Applications of Derivatives

(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.

(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 $\pm 1$.

(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 $-\frac{1}{2}$.)

(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.

(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.)

(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.

(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.

(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.

(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.

(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.

(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.