1.4 Domain and Range
SLO F-IF
Interpreting Functions. Understand what a function is and use function notation;
Learning Objectives
By the end of this section, you will be able to:
- describe the domain of a function as the set of its allowed inputs, and the range as the set of its resulting outputs;
- find a domain and range from a table, a list of ordered pairs, or a graph, by sweeping left-to-right and bottom-to-top;
- write a domain in interval notation, and say when interval notation is the wrong tool;
- distinguish the domain of a rule from the domain of a situation, and name the constraint the situation adds;
- identify the intercepts, maximum, and minimum of a graph, and say what each one means about the situation;
- use function notation to compare two functions drawn on the same axes.
1.4.1 The Water Park
Aly and Dayne work at a water park. At the end of each month they have to drain the small pool at the bottom of the ride they supervise, and each of them uses a pump to do it. One graph shows both drains at once: the amount of water in Aly's pool, \(a(x)\), and the amount in Dayne's pool, \(d(x)\), both measured against the time \(x\) in minutes since the pumps were switched on.
Figure 1.4.1 — Two pools draining on one pair of axes: the steeper line starts higher and still empties first.
Before any equations, look at the picture and say what you see. There is more in it than there appears to be:
- Both pools are draining — each graph falls as time goes on.
- Each one falls at a constant rate, because each graph is a straight segment, not a curve.
- The two rates are not the same. Aly's segment is steeper, so Aly's pool is draining faster.
- Dayne's pool starts with less water in it — his segment begins lower on the vertical axis.
- The two segments cross, so at one particular moment the two pools hold exactly the same amount of water.
- After the crossing, the order flips: Aly started with more and ends with less, because Aly's pool empties first.
- Both eventually reach zero. Both pools do get empty; they just do not get there at the same time.
Every observation above came from the shape of two lines, with no scale on either axis. Most of this section is a matter of attaching numbers to things you have already noticed.
Every one of those observations came from the shape of two lines, before a single number was attached to an axis. That is worth noticing, because the rest of this section is mostly a matter of attaching numbers to observations you have already made.
Dayne's pump, from a rate
Dayne worked out what his pump does: it removes water at 1000 gallons per minute, and it takes 24 minutes to empty his pool. Those two facts are enough to write the function.
Dayne's pump removes 1000 gallons a minute and empties the pool in 24 minutes. Write \(d(x)\), the gallons left after \(x\) minutes, and check it at both ends.
Solution
Step 1 — how much was there to begin with? If the pump removes 1000 gallons every minute and finishes in 24 minutes, the pool must have started with
$$1000 \cdot 24 = 24{,}000 \text{ gallons.}$$Step 2 — how much has gone? After \(x\) minutes the pump has removed \(1000x\) gallons, so what remains is
$$d(x) = 24{,}000 - 1000x$$Step 3 — say what each piece means.
| Piece | What it means |
|---|---|
| \(24{,}000\) | the amount of water in Dayne's pool at the moment the pump started |
| \(1000\) | the rate: gallons removed each minute |
| \(-1000x\) | the total removed after \(x\) minutes — the minus sign is the draining |
| \(d(x)\) | the amount of water left in the pool after \(x\) minutes |
Step 4 — check both ends.
$$d(0) = 24{,}000 - 0 = 24{,}000 \qquad d(24) = 24{,}000 - 1000(24) = 0$$At the start the pool is full; after 24 minutes it is empty.
Answer: \(d(x) = 24{,}000 - 1000x\), and both endpoints agree with the story, so the rule is right.
Aly's pump, from the graph
Aly's pump does not come with a stated rate. It comes with a graph, and reading a rule off a graph is a skill this chapter has been building since §1.2.
Aly's segment starts at 28,000 gallons when \(x = 0\) and reaches zero at 20 minutes. Write \(a(x)\).
Solution
Step 1 — find the rate. Two points are enough for a straight line, and the two easiest to read are the ends. Aly's pump removed 28,000 gallons in 20 minutes:
$$\frac{28{,}000 \text{ gallons}}{20 \text{ minutes}} = 1400 \text{ gallons per minute}$$Step 2 — same shape as Dayne's. Start full, subtract the rate times the time:
$$a(x) = 28{,}000 - 1400x$$Step 3 — check the ends.
$$a(0) = 28{,}000 \qquad a(20) = 28{,}000 - 1400(20) = 28{,}000 - 28{,}000 = 0$$Answer: \(a(x) = 28{,}000 - 1400x\).
Now the observations from the picture have numbers on them. Aly's pool started with 4000 more gallons and drained 400 gallons per minute faster, which is why it finished four minutes sooner. Keep both rules where you can see them; the rest of the section keeps returning to them.
A third pump empties a 30,000-gallon pool at 1500 gallons a minute.
a) Write the rule for the gallons remaining after \(x\) minutes.
b) How long does it take to empty?
c) How much is left after 12 minutes?
Solution
a) Start full, subtract the rate times the time:
$$p(x) = 30{,}000 - 1500x$$b) The pool is empty when \(p(x) = 0\):
$$30{,}000 - 1500x = 0 \quad\Longrightarrow\quad 1500x = 30{,}000 \quad\Longrightarrow\quad x = 20$$c) \(p(12) = 30{,}000 - 1500(12) = 30{,}000 - 18{,}000 = 12{,}000\).
Answer: a) \(p(x) = 30{,}000 - 1500x\). b) 20 minutes. c) 12,000 gallons.
1.4.2 Domain: The Set of Allowed Inputs
The domain of a function is the set of all of its possible input values.
