ and suppose $p$ had three distinct real roots $r_1<r_2<r_3$. A polynomial is continuous and differentiable everywhere, and $p(r_1)=p(r_2)=0$, so Rolle's Theorem gives a $c_1$ in $(r_1,r_2)$ with $p^{\prime }(c_1)=0$. Likewise $p(r_2)=p(r_3)=0$ gives a $c_2$ in $(r_2,r_3)$ with $p^{\prime }(c_2)=0$. Since $c_1<r_2<c_2$, these are two distinct zeros of $p^{\prime }$.

   But $p^{\prime }(x)=4x^3+a$, and $4x^3+a=0$ means $x^3=-\frac{a}{4}$, which has exactly one real solution, $x=\sqrt[3]{-a/4}$, because $x\mapsto x^3$ is one-to-one on $\mathbb{R}$. Two distinct zeros of $p^{\prime }$ is impossible, so three distinct roots of $p$ is impossible: $p$ has at most two real roots. $■$

   The bound is attained: with $a=0$, $b=-1$ the equation $x^4=1$ has the two real roots $\pm 1$.

3. 3.Computation · 7 min

   For each function, find the critical numbers, the intervals on which it increases and decreases, and every local extremum. Write out an explicit sign chart for the derivative, with one test value per interval, and classify each critical number with the First Derivative Test.

   1. (a)$f(x)=\frac{x}{x^2+1}$
   2. (b)$g(x)=3x^{2/3}-x$
   3. (c)Explain why the local maximum you found in (a) is in fact the absolute maximum of $f$ on the whole real line.
   Show the solution to question 3

   (a) Quotient rule, then the sign chart.

   $$f^{\prime }(x)=\frac{(x^2+1)\cdot 1-x\cdot 2x}{(x^2+1{)}^2}=\frac{1-x^2}{(x^2+1{)}^2}.$$

   The denominator is always positive, so $f^{\prime }$ exists everywhere and its sign is the sign of $1-x^2=(1-x)(1+x)$. Critical numbers: $x=-1$ and $x=1$.

   Sign chart. On $(-\infty ,-1)$ take $x=-2$: $1-4=-3<0$. On $(-1,1)$ take $x=0$: $1>0$. On $(1,\infty )$ take $x=2$: $-3<0$. So $f$ decreases on $(-\infty ,-1]$, increases on $[-1,1]$, and decreases on $[1,\infty )$.

   First Derivative Test: at $x=-1$ the sign goes $-$ to $+$, a local minimum, $f(-1)=-\frac{1}{2}$. At $x=1$ it goes $+$ to $-$, a local maximum, $f(1)=\frac{1}{2}$.

   (b) Watch for the point where $g^{\prime }$ fails to exist.

   $$g^{\prime }(x)=2x^{-1/3}-1=\frac{2}{x^{1/3}}-1=\frac{2-x^{1/3}}{x^{1/3}}.$$

   $g^{\prime }(x)=0$ when $x^{1/3}=2$, that is $x=8$. $g^{\prime }$ does not exist at $x=0$, and $g(0)=0$ is defined, so $0$ is in the domain and counts. Critical numbers: $x=0$ and $x=8$.

   Sign chart. On $(-\infty ,0)$ take $x=-1$: $\frac{2}{-1}-1=-3<0$. On $(0,8)$ take $x=1$: $2-1=1>0$. On $(8,\infty )$ take $x=27$: $\frac{2}{3}-1=-\frac{1}{3}<0$. So $g$ decreases on $(-\infty ,0]$, increases on $[0,8]$, decreases on $[8,\infty )$.

   At $x=0$: $-$ to $+$, a local minimum, $g(0)=0$. This is a cusp — $g^{\prime }(x)\to -\infty$ as $x\to 0^{-}$ and $g^{\prime }(x)\to +\infty$ as $x\to 0^{+}$ — which Fermat's Theorem could never have located. At $x=8$: $+$ to $-$, a local maximum, $g(8)=3\cdot 4-8=4$.

   (c) Work on the interval $I=[-1,\infty )$ first. There $f^{\prime }>0$ for $-1<x<1$ and $f^{\prime }<0$ for $x>1$, so by the First Derivative Test for Absolute Extrema, $f(1)=\frac{1}{2}$ is the absolute maximum of $f$ on $I$. For the remaining $x<-1$, the numerator $x$ is negative and the denominator positive, so $f(x)<0<\frac{1}{2}$. Hence $f(x)\le \frac{1}{2}$ for every real $x$, with equality only at $x=1$. (By the same argument applied to $-f$, or by the oddness $f(-x)=-f(x)$, the absolute minimum is $-\frac{1}{2}$.)

4. 4.Computation · 6 min

   Find the intervals of concavity and every inflection point of each function. For each number where $f^{\prime \prime }$ is zero or undefined, state explicitly whether $f^{\prime \prime }$ changes sign there — a zero of $f^{\prime \prime }$ is only a candidate.

   1. (a)$f(x)=x^4-6x^2+8x$
   2. (b)$g(x)=x{e}^{2x}$
   3. (c)$h(x)=3x^5-10x^4$
   Show the solution to question 4

   (a) $f^{\prime }(x)=4x^3-12x+8$ and $f^{\prime \prime }(x)=12x^2-12=12(x-1)(x+1)$, which exists everywhere and vanishes at $x=\pm 1$.

