Corrected exercises: complete curve sketching (201-103)

$f'(x)=6x-12$, zero at $x=2$: a single critical number. $f''(x)=6$ never vanishes and stays positive: the curve is concave up everywhere, with no inflection point. We recover the vertex of the parabola, $x=-\frac{b}{2a}=\frac{12}{6}=2$.

$f'(x)=6x^{2}-6x-36=6(x^{2}-x-6)=6(x-3)(x+2)$: critical numbers $-2$ and $3$. $f''(x)=12x-6$ vanishes at $x=\frac{1}{2}$ and goes from negative to positive: inflection point at $x=0.5$. Check: for a cubic, the inflection point is midway between the two critical numbers, $\frac{-2+3}{2}=0.5$.

$f'(x)=3x^{2}+6x+6=3(x^{2}+2x+2)$, with discriminant $4-8=-4<0$: $f'$ never vanishes and stays positive, since $f'(x)=3\left((x+1)^{2}+1\right)$. No critical number, $f$ is increasing on $\mathbb{R}$. Yet $f''(x)=6x+6$ vanishes and changes sign at $x=-1$: inflection point at $x=-1$. Pitfall: believing that a function with no critical number has no inflection point.

$f'(x)=1.5x^{2}-4x+1.5$, with discriminant $16-4\times 1.5\times 1.5=16-9=7$. So $x=\frac{4\pm\sqrt{7}}{2\times 1.5}=\frac{4\pm\sqrt{7}}{3}$: $x_{1}\approx 0.451$ and $x_{2}\approx 2.215$. Along the way, $f''(x)=3x-4$ puts the inflection point at $x=\frac{4}{3}$, midway between the two. Pitfall: dividing by $2$ instead of $2a=3$.

$f'(x)=12x^{3}-12x^{2}-24x=12x(x^{2}-x-2)=12x(x-2)(x+1)$: critical numbers $-1$, $0$ and $2$. $f''(x)=36x^{2}-24x-24=12(3x^{2}-2x-2)$, with roots $x=\frac{2\pm\sqrt{28}}{6}=\frac{1\pm\sqrt{7}}{3}$, simple roots and therefore a sign change: inflection points at $x\approx -0.549$ and $x\approx 1.215$. Pitfall: dividing by $x$ instead of factoring it out, and losing the critical number $0$.

$f'(x)=4x^{3}-12x^{2}+12x=4x(x^{2}-3x+3)$; the quadratic has discriminant $9-12<0$, so there is a single critical number, $x=0$. $f''(x)=12x^{2}-24x+12=12(x-1)^{2}$ vanishes at $x=1$ but stays POSITIVE on both sides: the concavity does not change, so there is no inflection point. Pitfall: declaring an inflection point wherever $f''=0$, when it is the sign change that counts.

Domain $\mathbb{R}\setminus\{0\}$. $f'(x)=2-\frac{18}{x^{2}}=\frac{2(x^{2}-9)}{x^{2}}$, zero at $x=-3$ and $x=3$; it does not exist at $0$, but $0$ is not in the domain, so it is not a critical number. $f''(x)=\frac{36}{x^{3}}$ never vanishes; it changes sign at $0$, outside the domain, so there is no inflection point. Smaller critical number: $-3$.

$f'(x)=\frac{-12x}{(x^{2}+3)^{2}}$, zero at $x=0$ only. Quotient rule for $f''$: $f''(x)=\frac{-12(x^{2}+3)^{2}+12x\cdot 2(x^{2}+3)\cdot 2x}{(x^{2}+3)^{4}}=\frac{-12(x^{2}+3)+48x^{2}}{(x^{2}+3)^{3}}=\frac{36(x^{2}-1)}{(x^{2}+3)^{3}}$, after cancelling a factor $x^{2}+3$. It vanishes and changes sign at $x=-1$ and $x=1$: the positive inflection point is at $x=1$. Pitfall: forgetting the factor $2x$ from the chain rule when differentiating $(x^{2}+3)^{2}$.

Domain $\mathbb{R}\setminus\{2\}$. $f'(x)=\frac{2x(x-2)-x^{2}}{(x-2)^{2}}=\frac{x^{2}-4x}{(x-2)^{2}}=\frac{x(x-4)}{(x-2)^{2}}$: critical numbers $0$ and $4$, the value $2$ being outside the domain. Writing $f(x)=x+2+\frac{4}{x-2}$, we get $f''(x)=\frac{8}{(x-2)^{3}}$, never zero: no inflection point.

Domain $(-\infty, 6]$. Product rule: $f'(x)=\sqrt{6-x}+x\cdot\frac{-1}{2\sqrt{6-x}}=\frac{2(6-x)-x}{2\sqrt{6-x}}=\frac{12-3x}{2\sqrt{6-x}}$. It vanishes at $x=4$ and does not exist at $x=6$, which IS in the domain: two critical numbers, $4$ and $6$. Next, $f''(x)=\frac{3x-24}{4(6-x)^{\frac{3}{2}}}<0$ on the domain: the curve is concave down, with no inflection point. Pitfall: forgetting the endpoint $6$, where the tangent line becomes vertical.

