Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: functions and their domain (201-103)

Q: How do you find the domain of a function in calculus 201-103?

Look at the symbols, not the meaning. Only three forms impose a condition: a fraction bar requires a nonzero denominator, a square root requires a radicand that is positive or zero, a logarithm requires a strictly positive argument. Write all the conditions, solve them separately, then take their intersection. Everything else, polynomials and exponentials included, exists everywhere.

Q: Why is the domain of a sum an intersection and not a union?

Because a sum needs both of its terms at the same time. If the first function exists from 2 onward and the second up to 5, their sum has a value only between 2 and 5, where both respond. The word union is used only for a piecewise-defined function, where each piece is a different formula on a different zone.

Q: Why does a root give a non-strict inequality and a logarithm a strict one?

Because the square root of zero exists and equals zero, whereas the logarithm of zero does not exist. The condition for the root is therefore radicand greater than or equal to zero, with a closed bracket, and the one for the logarithm is argument strictly greater than zero, with an open bracket. This is not a convention, you can check it on the calculator.

Q: Why does the average cost not exist when the quantity is zero?

The average cost is the total cost divided by the quantity produced. At zero units, this division is not defined, even though the total cost exists perfectly well there and equals the fixed cost. The domain of the average cost is therefore strictly positive. Economically, the average cost of zero units means nothing: you cannot spread an expense over nothing.

Q: How do you recognize a linear, exponential or logistic model in a table?

Calculate two more columns. If the differences from one row to the next are constant, the model is linear. If the ratios are constant, it is exponential. If the values rise faster and faster and then slow toward a ceiling, it is logistic. And if the differences of the differences are constant, the model is quadratic, not exponential.

Q: What is the domain of validity of a model?

It is the window of data on which the model was fitted, and it is almost always narrower than the mathematical domain. An exponential fitted over ten years can still be computed for two hundred years, but the result no longer means anything. In Social Science, stating this window is part of the answer and earns marks.

This is the first chapter of the course, and the one whose errors cost the most in December. A wrong domain does not make you fail the question where it is asked: it makes you fail the curve sketching three chapters later, when $x=7$ will be treated as an asymptote although the function is simply absent there.

This sheet does not repeat the course. It takes the three restrictions, the four ways of combining two functions and the three models of the Social Science program, and says for each which move earns marks and which line loses them.

It is written for CEGEP students in the Social Science program (profile with mathematics) who are starting Differential Calculus for Social Science, 201-103.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The domain is built condition by condition BEFORE any calculation: a skipped restriction sets off no alarm, it produces a number, and a wrong number then travels through the whole paper.

This chapter is part of 201-103, Differential Calculus for Social Science

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Domain of a function
Vertical line test
Piecewise-defined function
Operations on functions
Composition of functions
Inverse function
Average cost
Exponential model
Logistic model
Horizontal asymptote
Domain of validity of a model

The essentials

What a function is, and what it is not

• A function assigns to each input AT MOST one output. An input with no output is allowed, it is simply outside the domain; an input with two outputs is forbidden.
• On a graph, the test is a vertical line: if a vertical line cuts the curve twice, it is not a function.
• The domain is the set of inputs that HAVE an output. The range is the set of outputs actually reached.
• A piecewise-defined function is still ONE function: its domain is the union of the pieces, and this is the only place in the chapter where the word union is correct.

The same vertical line meets the left-hand curve at a single point and the right-hand figure at two: on the right, an input would have two outputs, which no function does.

Write “ $D_f=$ ” before the interval, not “ $x\ge 2$ ” alone: the marker looks for a SET, not a condition.