   Sign chart for $f^{\prime \prime }$: at $x=-2$, $12(-3)(-1)=36>0$; at $x=0$, $-12<0$; at $x=2$, $36>0$. By the Concavity Test, $f$ is concave upward on $(-\infty ,-1)$ and on $(1,\infty )$, and concave downward on $(-1,1)$. The sign changes at both candidates, so both are inflection points: $f(-1)=1-6-8=-13$ and $f(1)=1-6+8=3$, giving $(-1,-13)$ and $(1,3)$.

   (b) Product rule twice: $g^{\prime }(x)=e^{2x}+2x{e}^{2x}=(1+2x)e^{2x}$, and

   $$g^{\prime \prime }(x)=2e^{2x}+2(1+2x)e^{2x}=(4+4x)e^{2x}=4(x+1)e^{2x}.$$

   Since $e^{2x}>0$ always, the sign of $g^{\prime \prime }$ is the sign of $x+1$: negative for $x<-1$, positive for $x>-1$. So $g$ is concave downward on $(-\infty ,-1)$, concave upward on $(-1,\infty )$, and the sign change at $x=-1$ gives the single inflection point $\left(-1,-e^{-2}\right)$.

   (c) $h^{\prime }(x)=15x^4-40x^3$ and $h^{\prime \prime }(x)=60x^3-120x^2=60x^2(x-2)$. Candidates: $x=0$ and $x=2$.

   Sign chart: the factor $60x^2$ is positive except at $0$, so the sign of $h^{\prime \prime }$ is the sign of $x-2$ on either side of $0$. At $x=-1$: $60\cdot 1\cdot (-3)<0$. At $x=1$: $60\cdot 1\cdot (-1)<0$. At $x=3$: $60\cdot 9\cdot 1>0$.

   At $x=0$ the sign is negative on both sides — no change — so $(0,0)$ is not an inflection point even though $h^{\prime \prime }(0)=0$. At $x=2$ the sign changes from $-$ to $+$, so $(2,h(2))=(2,96-160)=(2,-64)$ is the only inflection point. $h$ is concave downward on $(-\infty ,2)$ and concave upward on $(2,\infty )$.

   What does happen at $x=0$: $h^{\prime }(x)=5x^3(3x-8)$ is positive just left of $0$ and negative just right of $0$, so by the First Derivative Test $h$ has a local maximum there, $h(0)=0$ — a critical number where the Second Derivative Test would have been silent.

5. 5.Critique · 9 min

   A student is asked to find and classify the local extrema, the inflection points, and the absolute extrema of $f(x)=3x^5-5x^3$ on the real line, and submits the following.

   Step 1. $f^{\prime }(x)=15x^4-15x^2=15x^2(x-1)(x+1)$, so the critical numbers are $x=-1$, $x=0$ and $x=1$.

   Step 2. $f^{\prime \prime }(x)=60x^3-30x$. Since $f^{\prime \prime }(-1)=-30<0$, there is a local maximum at $x=-1$ with value $f(-1)=2$; since $f^{\prime \prime }(1)=30>0$, there is a local minimum at $x=1$ with value $f(1)=-2$.

   Step 3. $f^{\prime \prime }(0)=0$, so by the Second Derivative Test $f$ has neither a maximum nor a minimum at $x=0$.

   Step 4. Inflection points occur where $f^{\prime \prime }=0$. Since $f^{\prime \prime }(0)=0$, the only inflection point is $(0,0)$.

   Step 5. The local maximum value $2$ is the largest value $f$ takes, so the absolute maximum of $f$ is $2$.

   Steps 1 and 2 are correct. Each of Steps 3, 4 and 5 is not.

   1. (a)For each of Steps 3, 4 and 5, state exactly what is wrong and name the theorem or hypothesis that has been misused. (One of the three reaches a true conclusion by an invalid argument; say which.)
   2. (b)Classify $x=0$ correctly, with a valid argument.
   3. (c)Find every inflection point of $f$, and state — with justification — the absolute maximum and minimum of $f$ on $\mathbb{R}$, if they exist.
   Show the solution to question 5

   (a) Three separate errors.

   Step 3 draws a conclusion the test does not offer. The Second Derivative Test has exactly two clauses, $f^{\prime \prime }(c)>0$ and $f^{\prime \prime }(c)<0$. When $f^{\prime \prime }(c)=0$ it is inconclusive — it says nothing at all, in either direction. The student has read “the test does not say maximum or minimum” as “the test says neither”. It so happens that $x=0$ is not an extremum (see (b)), so this is the step with a true conclusion and no valid argument.

   Step 4 confuses a necessary condition with a sufficient one, and then mis-solves the equation. An inflection point requires $f^{\prime \prime }$ to actually change sign; $f^{\prime \prime }(c)=0$ is only where to look, exactly as $f^{\prime }(c)=0$ is for extrema. Worse, the student solved $f^{\prime \prime }(0)=0$ instead of $f^{\prime \prime }(x)=0$: the equation $60x^3-30x=30x(2x^2-1)=0$ has three solutions, $x=0$ and $x=\pm \frac{1}{\sqrt{2}}$, and two candidates were never examined.

