Corrected exercises: functions and their domain (201-103)

No restriction: the only denominator is the number $5$, which is never zero. The domain is $\mathbb{R}=(-\infty, +\infty)$. Pitfall: seeing a fraction and looking for a value to exclude.

Condition $2x-7\neq 0$, that is $x\neq\frac{7}{2}$. Domain $\mathbb{R}\setminus\left\{\frac{7}{2}\right\}$; the excluded value is $\frac{7}{2}=3.5$. The numerator plays no role: $x=-4$ makes $g$ zero, it does not forbid it.

Condition $x^{2}+x-12\neq 0$. Factor: $x^{2}+x-12=(x+4)(x-3)$, zero at $x=-4$ and at $x=3$. Domain $\mathbb{R}\setminus\{-4, 3\}$. Pitfall: excluding $x=0$, which makes the numerator $5x$ zero, not the denominator.

Condition $2x-5\ge 0$, that is $x\ge\frac{5}{2}$. Domain $[2.5, +\infty)$, CLOSED bracket at $2.5$ since $\sqrt{0}=0$ exists.

Condition $9-4x\ge 0$, that is $-4x\ge -9$, so $x\le\frac{9}{4}$: we divide by a NEGATIVE number, so the inequality reverses. Domain $(-\infty, 2.25]$. Pitfall: keeping the direction and writing $[2.25, +\infty)$, which is almost exactly the forbidden values.

Two conditions in one: the root requires $x+6\ge 0$, the division requires $\sqrt{x+6}\neq 0$, so $x+6>0$ and $x>-6$. Domain $(-6, +\infty)$, OPEN bracket at $-6$. Pitfall: the closed bracket of the root alone.

Condition $12+x-x^{2}\ge 0$. Factor: $12+x-x^{2}=-(x^{2}-x-12)=-(x-4)(x+3)=(4-x)(x+3)$. The roots are $-3$ and $4$; since the coefficient of $x^{2}$ is negative, the quadratic is positive BETWEEN the roots, which $x=0$ confirms, where it equals $12$. Domain $[-3, 4]$. Pitfall: the reflex “outside the roots”, which only holds when the coefficient of $x^{2}$ is positive.

Root: $x+2\ge 0$, that is $x\ge -2$. Denominator: $x\neq 3$, and $3$ is indeed in $[-2, +\infty)$, so it must be removed. Domain $[-2, 3)\,\cup\,(3, +\infty)$: the smallest value in the domain is $-2$, the excluded value is $3$.

Numerator: $x+1\ge 0$, that is $x\ge -1$. Denominator: the root requires $7-x\ge 0$ and the division requires $\sqrt{7-x}\neq 0$, so $7-x>0$, that is $x<7$. Keep both: $[-1, 7)$, closed on the left, open on the right. Pitfall: treating both roots the same way, when only the one in the denominator becomes a strict inequality.

Condition $8-2x>0$, that is $-2x>-8$, so $x<4$, the inequality reversing when we divide by $-2$. Domain $(-\infty, 4)$, open at $4$ since $\ln 0$ does not exist.

Logarithm: $x-1>0$, that is $x>1$. Denominator: $\ln(x-1)\neq 0$; a logarithm equals $0$ when its argument equals $1$, so $x-1\neq 1$, that is $x\neq 2$. Domain $(1, 2)\,\cup\,(2, +\infty)$: left endpoint $1$, excluded value $2$. Pitfall: excluding only $x=1$, by confusing “the logarithm is zero” with “the logarithm does not exist”.

Treat each piece ON ITS OWN INTERVAL. For $x<1$, $\sqrt{x+5}$ requires $x\ge -5$: keep $[-5, 1)$. For $x\ge 1$, $\frac{6}{x+3}$ forbids only $-3$, which is not in this interval: keep $[1, +\infty)$. The domain is the union, $[-5, +\infty)$. Pitfall: removing $-3$, when at $x=-3$ it is the root that applies, and it equals $\sqrt{2}$.

Yes. The VERTICAL LINE TEST consists of sweeping a vertical line across the graph: if no position of that line meets the curve more than once, each input value receives at most one output, which is exactly the definition of a function. Here the curve is a parabola with a vertical axis, so each year receives a single rate. This test checks neither that the curve is continuous nor that it is increasing: it checks the UNIQUENESS of the output, nothing else.