The only three restrictions

• A denominator is never zero: condition $\neq 0$ , to be SOLVED, not copied.
• An even-index root takes no negative: condition $\ge 0$ , NON-STRICT inequality, since $\sqrt{0}=0$ exists.
• A logarithm takes only strictly positive values: condition $>0$ , STRICT inequality, since $\ln 0$ does not exist.
• Everything else exists everywhere: a polynomial, an exponential $b^{x}$ , a sine, an absolute value have no restriction.
• A root IN the denominator combines two restrictions that merge into a single, strict one: $\dfrac{1}{\sqrt{x-2}}$ requires $x>2$ , not $x\ge 2$ .

Three restrictions, three endpoints:

\frac{1}{x}

skips

0

;

\sqrt{x}

STARTS at

0

(solid point,

\sqrt{0}=0

);

\ln x

never reaches

0

and plunges along the vertical axis.

Three restrictions, not four. Looking for a restriction on an exponential or a polynomial wastes time and introduces false conditions.

The three families of models in the course

• CONSTANT increase from one period to the next: linear model $ax+b$ .
• Constant growth rate, hence a constant ratio from one period to the next: exponential model $a\,b^{t}$ .
• Growth that speeds up then slows toward a ceiling: logistic model.
• An exponential $a\,b^{t}$ with $a>0$ NEVER reaches zero: its asymptote is $y=0$ and it never touches it.
• A logistic with ceiling $M$ has TWO horizontal asymptotes, $y=0$ and $y=M$ . That is what sets it apart from an exponential on a graph.

These three models have $\mathbb{R}$ as their mathematical domain and a much narrower window as their domain of VALIDITY. Both must be written out.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The domain of a combination, operation by operation

Read this as soon as the function in the problem is built from two others. Here $f(x)=\sqrt{x-2}$ , defined on $[2, +\infty)$ , and $g(x)=\ln(5-x)$ , defined on $(-\infty, 5)$ . The last column gives the resulting domain; the red cell flags a rule that does not exist.

Combination	Condition to write	Resulting domain
$f+g$	$x\in D_f$ AND $x\in D_g$	$[2, 5)$
Example: $\sqrt{x-2}+\ln(5-x)$ equals $1+\ln 2\approx 1.693$ at $x=3$ , and does not exist at $x=7$ .
$f\times g$	the same intersection	$[2, 5)$
Example: $\sqrt{x-2}\,\ln(5-x)$ equals $\sqrt{2}\times 0=0$ at $x=4$ , since $\ln 1=0$ .
$\dfrac{f}{g}$	the intersection, THEN $g(x)\neq 0$	$[2, 4)\,\cup\,(4, 5)$
Example: $\dfrac{\sqrt{x-2}}{\ln(5-x)}$ also loses $x=4$ , since $\ln 1=0$ makes the denominator zero there.
$f\circ g$	$x\in D_g$ AND $g(x)\in D_f$	$(-\infty, -2.389]$
Example: $\sqrt{\ln(5-x)-2}$ requires $\ln(5-x)\ge 2$ , so $x\le 5-e^{2}\approx -2.389$ .
$D_{f+g}=D_f\cup D_g$	the union of the domains	does not exist no such rule
Example: At $x=7$ : $\sqrt{5}\approx 2.236$ exists, $\ln(-2)$ does not, so the sum does not exist. What to do: Intersect the two domains, never unite them. A sum needs BOTH of its terms.

The composition row is the only one that asks for two conditions of different natures: one on $x$ , one on $g(x)$ . It is also the only one the paper handles halfway.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Taking the union of the domains instead of their intersection

2 points, and the wrong domain spreads to the whole question

What not to write

“ $f$ exists for $x\ge 2$ and $g$ for $x<5$ , so $f+g$ exists for every $x$ .”

What to write

“ $f+g$ exists where BOTH exist: $D_{f+g}=D_f\cap D_g=[2, 5)$ .”

The bottom line sticks out neither on the left nor on the right: it stops where the shorter of the two stops, and that is exactly what intersection means.

Why: A sum needs both of its terms. The word union is only for piecewise-defined functions, where the pieces are different formulas on different zones.