The only forbidden value is $d=0$, so the algebraic domain is $\mathbb{R}\setminus\{0\}$. Since a density is positive, the domain of validity is $(0, +\infty)$: that is the one we study.

As $d\to 0^{+}$, the term $\dfrac{4{,}000}{d}$ tends to $+\infty$ and the other two stay bounded, so $g(d)\to+\infty$. The line $d=0$ is a VERTICAL ASYMPTOTE. The sign is indeed $+\infty$ because $d$ approaches zero through positive values, the only side that exists in this context.

As $d\to+\infty$, $\dfrac{4{,}000}{d}\to 0$ but $0.9d\to+\infty$, so $g(d)\to+\infty$. There is NO horizontal asymptote: a horizontal asymptote requires a finite limit, which is not the case here.

We calculate the difference between the curve and the line: $g(d)-(0.9d+120)=\dfrac{4{,}000}{d}$. This difference tends to 0 as $d\to+\infty$, which is exactly the definition of an OBLIQUE asymptote. Since $\dfrac{4{,}000}{d}>0$ on the domain, the curve stays ABOVE its asymptote and approaches it from above. The general method is this: separate the polynomial part of degree 1 from the remainder, and if the remainder tends to 0, the polynomial part is the asymptote.

The vertical asymptote says that at very low density, the cost per dwelling explodes: each isolated house requires its own length of water main, road and power line, and there is nobody to share it. The oblique asymptote says that at very high density the cost per dwelling does not level off, it STARTS RISING AGAIN with a slope of 90 cents per dwelling per hectare: you have to build upward, dig down, reinforce the networks. A curve that rises at both ends necessarily has a minimum in between, and that is the subject of the next exercise.

We rewrite $g(d)=4{,}000\,d^{-1}+0.9d+120$, so $g'(d)=-4{,}000\,d^{-2}+0.9=0.9-\dfrac{4{,}000}{d^{2}}$, in dollars per dwelling per hectare.

$0.9-\dfrac{4{,}000}{d^{2}}=0$ gives $d^{2}=\dfrac{4{,}000}{0.9}\approx 4{,}444.4$, so $d=\sqrt{4{,}444.4}\approx 66.7$ dwellings per hectare. The negative root is rejected by the domain. We keep about 67 dwellings per hectare.

For $d<66.7$, $d^{2}$ is small so $\dfrac{4{,}000}{d^{2}}>0.9$ and $g'(d)<0$: the function DECREASES. For $d>66.7$, the inequality reverses and $g'(d)>0$: the function INCREASES. The critical point is therefore a minimum, and it is ABSOLUTE on the domain since the function strictly decreases before it and strictly increases after it.

$g(66.7)=\dfrac{4{,}000}{66.67}+0.9\times 66.67+120=60+60+120=240$ dollars per dwelling. The two variable terms are exactly 60 each, which is no accident: at the minimum, the shared-network term and the build-upward term balance. Second derivative test: $g''(d)=\dfrac{8{,}000}{d^{3}}>0$ on the whole domain, so the curve is concave up everywhere and the critical point is indeed a minimum.

Table: on $(0, 66.7)$, $g'<0$ and $g$ decreases from $+\infty$ to 240; at $d\approx 66.7$, $g'=0$ and $g$ reaches its minimum of 240 dollars; on $(66.7, +\infty)$, $g'>0$ and $g$ increases from 240 toward $+\infty$ along its oblique asymptote. Management reading: there is an OPTIMAL service density, about 67 dwellings per hectare, and moving away from it costs more in both directions. Since the curve is fairly flat around the minimum, with $g(50)=245$ and $g(90)\approx 245$ as well, the right recommendation is not one number but a RANGE, from 50 to 90 dwellings per hectare, which is far more robust information than an optimum to the nearest tenth.

$V'(t)=3t^{2}-48t+180=3\,(t^{2}-16t+60)=3\,(t-6)(t-10)$, the factorization coming from $6+10=16$ and $6\times 10=60$. Differentiating and then factoring is what makes the sign study immediate; keeping the expanded form forces a discriminant table.

The critical points are $t=6$ and $t=10$. The product $(t-6)(t-10)$ is POSITIVE outside the roots and NEGATIVE between them, and the factor 3 changes nothing. So $V'>0$ on $[0, 6)$, $V'<0$ on $(6, 10)$ and $V'>0$ on $(10, 16]$.

$V$ increases on $[0, 6]$, decreases on $[6, 10]$, then increases on $[10, 16]$. At $t=6$, the derivative goes from positive to negative: it is a local MAXIMUM. At $t=10$, it goes from negative to positive: it is a local MINIMUM. This is the first derivative test, and it needs no second derivative.

$V(6)=216-864+1{,}080+400=832$ and $V(10)=1{,}000-2{,}400+1{,}800+400=800$. The gap between the peak and the trough is only 32 points: the reform produced a peak followed by a slight dip, and not the collapse that a badly framed graph might suggest.