   Step 5 confuses local with absolute. A local maximum only beats its neighbours. The Extreme Value Theorem, which is what guarantees absolute extrema, needs a closed interval, and $\mathbb{R}$ is not one. Indeed $f(2)=96-40=56>2$, and $f(x)\to +\infty$ as $x\to \infty$, so no value of $f$ is the largest.

   (b) Use the First Derivative Test, which always decides. $f^{\prime }(x)=15x^2(x^2-1)$. For $0<|x|<1$ the factor $15x^2$ is positive and $x^2-1$ is negative, so $f^{\prime }<0$ on $(-1,0)$ and on $(0,1)$. There is no sign change at $0$, so $f$ has no local extremum there: it is decreasing through the origin, with a momentarily horizontal tangent.

   (c) Inflection points. $f^{\prime \prime }(x)=30x\left(2x^2-1\right)$, zero at $x=-\frac{1}{\sqrt{2}}$, $0$, $\frac{1}{\sqrt{2}}$. Sign chart, using the factors $x$ and $2x^2-1$ (the latter negative exactly for $|x|<\frac{1}{\sqrt{2}}$):

   on $\left(-\infty ,-\frac{1}{\sqrt{2}}\right)$: $(-)(+)=-$; on $\left(-\frac{1}{\sqrt{2}},0\right)$: $(-)(-)=+$; on $\left(0,\frac{1}{\sqrt{2}}\right)$: $(+)(-)=-$; on $\left(\frac{1}{\sqrt{2}},\infty \right)$: $(+)(+)=+$.

   The sign changes at all three candidates, so all three are inflection points. Their heights: $f\left(\frac{1}{\sqrt{2}}\right)=\frac{3}{4\sqrt{2}}-\frac{5}{2\sqrt{2}}=\frac{3-10}{4\sqrt{2}}=-\frac{7\sqrt{2}}{8}$, and $f$ is odd, so

   $$\left(-\frac{1}{\sqrt{2}},\frac{7\sqrt{2}}{8}\right),(0,0),\left(\frac{1}{\sqrt{2}},-\frac{7\sqrt{2}}{8}\right).$$

   Absolute extrema. None. $f(x)=x^3(3x^2-5)\to +\infty$ as $x\to +\infty$ and $\to -\infty$ as $x\to -\infty$, so $f$ is unbounded in both directions. (On a closed interval the story changes: on $[-2,2]$, comparing $f(\pm 2)=\pm 56$, $f(\pm 1)=\mp 2$ and $f(0)=0$, the absolute maximum is $56$ at $x=2$ and the minimum $-56$ at $x=-2$ — both at endpoints.)

6. 6.Sketch · 10 min

   Let $f(x)=\frac{x}{x^2-1}$. Every feature you draw must be backed by a computation on the page.

   1. (a)State the domain, test for symmetry, find the intercepts, and find every asymptote — each vertical asymptote with both one-sided limits, each horizontal asymptote with the limit that produces it.
   2. (b)Compute $f^{\prime }$ and show that $f$ has no critical numbers. On which intervals is $f$ decreasing? Then explain why the sentence “$f$ is decreasing on its whole domain” is nevertheless false, using two specific values of $f$.
   3. (c)Compute $f^{\prime \prime }$, find the intervals of concavity, and find every inflection point.
   4. (d)Sketch $y=f(x)$ for $-3\le x\le 3$, drawing every asymptote dashed, marking the inflection point and the points $\left(\pm 2,\pm \frac{2}{3}\right)$, and showing from which side each outer branch approaches $y=0$.
   Show the solution to question 6

   (a) Domain: $x^2-1=(x-1)(x+1)\ne 0$, so all $x\ne \pm 1$. Symmetry: $f(-x)=\frac{-x}{x^2-1}=-f(x)$, so $f$ is odd and the graph is symmetric about the origin. Intercepts: $f(x)=0$ only when the numerator $x=0$, and $f(0)=0$, so the origin is the only intercept.

   Vertical asymptotes. At $x=1$ the numerator is $1\ne 0$ while the denominator vanishes. As $x\to 1^{-}$, $x-1\to 0^{-}$ and $x+1\to 2$, so the denominator $\to 0^{-}$ and $f(x)\to -\infty$; as $x\to 1^{+}$ the denominator $\to 0^{+}$ and $f(x)\to +\infty$. At $x=-1$ the numerator is $-1$. As $x\to -1^{-}$, $x+1\to 0^{-}$ and $x-1\to -2$, so the denominator $\to 0^{+}$ and $f(x)\to -\infty$; as $x\to -1^{+}$ the denominator $\to 0^{-}$ and $f(x)\to +\infty$. (The second pair also follows from the first by oddness.)

   Horizontal asymptote. Dividing by $x^2$, $f(x)=\frac{1/x}{1-1/x^2}\to \frac{0}{1}=0$ as $x\to \pm \infty$, so $y=0$ is the horizontal asymptote in both directions.

   (b) Quotient rule:

   $$f^{\prime }(x)=\frac{(x^2-1)\cdot 1-x\cdot 2x}{(x^2-1{)}^2}=\frac{-(x^2+1)}{(x^2-1{)}^2}.$$

   The numerator is strictly negative for every $x$ and the denominator is strictly positive on the domain, so $f^{\prime }(x)<0$ at every point of the domain: $f^{\prime }$ is never zero, and the only places $f^{\prime }$ fails to exist, $x=\pm 1$, are not in the domain, so they are not critical numbers. There are none.