2. Giving the same inequality to the root and to the logarithm

1 point per endpoint, and two wrong endpoints per paper on average

What not to write

“ $\sqrt{x-2}$ exists for $x>2$ and $\ln(5-x)$ for $5-x\ge 0$ .”

What to write

“ $\sqrt{x-2}$ exists for $x\ge 2$ , since $\sqrt{0}=0$ ; $\ln(5-x)$ requires $5-x>0$ , since $\ln 0$ does not exist.”

The root accepts its endpoint: CLOSED bracket at

2

, since

\sqrt{0}=0

. The logarithm refuses it: OPEN bracket at

5

, since

\ln 0

does not exist.

Why: The difference is not a convention: $0$ has a square root and has no logarithm. Each endpoint is therefore justified by a value, not by a habit.

3. Applying only one restriction when the expression contains two

the whole question, since the announced domain contains a point where the function does not exist

What not to write

“ $\dfrac{\sqrt{x-2}}{x-5}$ : we need $x\ge 2$ .”

What to write

“We need $x\ge 2$ for the root AND $x\neq 5$ for the denominator, so $D_f=[2, 5)\,\cup\,(5, +\infty)$ .”

The root imposes

x\ge 2

, the denominator removes

5

: the domain is

[2, 5)\,\cup\,(5, +\infty)

, an interval with a hole at

5

(red).

Why: Each FORM brings its own condition, independently of the others. List the forms before solving anything, otherwise the first one found makes you forget the second.

4. Simplifying before stating the domain

1 point here, and the error comes back in curve sketching

What not to write

“ $\dfrac{x^{2}-9}{x-3}=x+3$ , so the domain is $\mathbb{R}$ .”

What to write

“ $\dfrac{x^{2}-9}{x-3}=x+3$ for $x\neq 3$ : the domain is read off the STARTING expression, $D_f=\mathbb{R}\setminus\{3\}$ .”

\frac{x^{2}-9}{x-3}

coincides with the line

y=x+3

, except at

x=3

, where the fraction does not exist: a HOLE (open circle), which simplifying made disappear from the writing.

Why: Simplifying makes the factor disappear, not the restriction. It is the same move that, later, announces a vertical asymptote where there is only a hole.

5. The average cost evaluated at $q=0$

2 points, and an absurd interpretation sentence on top

What not to write

“The starting average cost is $\overline{C}(0)=\dfrac{C(0)}{0}$ , that is, the fixed cost.”

What to write

“ $C(0)=300$ dollars is the fixed cost, but $\overline{C}(q)=\dfrac{C(q)}{q}$ is not defined at $q=0$ : $D_{\overline{C}}=\,(0, +\infty)$ .”

The solid point on the vertical axis is

C(0)=300

, the fixed cost. The second curve climbs without end as

q

approaches

0

\overline{C}

does not exist there.

Why: Dividing by the quantity produced removes zero from the domain, even when the starting function is perfectly defined there. Economically, the average cost of zero units means nothing.

6. Composing while requiring only one of the two conditions

3 points, and a domain that contains almost everything when it contains almost nothing

What not to write

“ $(f\circ g)(x)=\sqrt{\ln(5-x)-2}$ exists as soon as $x<5$ .”

What to write

“We need $x<5$ for $g(x)$ to exist, AND $\ln(5-x)\ge 2$ for the root to exist, so $x\le 5-e^{2}\approx -2.389$ .”

The condition

x<5

(orange) only lets

x

into

g

. The root also requires

\ln(5-x)\ge 2

: what remains is

x\le 5-e^{2}\approx -2.39

(green), a small part of the first interval.

Why: Composing means passing an output through a second machine: the input must be accepted by the first and its output must be accepted by the second.

7. Swapping $x$ and $y$ without swapping the domain and the range

half the marks of the question, and the error comes back every semester

What not to write

“ $f(x)=\sqrt{x-2}$ has inverse $f^{-1}(x)=x^{2}+2$ , defined on $\mathbb{R}$ .”