NO, and that is the trap of the exercise. We must compare with the values at the ENDPOINTS: $V(0)=400$ and $V(16)=4{,}096-6{,}144+2{,}880+400=1{,}232$. The absolute maximum is therefore $1{,}232$, reached at $t=16$, and not 832; the absolute minimum is 400, reached at $t=0$, and not 800. Neither of the two LOCAL extrema is absolute. The word “local” is not a vocabulary ornament: it means that the comparison was made only with the immediate neighbours, and exercise 7 will draw the complete method from it.

$V'(t)=3t^{2}-48t+180$, so $V''(t)=6t-48=6\,(t-8)$. It is NEGATIVE on $[0, 8)$ and POSITIVE on $(8, 16]$. The curve is therefore concave down before $t=8$, and concave up after.

$V''$ vanishes at $t=8$ AND changes sign there, so $t=8$ is an INFLECTION POINT. Its y-coordinate is $V(8)=512-1{,}536+1{,}440+400=816$. Both conditions are necessary: a second derivative that vanishes without changing sign gives no inflection point, as part d) will show.

$V''(6)=6\times(-2)=-12<0$, so the curve is concave down at that point: $t=6$ is a local MAXIMUM. $V''(10)=6\times 2=12>0$, concave up: $t=10$ is a local MINIMUM. We recover the conclusions of exercise 3 in two evaluations, with no sign chart at all. That is the point of the test, and it is why we use it when the second derivative is easy to calculate.

For $f(x)=x^{4}$: $f'(x)=4x^{3}$ vanishes at 0, and $f''(x)=12x^{2}$ is ALSO 0 there. The test is inconclusive. Yet $f(x)=x^{4}>0$ for every $x\ne 0$, so $x=0$ is a minimum. For $h(x)=x^{3}$: $h'(x)=3x^{2}$ and $h''(x)=6x$ both vanish at 0, the test is inconclusive again, and this time $x=0$ is NOT an extremum, since $h$ is strictly increasing. Two opposite situations, one and the same verdict from the test: it is SILENT.

The rule: when $f''(a)=0$ at a critical point, the second derivative test says nothing, in either direction. You must then go back to the FIRST DERIVATIVE TEST, which examines the SIGN of $f'$ on either side of the point and which always reaches a conclusion. For $x^{4}$, $f'$ goes from negative to positive, so a minimum; for $x^{3}$, $f'$ stays positive on both sides, so no extremum. The first derivative test is slower and never silent; the second derivative test is fast and fallible. On an exam, start with the second and fall back on the first as soon as you get zero.

$N(0)=\dfrac{9{,}000}{1+29}=\dfrac{9{,}000}{30}=300$ adoptions at launch. As $t\to+\infty$, $e^{-0.5t}\to 0$, so the denominator tends to 1 and $N(t)\to 9{,}000$. The line $y=9{,}000$ is the horizontal asymptote, that is, the size of the attainable market.

We differentiate $N(t)=9{,}000\,\bigl(1+29e^{-0.5t}\bigr)^{-1}$ by the chain rule: $N'(t)=-9{,}000\,\bigl(1+29e^{-0.5t}\bigr)^{-2}\times 29\times(-0.5)\,e^{-0.5t}$. The two minus signs cancel and $9{,}000\times 29\times 0.5=130{,}500$, hence the result. The numerator is strictly positive and the denominator is a square, so $N'>0$ everywhere: cumulative adoptions can only grow, which is the nature of a cumulative total.

$29\,e^{-0.5t}=1$ gives $e^{-0.5t}=\dfrac{1}{29}$, so $-0.5t=-\ln 29$ and $t=2\ln 29\approx 2\times 3.367=6.73$ months. The y-coordinate is then $N=\dfrac{9{,}000}{1+1}=4{,}500$, exactly HALF the ceiling. This is a general property of the logistic model: its inflection point is always at half height.

$N'(6.73)=\dfrac{130{,}500\times\frac{1}{29}}{(1+1)^{2}}=\dfrac{4{,}500}{4}=1{,}125$ adoptions per month. This is the MAXIMUM rate: before it, the spread speeds up; after it, it slows down while remaining positive.

It is not the ceiling, which everyone can guess, it is the DATE of the inflection point. It says when the campaign stops becoming more effective: through the seventh month, each month recruits more than the previous one, and the promotional budget justifies itself; after that, the same effort produces less and less, because there are fewer and fewer people left to convince. The decision to cut or redirect spending is therefore made around $t=6.7$, and not when the curve looks flat, which would come too late. Note also that this model is an assumption, not an observation: the ceiling of 9,000 and the coefficient $0.5$ were fitted on the first months, so the date of the inflection point must be recalculated as data arrive.

The product $(t-2)(t-7)$ is positive outside the roots and negative between them, and the factor $0.2$ changes nothing. So $f'>0$ on $[0, 2)$, $f'<0$ on $(2, 7)$ and $f'>0$ on $(7, 10]$. The indicator $f$ INCREASES on $[0, 2]$, DECREASES on $[2, 7]$, then increases again on $[7, 10]$.

At $t=2$, $f'$ goes from positive to negative: $f$ reaches a local MAXIMUM there. At $t=7$, $f'$ goes from negative to positive: $f$ reaches a local MINIMUM there. These two conclusions are read solely from the SIGN changes of the given curve, and not from its height.