   By the Increasing/Decreasing Test, $f$ is decreasing on each of the three intervals $(-\infty ,-1)$, $(-1,1)$ and $(1,\infty )$ separately. It is not decreasing on their union: $\frac{1}{2}<2$ but

   $$f\left(\frac{1}{2}\right)=\frac{1/2}{1/4-1}=-\frac{2}{3}<\frac{2}{3}=f(2).$$

   The test's hypothesis is “on an interval”, and the domain is three intervals with two gaps; the Mean Value Theorem cannot be applied across a gap, so nothing links the values on one side to those on the other.

   (c) Write $f^{\prime }(x)=-(x^2+1)(x^2-1{)}^{-2}$ and differentiate:

   $$f^{\prime \prime }(x)=-2x(x^2-1{)}^{-2}+(x^2+1)\cdot 2(x^2-1{)}^{-3}\cdot 2x=\frac{-2x(x^2-1)+4x(x^2+1)}{(x^2-1{)}^3}=\frac{2x^3+6x}{(x^2-1{)}^3}=\frac{2x(x^2+3)}{(x^2-1{)}^3}.$$

   Since $x^2+3>0$, the sign of $f^{\prime \prime }$ is the sign of $x$ times the sign of $x^2-1$. On $(-\infty ,-1)$: $(-)(+)=-$, concave down. On $(-1,0)$: $(-)(-)=+$, concave up. On $(0,1)$: $(+)(-)=-$, concave down. On $(1,\infty )$: $(+)(+)=+$, concave up.

   $f^{\prime \prime }$ vanishes only at $x=0$, where it changes sign, so $(0,0)$ is an inflection point. The concavity also flips across $x=\pm 1$, but those are not in the domain and are not inflection points — there is no point of the curve there.

   (d) The sketch. Three branches, decreasing on each. Left branch, $x<-1$: enters from the far left just below $y=0$ (values are negative there: $f(-3)=-\frac{3}{8}$, $f(-2)=-\frac{2}{3}$), bends concave down, and plunges to $-\infty$ at the dashed line $x=-1$. Middle branch, $-1<x<1$: descends from $+\infty$ just right of $x=-1$, concave up, through $\left(-\frac{1}{2},\frac{2}{3}\right)$, then straight through the origin with slope $f^{\prime }(0)=-1$ — the inflection point, mark it — now concave down, through $\left(\frac{1}{2},-\frac{2}{3}\right)$, and down to $-\infty$ at $x=1$. Right branch, $x>1$: descends from $+\infty$, concave up, through $\left(2,\frac{2}{3}\right)$ and $\left(3,\frac{3}{8}\right)$, settling onto $y=0$ from above. Draw $x=-1$, $x=1$ and $y=0$ dashed. Check the odd symmetry at the end: rotating the picture a half-turn about the origin should map it onto itself.

7. 7.Computation · 10 min

   Two optimization problems, one with a closed domain and one with an open domain. In each, write down the objective, the constraint, and the domain before you differentiate, and name the result that certifies your extremum is absolute rather than merely local.

   1. (a)A rectangle has its base on the $x$-axis and its two upper vertices on the parabola $y=12-x^2$. Find the dimensions of the rectangle of largest area.
   2. (b)An open-topped box with a square base is to hold $32$ cubic units. Find the dimensions that use the least material for the base and the four sides, and the amount of material used.
   Show the solution to question 7

   (a) Set up. By symmetry of the parabola about the $y$-axis, put the upper vertices at $(\pm x,12-x^2)$ with $x>0$. Then the width is $2x$, the height is $12-x^2$, and the objective is

   $$A(x)=2x\left(12-x^2\right)=24x-2x^3.$$

   Constraint: the vertices must be above the axis, $12-x^2\ge 0$. Domain: $0\le x\le \sqrt{12}=2\sqrt{3}$, a closed interval, so the Closed Interval Method applies ($A$ is a polynomial, hence continuous).

   Solve. $A^{\prime }(x)=24-6x^2=6\left(4-x^2\right)$, zero at $x=2$ (the root $x=-2$ is outside the domain). Compare: $A(0)=0$, $A(2\sqrt{3})=0$, and $A(2)=2\cdot 2\cdot (12-4)=32$. The maximum area is $32$, attained at $x=2$.

   Answer the question asked. The rectangle is $2x=4$ wide and $12-4=8$ tall. Sanity check: $4\times 8=32$.

   (b) Set up. Let the base have side $x$ and the box height $h$. Constraint: $x^{2}h=32$, so $h=\frac{32}{x^2}$. The material is the base plus four sides:

   $$S=x^2+4xh=x^2+4x\cdot \frac{32}{x^2}=x^2+\frac{128}{x}.$$

   Domain: $x>0$, an open interval — no endpoints to test, so the Closed Interval Method is unavailable.