The fitted domain is $[0, 12]$, the thirteen years from 2012 to 2024. For the range, we need the vertex: $t=-\dfrac{b}{2a}=\dfrac{0.8}{2\times 0.05}=8$, and $T(8)=0.05\times 64-6.4+8=3.2-6.4+8=4.8$. At the endpoints, $T(0)=8$ and $T(12)=0.05\times 144-9.6+8=7.2-9.6+8=5.6$. The maximum is therefore 8, reached at $t=0$, and the range is $[4.8, 8]$. You must compare BOTH endpoints with the vertex: settling for $T(12)$ would give a wrong range.

Solve $0.05t^{2}-0.8t+8=5.6$, that is $0.05t^{2}-0.8t+2.4=0$. Multiplying by 20: $t^{2}-16t+48=0$, so $t=\dfrac{16\pm\sqrt{256-192}}{2}=\dfrac{16\pm 8}{2}$, that is $t=4$ and $t=12$. TWO years, 2016 and 2024, give the same rate. Direct consequence: $T$ is not one-to-one on $[0, 12]$, so it has NO inverse there. Knowing the rate does not let you recover the year, which is the normal situation for an indicator that falls and then rises.

The INDEPENDENT variable is $t$, time, in years since 2012: it is the one we choose. The DEPENDENT variable is $T$, the unemployment rate, as a percentage of the labour force: it is the one we read. In function notation, the independent variable is always the one written in parentheses.

$T(30)=0.05\times 900-0.8\times 30+8=45-24+8=29$. The model announces 29% unemployment. This number means nothing, for two reasons. First, $t=30$ is OUTSIDE the fitted domain: the model was built on thirteen years, it says nothing about the thirtieth. Second, its very shape condemns it, a parabola opening upward and growing without bound: it will eventually predict 200% unemployment, which is impossible. A model always has two domains, the ALGEBRAIC domain, here $\mathbb{R}$, and the domain of VALIDITY, here $[0, 12]$. Confusing them is the costliest error of the course, because it produces no error message: it produces a number.

The only risk is in the denominator. $x^{2}-9=0$ gives $x=3$ or $x=-3$. The domain is therefore $\mathbb{R}\setminus\{-3, 3\}$, that is $(-\infty, -3)\,\cup\,(-3, 3)\,\cup\,(3, +\infty)$. The numerator $2x+5$ restricts nothing: a zero numerator gives a zero value, not a forbidden value. Confusing “the numerator is zero” with “the function does not exist” is the first pitfall of the chapter, and it still costs marks in curve sketching, where $x=-2.5$ will be a ZERO and not an asymptote.

The square root requires $12-3x\ge 0$, so $3x\le 12$ and $x\le 4$. The domain is $(-\infty, 4]$. The inequality is NON-STRICT, because $\sqrt{0}=0$ exists. The direction of the inequality does not change here because we divide by $+3$; with a negative coefficient, it would have had to be reversed.

The logarithm requires a STRICTLY positive argument: $x-2>0$, so $x>2$. The domain is $(2, +\infty)$, open bracket. This is the difference to hold on to between the root and the logarithm: $\sqrt{0}$ exists, $\ln 0$ does not.

Two conditions to satisfy AT THE SAME TIME. The root in the numerator requires $x-1\ge 0$, so $x\ge 1$. The denominator requires $x\ne 5$. The domain is the intersection: $[1, 5)\,\cup\,(5, +\infty)$. The classic error is to handle only one: you read the root, forget the denominator, and then differentiate a function that does not exist at 5.

He is wrong for INTEGER powers: $(-2)^{3}=-8$ is perfectly computable. He is right, and that is the whole point of the remark, for FRACTIONAL powers with an even denominator, since $x^{1/2}$ is exactly $\sqrt{x}$: that is the even-root restriction, already counted, not a fourth one. The three-item list, zero denominator, even root of a negative, logarithm of a number that is not strictly positive, is enough for all of 201-103.

$x=12$ falls in the first bracket: $C(12)=45$ dollars. $x=35$ falls in the second: $C(35)=45+7\times(35-20)=45+105=150$ dollars. $x=60$ falls in the third: $C(60)=250$ dollars. The one move to take care over is to READ the condition before choosing the formula; applying the second formula to $x=60$ would give $45+280=325$ dollars, an amount nobody pays.