What to write

“ $f^{-1}(x)=x^{2}+2$ on $[0, +\infty)$ only: the domain of $f^{-1}$ is the RANGE of $f$ , which is $[0, +\infty)$ .”

f

(blue) and

f^{-1}

(green) are symmetric about

y=x

, provided

x\ge 0

is kept for

f^{-1}

. The red branch,

x<0

, would return values that

f

never takes.

Why: An inverse is not a formula, it is a formula WITH its domain. Without the restriction, $x^{2}+2$ would return $6$ for $x=-2$ , an input that $f$ never produced.

8. Extending a model beyond the window on which it was fitted

1 to 2 interpretation marks, systematically removed in Social Science

What not to write

“The model $P(t)=1200\times 1.04^{t}$ fitted over ten years gives $P(200)\approx 3.1$ million inhabitants.”

What to write

“The model was fitted on $0\le t\le 10$ . Beyond that, it can still be computed but is no longer valid: its mathematical domain is $\mathbb{R}$ , its domain of validity is $[0, 10]$ .”

The model was fitted on

[0, 10]

(green). The formula keeps giving numbers beyond that (red), but nothing guarantees that they still describe the population.

Why: This is the point of method specific to the program: a model is a fit to data, not a law. Stating its window of validity is part of the expected answer.

Which method to choose

Which condition to write, according to the FORM of the expression

We do not read the meaning of the function, we look at the symbols present. A form spotted, a condition written.

If a fraction bar → write denominator $\neq 0$ , then SOLVE the equation denominator $=0$

Example: $\dfrac{3}{x^{2}-9}$ : we remove $-3$ and $3$

the most frequent error is to copy the condition without solving it
If a square root, or any even-index root → write radicand $\ge 0$ , NON-STRICT inequality

Example: $\sqrt{2x-6}$ : $x\ge 3$
If an $\ln$ or a $\log$ → write argument $>0$ , STRICT inequality

Example: $\ln(2x-6)$ : $x>3$
If a root IN the denominator → the two conditions merge into a single, strict one

Example: $\dfrac{1}{\sqrt{2x-6}}$ : $x>3$ , not $x\ge 3$
If a logarithm IN the denominator → argument $>0$ , AND argument $\neq 1$ , since $\ln 1=0$

Example: $\dfrac{1}{\ln(8-x)}$ : $x<8$ and $x\neq 7$

this is the most forgotten condition in the whole chapter
If several of these forms in the same expression → one condition per form, solved separately, then INTERSECTION

Example: $\dfrac{\sqrt{x-2}}{x-5}$ : $[2, 5)\,\cup\,(5, +\infty)$
If a polynomial, an exponential $b^{x}$ , a sine, an absolute value → no condition, the domain is $\mathbb{R}$

Example: $3^{x}-x^{2}$ : $D=\mathbb{R}$

If no branch applies, there is no restriction: the domain is $\mathbb{R}$ , and it must be written rather than leaving the box empty. Then you still intersect with the domain of the context, which is not a matter of symbols.

Which model, according to what the data table shows

We calculate two more columns before choosing: the successive DIFFERENCES and the successive RATIOS.

The logistic is the only one of the three to flatten under the dotted line: it is this ceiling, not the speed of the climb, that sets it apart from the exponential.

If the successive differences are constant → linear model $ax+b$ , with $a$ the common difference

Example: $12$ , $17$ , $22$ , $27$ : difference $5$ , so $5t+12$
If the successive ratios are constant → exponential model $a\,b^{t}$ , with $b$ the common ratio

Example: $400$ , $440$ , $484$ : ratio $1.1$ , so $400\times 1.1^{t}$
If the differences grow then shrink, and the values level off → logistic model, the ceiling is read from the last value reached

Example: a spread that goes from $2$ to $50$ then stalls near $60$
If the differences OF the differences are constant → quadratic model, not exponential

Example: $1$ , $4$ , $9$ , $16$ : differences $3$ , $5$ , $7$ , so $t^{2}$

this is the costliest confusion: a parabola climbs fast without being an exponential

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a domain, condition by condition

When to use it: As soon as the problem says “find the domain” or “for which values does this model make sense”.