   Solve.

   $$S^{\prime }(x)=2x-\frac{128}{x^2}=\frac{2x^3-128}{x^2}=\frac{2\left(x^3-64\right)}{x^2},$$

   zero exactly when $x^3=64$, that is $x=4$: the only critical number. For $0<x<4$, $x^3-64<0$ so $S^{\prime }<0$; for $x>4$, $S^{\prime }>0$. The derivative is negative everywhere left of the single critical number and positive everywhere right of it, so by the First Derivative Test for Absolute Extrema, $S(4)$ is the absolute minimum on $(0,\infty )$.

   Answer. $x=4$ and $h=\frac{32}{16}=2$: a $4\times 4\times 2$ box, using $S(4)=16+32=48$ square units. Sanity check: base $16$ plus four sides of $4\cdot 2=8$ each gives $16+32=48$. The height is half the base side — the open-topped counterpart of the closed can whose height equals its diameter.

8. 8.Computation · 9 min

   In every part: name the given rate and the wanted rate with units, write an equation between the quantities, differentiate it with respect to $t$, and only then substitute the instantaneous values.

   1. (a)A spherical balloon leaks air at $4\pi {\text{cm}}^3/\text{s}$. How fast is its radius shrinking when the radius is $2$ cm, and how fast is its surface area $S=4\pi r^2$ shrinking at that same instant?
   2. (b)A water trough is $10$ m long, and its cross-section is an isosceles triangle with the vertex at the bottom, $2$ m across the top and $1$ m deep. Water is pumped in at $\frac{1}{2}{\text{m}}^3/\text{min}$. How fast is the water level rising when the water is $0.4$ m deep?
   3. (c)At what depth is the level in (b) rising at exactly $\frac{1}{10}$ m/min?
   Show the solution to question 8

   (a) Volume first. Given $\frac{dV}{dt}=-4\pi {\text{cm}}^3/\text{s}$ (negative: the volume is falling); wanted $\frac{dr}{dt}$ and $\frac{dS}{dt}$ when $r=2$.

   The quantities are tied by $V=\frac{4}{3}\pi r^3$. Differentiating with respect to $t$, using the chain rule on $r^3$,

   $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$$

   Now substitute $r=2$ and $\frac{dV}{dt}=-4\pi$: $-4\pi =16\pi \frac{dr}{dt}$, so $\frac{dr}{dt}=-\frac{1}{4}$ cm/s. The radius shrinks at $\frac{1}{4}$ cm/s.

   Then the surface. From $S=4\pi r^2$, $\frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi \cdot 2\cdot \left(-\frac{1}{4}\right)=-4\pi {\text{cm}}^2/\text{s}$. The area shrinks at $4\pi {\text{cm}}^2/\text{s}$.

   Sanity check: dividing the two rate equations gives $\frac{dS}{dt}=\frac{2}{r}\frac{dV}{dt}$, and with $r=2$ that is $\frac{dS}{dt}=\frac{dV}{dt}$ numerically — consistent.

   (b) Eliminate the width before differentiating. Let $h$ be the depth and $w$ the width of the water surface. Given $\frac{dV}{dt}=\frac{1}{2}{\text{m}}^3/\text{min}$; wanted $\frac{dh}{dt}$ when $h=\frac{2}{5}$.

   The water's cross-section is a triangle similar to the trough's end, so $\frac{w}{h}=\frac{2}{1}$ and $w=2h$. Its area is $\frac{1}{2}wh=\frac{1}{2}(2h)(h)=h^2$, and the volume of water is that area times the length:

   $$V=10h^2.$$

   Differentiate: $\frac{dV}{dt}=20h\frac{dh}{dt}$. Substitute $h=\frac{2}{5}$ and $\frac{dV}{dt}=\frac{1}{2}$:

   $$\frac{1}{2}=20\cdot \frac{2}{5}\cdot \frac{dh}{dt}=8\frac{dh}{dt},\frac{dh}{dt}=\frac{1}{16}\text{m/min}.$$

   (c) Keep the differentiated equation and now treat $h$ as the unknown: $\frac{1}{2}=20h\cdot \frac{1}{10}=2h$, so $h=\frac{1}{4}$ m.

   Sanity check: $\frac{dh}{dt}=\frac{1}{40h}$ is inversely proportional to the depth, so the level rises faster when the water is shallower and the surface narrower; a rate of $\frac{1}{10}$ exceeds $\frac{1}{16}$, and $\frac{1}{4}<\frac{2}{5}$ is indeed shallower.

9. 9.Computation · 9 min

   A particle moves along a straight line for $t\ge 0$ with acceleration $a(t)=12t-30$ (in $\text{m}/{\text{s}}^2$), initial velocity $v(0)=24$ m/s and initial position $s(0)=1$ m.

   1. (a)Find $v(t)$ and $s(t)$, and check your answer by differentiating.
   2. (b)When is the particle at rest, and during which time intervals is it moving in the positive direction?
   3. (c)During which time intervals is the particle speeding up? Justify each interval with the signs of $v$ and $a$. Then find the greatest speed the particle reaches while it is moving in the negative direction.
   4. (d)Find the total distance travelled during $0\le t\le 5$, and compare it with the displacement over the same interval.
   Show the solution to question 9

   (a) Antidifferentiate twice, fixing each constant with its initial condition. $v$ is an antiderivative of $a$: $v(t)=6t^2-30t+C$, and $v(0)=24$ gives $C=24$, so

   $$v(t)=6t^2-30t+24=6(t-1)(t-4).$$

   $s$ is an antiderivative of $v$: $s(t)=2t^3-15t^2+24t+D$, and $s(0)=1$ gives $D=1$, so $s(t)=2t^3-15t^2+24t+1$.