The domain is $[0, +\infty)$: you can use any positive amount. For the range, the first bracket gives only the value 45; the second sweeps continuously from $45$ excluded up to $C(50)=45+7\times 30=255$ included; the third gives only 250, already reached in the second. The range is therefore $[45, 255]$.

At $x=20$: the left-hand limit is 45, the right-hand limit is $45+7\times 0=45$, and $C(20)=45$. The three coincide, so $C$ is CONTINUOUS at 20. At $x=50$: the left-hand limit is $45+7\times 30=255$ and $C(50)=255$, but the right-hand limit is 250. The left-hand and right-hand limits differ, so $C$ is DISCONTINUOUS at 50, with a downward jump of 5 dollars.

Compare the formula of the second bracket with the unlimited rate: $45+7(x-20)>250$ gives $7(x-20)>205$, then $x-20>\dfrac{205}{7}\approx 29.29$, so $x>49.29$ GB approximately. Beyond $49.29$ GB, the subscriber would pay more under the bracket formula than for the unlimited plan. This is precisely the zone where the jump occurs, and that is why the company placed it there: above this threshold, the customer would ask for the unlimited plan anyway.

A tax schedule is CONTINUOUS by construction, and that continuity is a political decision: if the rate applied to all of your income as soon as you crossed a bracket, earning one more dollar could cost you hundreds of dollars net, which would discourage work at the edge of the bracket. That is why a marginal rate applies ONLY to the part of the income inside the bracket. A commercial plan has no such constraint: the downward jump BENEFITS the customer and serves as a lure toward the unlimited offer, and nobody complains about paying less. The continuity of a piecewise function is therefore not a mathematical property one puts up with, it reflects a choice.

$P(q)=R(q)-C(q)=60q-0.1q^{2}-(0.2q^{2}+18q+500)$. Distribute the minus sign over ALL THREE terms, which is forgotten every other time: $P(q)=60q-0.1q^{2}-0.2q^{2}-18q-500=-0.3q^{2}+42q-500$. The cheapest check is to evaluate at $q=0$: $P(0)=-500$, and we do recover the loss equal to the fixed cost when nothing is sold.

$\overline{C}(q)=\dfrac{0.2q^{2}+18q+500}{q}=0.2q+18+\dfrac{500}{q}$. The algebraic domain is $\mathbb{R}\setminus\{0\}$, and the domain of the context is $(0, +\infty)$. That is the point of the chapter: $C$ exists at $q=0$, where it equals 500 dollars, but its AVERAGE cost does not exist at 0. The division removed a value that neither the numerator nor the denominator posed as a problem before the operation.

$P(50)=-0.3\times 2{,}500+42\times 50-500=-750+2{,}100-500=850$ dollars. $C(50)=0.2\times 2{,}500+18\times 50+500=500+900+500=1{,}900$ dollars, so $\overline{C}(50)=\dfrac{1{,}900}{50}=38$ dollars per unit. Finally $R(50)=60\times 50-0.1\times 2{,}500=3{,}000-250=2{,}750$ dollars, and the ratio is $\dfrac{2{,}750}{1{,}900}\approx 1.45$: each dollar spent brings in about 1.45 dollars. We can cross-check: $R(50)-C(50)=2{,}750-1{,}900=850$, which confirms $P(50)$.

$R-C$ is the PROFIT, the most useful of the four. $\dfrac{R}{C}$ is a profitability ratio with no unit, perfectly interpretable. $R+C$ has no direct meaning, since we do not add what comes in and what goes out, though the sum does appear in certain gross-flow calculations. $R\times C$ has NO meaning: its unit would be the square dollar. Algebra allows all four operations, the context keeps only two or three; knowing which of the two is in charge is exactly what the course teaches.

For the sum, the difference and the product, the domain is the INTERSECTION of the two domains: the value must be allowed for both functions at once. For the quotient $\dfrac{f}{g}$, it is that same intersection WITHOUT the values where $g$ is zero. The rule never says “union”: a value forbidden for one function stays forbidden for the operation, even if the other can live with it.