1 List the FORMS present: fraction bar, even root, logarithm. One form, one condition, and write them all down before solving a single one.
2 Write each condition with the right inequality: NON-STRICT for the root, STRICT for the logarithm, $\neq$ for the denominator.
3 Solve each condition separately, on its own line.
4 Take the INTERSECTION of the solutions and write it in interval notation, with the right brackets.
5 Finally remove the values that make a denominator zero, including when that denominator is a logarithm, which is zero at $1$ .
6 Intersect with the domain of the CONTEXT when there is one: a quantity, a headcount, a duration are not negative.

Concluding sentence

“The conditions $x-2\ge 0$ , $8-x>0$ and $\ln(8-x)\neq 0$ give $x\ge 2$ , $x<8$ and $x\neq 7$ respectively. So $D_f=[2, 7)\,\cup\,(7, 8)$ .”

The trap: Skipping the line of conditions and giving the interval directly. A correct domain without its written conditions loses half the marks, because the marking scheme pays for the reasoning.

Marking: 1 mark per condition written and solved, 1 mark for the intersection, 1 mark for the interval notation.

Building an inverse and interpreting it

When to use it: As soon as the problem turns the question around: we know the output, we look for the input. “What price must be set to sell 300 units?”

1 First check that the function is one-to-one on the domain considered, and SAY SO: a quantity that falls then rises has no inverse over its whole period.
2 Write $y=f(x)$ , then isolate $x$ in terms of $y$ , one inverse operation at a time.
3 Swap the names of the variables to write $f^{-1}(x)$ , and not before: swapping too early is the source of sign errors.
4 Give the DOMAIN of $f^{-1}$ , which is the range of $f$ , and not $\mathbb{R}$ by default.
5 Check on one value: $f^{-1}(f(a))=a$ .
6 Say in words what the inverse calculates, in the units of the problem.

Concluding sentence

“The function $f$ is strictly decreasing on $[0, 50]$ , hence one-to-one: it has an inverse. From $y=100-2x$ we get $x=\dfrac{100-y}{2}$ , so $f^{-1}(x)=\dfrac{100-x}{2}$ , defined on $[0, 100]$ , the range of $f$ . It gives the quantity sold from the price.”

The trap: Announcing an inverse without having justified that the function is one-to-one. This is the line the marker looks for first, because it proves that you know under what condition an inverse exists.

Marking: 1 mark for one-to-one, 2 marks for isolating $x$ , 1 mark for the domain, 1 mark for the interpretation.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Test an excluded value

Take a value that the domain you found rejects and enter it in the calculator. It MUST return a domain error. If it returns a number, the condition you wrote was too wide.

For $D_f=[2, 7)\,\cup\,(7, 8)$ , enter $x=7$ : the TI shows “ERR:DIVIDE BY 0”, the Casio “Math ERROR”. A displayed number means that the condition $\ln(8-x)\neq 0$ was forgotten.
Test an interior value

Take a value well inside the domain you found. It must return a number. The check catches the opposite error, a domain that is too narrow.

At $x=5$ , $\dfrac{\sqrt{5-2}}{\ln(8-5)}=\dfrac{\sqrt{3}}{\ln 3}\approx 1.577$ . An error here would say that an endpoint was put on the wrong side.
The two endpoints, one by one

Each endpoint of the domain is tested at the value itself, not next to it: that is what decides between an open and a closed bracket.

$\sqrt{2-2}=0$ exists, so $2$ is IN the domain, closed bracket. $\ln(8-8)$ does not exist, so $8$ is out, open bracket.
The check on the inverse

Apply $f$ then $f^{-1}$ to any value in the domain: you must land back on it. It is the cheapest check in the chapter and it catches almost every sign error.