   Check: $s^{\prime }(t)=6t^2-30t+24=v(t)$ and $s^{\prime \prime }(t)=12t-30=a(t)$.

   (b) At rest when $v=0$: $t=1$ s and $t=4$ s. Sign of $v=6(t-1)(t-4)$: positive on $[0,1)$, negative on $(1,4)$, positive on $(4,\infty )$. So the particle moves in the positive direction on $[0,1)$ and for $t>4$, and in the negative direction on $(1,4)$.

   (c) $a(t)=12t-30=6(2t-5)$ is negative for $t<\frac{5}{2}$ and positive for $t>\frac{5}{2}$. The particle speeds up exactly when $v$ and $a$ share a sign: on $\left(1,\frac{5}{2}\right)$ both are negative, and on $(4,\infty )$ both are positive. It slows down on $(0,1)$ and on $\left(\frac{5}{2},4\right)$, where the signs differ.

   Greatest speed while moving backward: on $(1,4)$ the speed is $|v|=-v(t)$, a continuous function on $[1,4]$ that is $0$ at both endpoints. Its derivative is $-a(t)$, zero only at $t=\frac{5}{2}$, so by the Closed Interval Method the maximum is at $t=\frac{5}{2}$:

   $$\text{speed}=-v\left(\frac{5}{2}\right)=-6\cdot \frac{3}{2}\cdot \left(-\frac{3}{2}\right)=\frac{27}{2}\text{m/s}.$$

   This is exactly the instant the acceleration changes sign — the particle stops speeding up and starts slowing down.

   (d) Split at the turning times. $s(0)=1$, $s(1)=2-15+24+1=12$, $s(4)=128-240+96+1=-15$, $s(5)=250-375+120+1=-4$.

   The three legs have lengths $|12-1|=11$, $|-15-12|=27$ and $|-4-(-15)|=11$, so the total distance is $11+27+11=49$ m. The displacement is $s(5)-s(0)=-4-1=-5$ m: the particle ends $5$ m behind where it started, having travelled $49$ m to get there.

10. 10.Proof · 12 min

    Let $f$ be twice differentiable on an interval $I$, let $a$ be a point of $I$, and let $L(x)=f(a)+f^{\prime }(a)(x-a)$ be the linearization of $f$ at $a$. Suppose $|f^{\prime \prime }(t)|\le M$ for every $t$ in $I$.

    1. (a)Prove that $|f(x)-L(x)|\le M(x-a{)}^2$ for every $x$ in $I$. (Consider $g(t)=f(t)-L(t)$ and apply the Mean Value Theorem twice: once to $g$, once to $g^{\prime }$.)
    2. (b)Use the linearization of $f(x)=x^{1/3}$ at $a=27$ to estimate $\sqrt[3]{27.3}$ as an exact fraction, and use (a) to give an explicit bound on the error of that estimate.
    3. (c)Decide, with proof, whether the estimate in (b) is too large or too small.
    Show the solution to question 10

    (a) Proof. Define $g(t)=f(t)-L(t)=f(t)-f(a)-f^{\prime }(a)(t-a)$ for $t$ in $I$. Then

    $$g(a)=0,g^{\prime }(t)=f^{\prime }(t)-f^{\prime }(a)\text{ so }{g}^{\prime }(a)=0,g^{\prime \prime }(t)=f^{\prime \prime }(t).$$

    Because $f$ is twice differentiable on $I$, both $g$ and $g^{\prime }$ are differentiable on $I$, hence continuous there, so the Mean Value Theorem may be applied to either on any closed subinterval of $I$.

    Fix $x$ in $I$. If $x=a$ both sides are $0$. Suppose $x>a$. The Mean Value Theorem for $g$ on $[a,x]$ gives a $c$ in $(a,x)$ with

    $$g(x)-g(a)=g^{\prime }(c)(x-a),\text{i.e.}g(x)=g^{\prime }(c)(x-a).$$

    The Mean Value Theorem for $g^{\prime }$ on $[a,c]$ gives a $d$ in $(a,c)$ with $g^{\prime }(c)-g^{\prime }(a)=g^{\prime \prime }(d)(c-a)$, i.e. $g^{\prime }(c)=f^{\prime \prime }(d)(c-a)$. Combining,

    $$g(x)=f^{\prime \prime }(d)(c-a)(x-a).$$

    Since $d$ is in $I$, $|f^{\prime \prime }(d)|\le M$; and $0<c-a<x-a$. Hence $|g(x)|\le M(c-a)(x-a)\le M(x-a{)}^2$.

    If $x<a$, the same two applications on $[x,a]$ and $[c,a]$ give $g(x)=f^{\prime \prime }(d)(c-a)(x-a)$ with $c$ in $(x,a)$ and $d$ in $(c,a)$; now both $c-a$ and $x-a$ are negative with $|c-a|<|x-a|$, and the same bound follows. In every case $|f(x)-L(x)|\le M(x-a{)}^2$. $■$

    (The constant can be sharpened to $\frac{M}{2}$ by a finer argument; the point here is the order $(x-a{)}^2$, which is what makes linear approximation work.)