$(f\circ g)(x)=f\!\left(\dfrac{1}{x-7}\right)=\sqrt{\dfrac{1}{x-7}-3}$. Two conditions pile up: first $x\ne 7$ so that the inside exists, then $\dfrac{1}{x-7}-3\ge 0$, that is $\dfrac{1}{x-7}\ge 3$. If $x<7$, the quotient is negative and cannot exceed 3: no solution. If $x>7$, we can multiply by $x-7>0$ without reversing the inequality: $1\ge 3(x-7)$, so $x-7\le \dfrac{1}{3}$ and $x\le \dfrac{22}{3}$. The domain is $\left(7, \dfrac{22}{3}\right]$, or about $(7, 7.33]$. You must separate the two sign cases: cross-multiplying without looking at the sign of the denominator is the error that gives a wrong domain.

$(g\circ f)(x)=g\!\left(\sqrt{x-3}\right)=\dfrac{1}{\sqrt{x-3}-7}$. Conditions: $x-3\ge 0$, so $x\ge 3$, and $\sqrt{x-3}\ne 7$, so $x-3\ne 49$ and $x\ne 52$. The domain is $[3, 52)\,\cup\,(52, +\infty)$. The two compositions have neither the same formula nor the same domain: composition is not commutative, and it is visible here at first glance.

$D(R(s))=0.08\times 1{,}800s-400=144s-400$. The coefficient 144 can be read: each extra dollar of hourly wage adds 144 dollars of leisure a year, because it adds 1,800 dollars of income, of which 8% goes to leisure. This is the chain rule, seen here on linear functions, before meeting it again as the chain rule for derivatives in the chapter on the derivative.

$D=144\times 16.50-400=2{,}376-400=1{,}976$ dollars a year. We can cross-check through the intermediate step: $R=1{,}800\times 16.50=29{,}700$ dollars, then $D=0.08\times 29{,}700-400=2{,}376-400=1{,}976$ dollars. The check by the long route is worth the minute it costs.

$144s-400>0$ gives $s>\dfrac{400}{144}\approx 2.78$ dollars an hour. The calculation is correct and useless: no legal hourly wage is below 2.78 dollars, so this threshold lies OUTSIDE the domain of validity of the model, which starts at the minimum wage. The algebraic domain of $s\mapsto 144s-400$ is all of $\mathbb{R}$, the domain of the context is about $[15.75, +\infty)$. Answering 2.78 dollars without this sentence is answering a question nobody is asking.

The model requires $q\ge 0$ and $p\ge 0$. From $90-1.5p\ge 0$ we get $p\le 60$. The domain of the context is therefore $[0, 60]$: beyond 60 dollars, the model would announce a negative number of meals, which does not exist. The algebraic domain would be all of $\mathbb{R}$; this is again the distinction of the chapter.

$q$ is linear with slope $-1.5$, strictly negative, so strictly DECREASING on $[0, 60]$, hence one-to-one: two different prices cannot give the same quantity. We can then isolate $p$ in $q=90-1.5p$: $1.5p=90-q$, so $p(q)=\dfrac{90-q}{1.5}=60-\dfrac{2q}{3}$. Checking that the function is one-to-one BEFORE isolating is not a formality: without it, the expression obtained would not be a function, as we saw in exercise 1 with the unemployment rate.

$q$ has domain $[0, 60]$, the prices, and range $[0, 90]$, the quantities: $q(0)=90$ and $q(60)=0$. $p$ has domain $[0, 90]$ and range $[0, 60]$. The domain of one is exactly the range of the other, and vice versa: this is the property that defines a pair of inverse functions, and it is the first thing to check when an inverse seems wrong.

$q(24)=90-1.5\times 24=90-36=54$ meals. Then $p(54)=60-\dfrac{2\times 54}{3}=60-36=24$ dollars. We land back on the starting price. This equality checks that $p$ was isolated correctly and that no sign or division error was made: a wrong inverse almost always fails this test, even though it may pass a simple unit-consistency check.

In economics, price has been plotted on the vertical axis since Alfred Marshall, at the end of the nineteenth century, and the habit has stuck: all the supply and demand curves in economics courses are drawn this way, and changing the axes would make the figures unreadable for those who read them. Mathematically, this amounts to plotting the INVERSE $p(q)$ rather than $q(p)$. The symmetry about the line $y=x$ remains true, but it is not visible on the figure, for two reasons: only one of the two curves is drawn, and the two axes have neither the same unit nor the same scale, whereas the symmetry only shows in an orthonormal frame. A property can be exact and invisible; it is the graph that misleads, not the property.