With $f(x)=100-2x$ and $f^{-1}(x)=\dfrac{100-x}{2}$ : $f(17)=66$ , then $\dfrac{100-66}{2}=17$ . Landing on $-17$ or on $34$ signals an operation inverted in the wrong order.
The domain of the context

Reread the problem and ask what the variable represents. A quantity, a headcount, a duration, a price are not negative, and the final domain is the intersection with that constraint.

A mathematical domain $\mathbb{R}\setminus\{0\}$ for an average cost becomes $(0, +\infty)$ once the context is read. Announcing $-5$ units produced costs the interpretation mark.

The typical problem, taken apart

The domain of a composite index, the three restrictions at once

A research report models an index of social tension by $f(x)=\dfrac{\sqrt{x-2}}{\ln(8-x)}$ , where $x$ is a density indicator measured between $2$ and $8$ .

Find the domain of definition of $f$ , then calculate $f(5)$ .

Step 1

The expression contains a square root, a logarithm and a fraction bar: three forms, so three conditions to write.

Why

Listing BEFORE solving is the move that earns marks: once the first condition is solved, we systematically forget to look for the others.

Step 2

Root: $x-2\ge 0$ , so $x\ge 2$ . The inequality is non-strict, since $\sqrt{0}=0$ exists.

Why

Justifying the non-strict inequality by a value, not by a habit, secures the closed bracket at $2$ .

Step 3

Logarithm: $8-x>0$ , so $x<8$ . The inequality is strict, since $\ln 0$ does not exist.

Why

This is the second endpoint, and it is of a different nature from the first. Two neighbouring restrictions have no reason to give the same type of bracket.

Step 4

Denominator: $\ln(8-x)\neq 0$ . Now $\ln u=0$ is equivalent to $u=1$ , so $8-x\neq 1$ , that is $x\neq 7$ .

Why

This is the step the paper skips. The reflex writes $8-x\neq 0$ , which gives back the condition of the logarithm and lets $x=7$ through, where the function does not exist.

Step 5

Intersection: $x\ge 2$ and $x<8$ give $[2, 8)$ , from which we remove $7$ . So $D_f=[2, 7)\,\cup\,(7, 8)$ .

Why

The hole appears only here, after the intersection. Removing $7$ too early, before having both endpoints, gives a wrong writing every other time.

Step 6

At $x=5$ : $f(5)=\dfrac{\sqrt{3}}{\ln 3}\approx\dfrac{1.732}{1.099}\approx 1.577$ .

Why

The requested value is calculated only after checking that $5$ does belong to the domain. Calculating first and checking afterward means accepting to hand in a number that does not exist.

Step 7

Check: at $x=7$ , the calculator returns an error; at $x=8$ too; at $x=2$ it returns $0$ .

Why

The three tests each confirm a different endpoint, and the $0$ obtained at $2$ confirms that the bracket is indeed closed there.

The conclusion, written out

“The domain of $f$ is $D_f=[2, 7)\,\cup\,(7, 8)$ , and $f(5)=\dfrac{\sqrt{3}}{\ln 3}\approx 1.577$ .”

The classic mistake on this problem: Writing $8-x\neq 0$ for the denominator instead of $8-x\neq 1$ . The announced domain is then all of $[2, 8)$ , it contains $x=7$ , and the function does not exist there.