    (b) The estimate. $f(27)=3$, and $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3}$ gives $f^{\prime }(27)=\frac{1}{3}\cdot \frac{1}{9}=\frac{1}{27}$. So

    $$L(x)=3+\frac{x-27}{27},\sqrt[3]{27.3}\approx L(27.3)=3+\frac{0.3}{27}=3+\frac{1}{90}=\frac{271}{90}.$$

    The bound. Take $I=[27,27.3]$. $f^{\prime \prime }(x)=-\frac{2}{9}{x}^{-5/3}$, so $|f^{\prime \prime }(t)|=\frac{2}{9}{t}^{-5/3}$, which decreases as $t$ grows; its largest value on $I$ is at $t=27$:

    $$M=\frac{2}{9}\cdot 27^{-5/3}=\frac{2}{9}\cdot \frac{1}{3^5}=\frac{2}{9\cdot 243}=\frac{2}{2187}.$$

    By (a), with $x-a=0.3=\frac{3}{10}$,

    $$\left|\sqrt[3]{27.3}-\frac{271}{90}\right|\le \frac{2}{2187}\cdot \frac{9}{100}=\frac{18}{218700}=\frac{1}{12150}<10^{-4}.$$

    So the fraction $\frac{271}{90}=3.0111\dots$ is correct to at least four decimal places.

    (c) Too large. The proof of (a) gave the signed identity $g(x)=f^{\prime \prime }(d)(c-a)(x-a)$. Here $x-a=0.3>0$, $c-a>0$, and $f^{\prime \prime }(d)=-\frac{2}{9}{d}^{-5/3}<0$ because $d>0$. So $g(27.3)=f(27.3)-L(27.3)<0$: the true value lies below the tangent line, and the estimate is too large. This is exactly the picture of concavity: $f^{\prime \prime }<0$ on $(0,\infty )$, so $x^{1/3}$ is concave downward there and its graph lies below every tangent.

    Sanity check: $3.01107^3\approx 27.29999$, so $\sqrt[3]{27.3}\approx 3.01107$, about $4\times 10^{-5}$ below $\frac{271}{90}\approx 3.01111$ and well inside the bound.

11. 11.Computation · 8 min

    Let $f(x)=x^3-5x$, whose roots are $0$, $\sqrt{5}$ and $-\sqrt{5}$. Keep every iterate as an exact fraction; no decimals are needed anywhere.

    1. (a)Write down Newton's iteration for $f$ and simplify it to the form $x_{n+1}=\frac{2x_n^3}{3x_n^2-5}$.
    2. (b)Start at $x_0=1$. Compute $x_1$ and $x_2$, describe what the method does from this start, and explain the behaviour geometrically in terms of tangent lines. Find also the two starting values for which $x_1$ does not exist at all.
    3. (c)Start instead at $x_0=\frac{1}{10}$ and compute $x_1$ exactly. The quadratic-convergence estimate says $|e_{n+1}|\approx \frac{|f^{\prime \prime }(r)|}{2|f^{\prime }(r)|}|e_n{|}^2$ near a root $r$; evaluate that constant at $r=0$ and reconcile it with your $x_1$.
    Show the solution to question 11

    (a) $f^{\prime }(x)=3x^2-5$, so Newton's step is

    $$x_{n+1}=x_n-\frac{x_n^3-5x_n}{3x_n^2-5}=\frac{x_n\left(3x_n^2-5\right)-\left(x_n^3-5x_n\right)}{3x_n^2-5}=\frac{3x_n^3-5x_n-x_n^3+5x_n}{3x_n^2-5}=\frac{2x_n^3}{3x_n^2-5}.$$

    (b) A two-cycle. From $x_0=1$: $x_1=\frac{2}{3-5}=-1$, and then $x_2=\frac{2(-1{)}^3}{3-5}=\frac{-2}{-2}=1$. The iterates are $1,-1,1,-1,\dots$ forever: the method neither converges nor diverges, it oscillates, and never comes near any of the three roots.

    Geometrically: $f(1)=-4$ and $f^{\prime }(1)=-2$, so the tangent at $(1,-4)$ is $y=-4-2(x-1)$, which meets the $x$-axis where $-2(x-1)=4$, i.e. at $x=-1$. Because $f$ is odd, $f(-1)=4$ and $f^{\prime }(-1)=-2$: the tangent at $(-1,4)$ is parallel to the first one and lands at $x=1$. The two tangent lines simply hand the point back and forth across the origin.

    $x_1$ fails to exist when $f^{\prime }(x_0)=0$, i.e. $3x_0^2=5$, so $x_0=\pm \sqrt{5/3}=\pm \frac{\sqrt{15}}{3}$. These are the local extrema of $f$, where the tangent is horizontal and never meets the axis — Newton's formula divides by zero.

    (c) Cubic, not quadratic. With $x_0=\frac{1}{10}$:

    $$x_1=\frac{2\cdot \frac{1}{1000}}{\frac{3}{100}-5}=\frac{\frac{1}{500}}{-\frac{497}{100}}=-\frac{1}{500}\cdot \frac{100}{497}=-\frac{1}{2485}.$$

    So one step takes the error from $|e_0|=\frac{1}{10}$ to $|e_1|=\frac{1}{2485}\approx 0.0004$.