Learn by heart

• THREE restrictions, not four: a zero denominator, an even root of a negative, a logarithm of a number that is not strictly positive.
• The root gives a NON-STRICT inequality, the logarithm a STRICT one. $\sqrt{0}$ exists, $\ln 0$ does not.
• $\ln u=0$ is equivalent to $u=1$ : a logarithm in the denominator also forbids the value that makes its argument equal to $1$ .
• Sum, difference, product: INTERSECTION of the domains. Quotient: that intersection, without the zeros of the denominator.
• Composition: TWO conditions, $x\in D_g$ and $g(x)\in D_f$ . Never just one.
• Inverse: domain and range SWAP, and it exists only where the function is one-to-one.
• The average cost $\overline{C}(q)=\dfrac{C(q)}{q}$ is NEVER defined at $q=0$ , even when $C$ is.
• Mathematical domain and domain of VALIDITY are two different answers, and the problem often expects both.

Frequently asked questions

How do you find the domain of a function in calculus 201-103?

Look at the symbols, not the meaning. Only three forms impose a condition: a fraction bar requires a nonzero denominator, a square root requires a radicand that is positive or zero, a logarithm requires a strictly positive argument. Write all the conditions, solve them separately, then take their intersection. Everything else, polynomials and exponentials included, exists everywhere.

Why is the domain of a sum an intersection and not a union?

Because a sum needs both of its terms at the same time. If the first function exists from 2 onward and the second up to 5, their sum has a value only between 2 and 5, where both respond. The word union is used only for a piecewise-defined function, where each piece is a different formula on a different zone.

Why does a root give a non-strict inequality and a logarithm a strict one?

Because the square root of zero exists and equals zero, whereas the logarithm of zero does not exist. The condition for the root is therefore radicand greater than or equal to zero, with a closed bracket, and the one for the logarithm is argument strictly greater than zero, with an open bracket. This is not a convention, you can check it on the calculator.

Why does the average cost not exist when the quantity is zero?

The average cost is the total cost divided by the quantity produced. At zero units, this division is not defined, even though the total cost exists perfectly well there and equals the fixed cost. The domain of the average cost is therefore strictly positive. Economically, the average cost of zero units means nothing: you cannot spread an expense over nothing.

How do you recognize a linear, exponential or logistic model in a table?

Calculate two more columns. If the differences from one row to the next are constant, the model is linear. If the ratios are constant, it is exponential. If the values rise faster and faster and then slow toward a ceiling, it is logistic. And if the differences of the differences are constant, the model is quadratic, not exponential.

What is the domain of validity of a model?

It is the window of data on which the model was fitted, and it is almost always narrower than the mathematical domain. An exponential fitted over ten years can still be computed for two hundred years, but the result no longer means anything. In Social Science, stating this window is part of the answer and earns marks.

Practise it

Corrected exercises: Functions and their domain, 201-103

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Next sheet Limits and continuity

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-103-functions-and-domain. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: functions and their domain (201-103)

The essentials

What a function is, and what it is not

The only three restrictions

The three families of models in the course

The rules in table form

The domain of a combination, operation by operation

The mistakes that cost marks

1. Taking the union of the domains instead of their intersection

2. Giving the same inequality to the root and to the logarithm

3. Applying only one restriction when the expression contains two

4. Simplifying before stating the domain

5. The average cost evaluated at q=0q=0q=0

6. Composing while requiring only one of the two conditions

7. Swapping xxx and yyy without swapping the domain and the range

8. Extending a model beyond the window on which it was fitted

Which method to choose

Which condition to write, according to the FORM of the expression

Which model, according to what the data table shows

How the answer is expected to be written

Writing up a domain, condition by condition

Building an inverse and interpreting it

Check before you hand in

The typical problem, taken apart

The domain of a composite index, the three restrictions at once

Learn by heart

Frequently asked questions

How do you find the domain of a function in calculus 201-103?

Why is the domain of a sum an intersection and not a union?

Why does a root give a non-strict inequality and a logarithm a strict one?

Why does the average cost not exist when the quantity is zero?

How do you recognize a linear, exponential or logistic model in a table?

What is the domain of validity of a model?

Corrected exercises: Functions and their domain, 201-103

See also

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5. The average cost evaluated at $q=0$

7. Swapping $x$ and $y$ without swapping the domain and the range