    The estimate's constant at $r=0$: $f^{\prime \prime }(x)=6x$, so $f^{\prime \prime }(0)=0$, while $f^{\prime }(0)=-5\ne 0$. The constant is $\frac{0}{2\cdot 5}=0$ — the proposition applies (its hypothesis $f^{\prime }(r)\ne 0$ holds) and predicts that the $|e_n{|}^2$ term vanishes. The actual behaviour is one order better: for small $x_n$ the iteration reads

    $$x_{n+1}=\frac{2x_n^3}{3x_n^2-5}\approx -\frac{2}{5}{x}_n^3,$$

    and indeed $-\frac{2}{5}\cdot \frac{1}{1000}=-\frac{1}{2500}$, very close to $-\frac{1}{2485}$. The root $0$ is also an inflection point of $f$, so the tangent line matches the curve to third order there and the errors cube at each step instead of squaring. Quadratic convergence is the worst case the proposition promises, not the best.

12. 12.Proof · 10 min

    Both parts concern a quantity and its rate of change. Name every theorem or corollary you use, and check the interval hypothesis wherever a corollary of the Mean Value Theorem is invoked.

    1. (a)Let $C(x)$ be a firm's cost of producing $x$ units, differentiable for $x>0$, and let $c(x)=\frac{C(x)}{x}$ be the average cost. Prove that at any critical number of $c$, marginal cost equals average cost: $C^{\prime }(x)=c(x)$.
    2. (b)Suppose in addition that $C^{\prime \prime }(x)>0$ for all $x>0$ and that $c$ has a critical number $x_0$. Prove that $x_0$ is the only critical number of $c$, and that $c(x_0)$ is the absolute minimum of average cost on $(0,\infty )$.
    3. (c)A population satisfies $P^{\prime }(t)=kP(t)$ for all real $t$, where $k$ is a constant. Prove that $P(t)=P(0)e^{kt}$ for all $t$ — a constant relative growth rate forces exponential growth and nothing else. (Consider $g(t)=P(t)e^{-kt}$.)
    Show the solution to question 12

    (a) Proof. For $x>0$, $c$ is a quotient of differentiable functions with nonzero denominator, so by the quotient rule

    $$c^{\prime }(x)=\frac{x{C}^{\prime }(x)-C(x)}{x^2}.$$

    Since $c^{\prime }$ exists at every $x>0$, a critical number of $c$ is a solution of $c^{\prime }(x)=0$, i.e. of $x{C}^{\prime }(x)-C(x)=0$. Dividing by $x>0$,

    $$C^{\prime }(x)=\frac{C(x)}{x}=c(x).■$$

    (b) Proof. Let $N(x)=x{C}^{\prime }(x)-C(x)$, so that $c^{\prime }(x)=\frac{N(x)}{x^2}$ and $c^{\prime }$ has the sign of $N$. Since $C^{\prime \prime }$ exists, $C^{\prime }$ is differentiable and so is $N$, with

    $$N^{\prime }(x)=C^{\prime }(x)+x{C}^{\prime \prime }(x)-C^{\prime }(x)=x{C}^{\prime \prime }(x)>0(x>0).$$

    By the Increasing/Decreasing Test on the interval $(0,\infty )$, $N$ is increasing there.

    Uniqueness. By (a), $N(x_0)=0$. If $N$ had a second zero $x_1\ne x_0$, say $x_0<x_1$, then increasing would force $N(x_0)<N(x_1)$, i.e. $0<0$ — impossible (and symmetrically if $x_1<x_0$). So $x_0$ is the only zero of $N$, hence the only critical number of $c$.

    Absolute minimum. Because $N$ is increasing with $N(x_0)=0$, we have $N(x)<0$ for $0<x<x_0$ and $N(x)>0$ for $x>x_0$. Therefore $c^{\prime }(x)<0$ for all $x<x_0$ in $(0,\infty )$ and $c^{\prime }(x)>0$ for all $x>x_0$. The function $c$ is continuous on $(0,\infty )$, so the First Derivative Test for Absolute Extrema applies and $c(x_0)$ is the absolute minimum of average cost. $■$

    The economic reading: while the next unit costs less than the running average ($C^{\prime }<c$), producing it pulls the average down; once it costs more, the average rises. The bottom is where the two agree.

    (c) Proof. Put $g(t)=P(t)e^{-kt}$ for all real $t$. By the product and chain rules,

    $$g^{\prime }(t)=P^{\prime }(t)e^{-kt}+P(t)\cdot \left(-k{e}^{-kt}\right)=(P^{\prime }(t)-kP(t))e^{-kt}=0\cdot e^{-kt}=0$$

    for every $t$, using the hypothesis $P^{\prime }=kP$. The domain of $g$ is the whole real line, which is an interval, so the corollary “zero derivative means constant” applies: $g(t)=g(0)$ for all $t$, and $g(0)=P(0)e^0=P(0)$. Hence $P(t)e^{-kt}=P(0)$, i.e.

    $$P(t)=P(0)e^{kt}.■$$

    The interval hypothesis is not decoration: had the population been observed only on two disjoint intervals of time, the constant $g$ could differ on the two pieces, and the two stretches of exponential growth would not need to line up.