Notice what the definition does not say. It does not say "all numbers." It says all the possible inputs, which leaves open the question of what makes an input possible. That question has to be answered fresh for every function.
For Dayne's pool the input is time in minutes since the pump started. Which times are possible?
| Candidate input | Possible? | Why |
|---|---|---|
| \(5\) | yes | Five minutes in, the pump is running and the pool holds \(d(5) = 19{,}000\) gallons. |
| \(8\tfrac{1}{2}\) | yes | Time does not have to land on a whole minute. At \(8.5\) minutes the pool holds \(15{,}500\) gallons. |
| \(-3\) | no | Three minutes before the pump started, the function has nothing to report. The story begins at \(x = 0\). |
| \(30\) | no | The pool was empty at 24 minutes. By 30 minutes there is no draining left to describe. |
The numbers \(-3\) and \(30\) are perfectly good numbers. Nothing is wrong with them arithmetically — \(24{,}000 - 1000(30)\) multiplies out without complaint. They are not in the domain because the situation does not extend that far. That is the whole idea of this section in two rows of a table.
So the domain of Dayne's draining function is every time from the start to the finish:
$$0 \le x \le 24$$And Aly's, by the same reasoning, is
$$0 \le x \le 20$$Two functions, one graph, two different domains. This is easy to miss and worth saying plainly: drawing two functions on the same pair of axes does not give them the same domain. Aly's function stops existing at 20 minutes while Dayne's runs on for four more.
Definition 1.4.1 — Domain: sweep left to right, and the band you lay down along the input axis is the set of allowed inputs.
For each function, state the domain and say what decided it.
a) The number of gallons left in the third pump's pool, \(p(x) = 30{,}000 - 1500x\).
b) \(g(x) = \dfrac{5}{x - 9}\), with no situation attached.
Solution
a) The story starts when the pump switches on and ends when the pool is empty, which Try It Now 1.4.1 found to be 20 minutes. So the domain is \(0 \le x \le 20\). The situation decided it — the arithmetic would happily accept \(-4\) or \(100\).
b) The only trouble is the denominator. \(x - 9 = 0\) when \(x = 9\), and dividing by zero has no value. So the domain is all real numbers except 9, written \(x \neq 9\). The arithmetic decided it — there is no situation here to consult.
Answer: a) \(0 \le x \le 20\), decided by the situation. b) \(x \neq 9\), decided by the rule.
1.4.3 Range: The Set of Resulting Outputs
The second definition is the mirror of the first.
The range of a function is the set of all of its possible output values.
Domain is what goes in; range is what comes out. If you have the domain and the rule, the range is not a new mystery — it is what you get by running every allowed input through the machine and collecting the results.
For Dayne's pool the output is an amount of water in gallons. The most it ever holds is 24,000, at the start. The least is 0, at the end. It falls steadily between those two without skipping anything, so every amount in between actually occurs at some moment:
$$0 \le d \le 24{,}000$$Aly's range is \(0 \le a \le 28{,}000\) by the same argument.
Definition 1.4.2 — Range: sweep bottom to top, and the band you lay down along the output axis is the set of resulting outputs.
Testing whether a number is in the range
Here is the move that makes ranges useful, and you will use it constantly: to test whether a number is in the range, work backwards and ask which input would produce it. If that input is in the domain, the number is in the range. If it is not, it is not.
Both tests below solve cleanly and hand back a number. What decides the question is whether that number is an input the function is allowed to take.
Is 2000 gallons a possible output for Dayne's pool? Is 25,000?
Solution
Part 1 — is 2000 in the range? Work backwards. Solve \(d(x) = 2000\):
$$24{,}000 - 1000x = 2000 \quad\Longrightarrow\quad 1000x = 22{,}000 \quad\Longrightarrow\quad x = 22$$The input 22 is in the domain \([0, 24]\), so yes — and you have learned something more specific than "yes." Dayne's pool holds 2000 gallons at exactly 22 minutes, two minutes before it runs dry.
Part 2 — is 25,000 in the range? Solve \(d(x) = 25{,}000\):
$$24{,}000 - 1000x = 25{,}000 \quad\Longrightarrow\quad 1000x = -1000 \quad\Longrightarrow\quad x = -1$$That input is one minute before the pump switched on, and it is not in the domain. So 25,000 is not in the range — which makes sense, since the pool never held more water than it started with.
Answer: 2000 is in the range (at \(x = 22\)); 25,000 is not.
Notice that the arithmetic answered happily both times. It was the domain that decided the question.
Using Aly's rule \(a(x) = 28{,}000 - 1400x\) with domain \([0, 20]\):
a) Is 7000 gallons in the range?
b) Is 30,000 gallons in the range?
Solution
a) Solve \(28{,}000 - 1400x = 7000\):
$$1400x = 21{,}000 \quad\Longrightarrow\quad x = 15$$15 is in \([0, 20]\), so yes — Aly's pool holds 7000 gallons at 15 minutes.
b) Solve \(28{,}000 - 1400x = 30{,}000\):
$$1400x = -2000 \quad\Longrightarrow\quad x = -\tfrac{10}{7} \approx -1.43$$Negative, so not in the domain. No — the pool never held 30,000 gallons.
Answer: a) yes, at 15 minutes. b) no.
1.4.4 Writing a Domain Down: Interval Notation
Inequalities like \(0 \le x \le 24\) are correct and clear. There is also a shorthand for them, and this section is the right place to learn it, because a draining pool produces exactly the kind of domain it was designed for.
Interval notation writes a set of numbers by giving its two endpoints, with a square bracket where the endpoint is included and a round bracket where it is not.
Dayne's domain, every time from 0 through 24 with both ends included, is written
$$[0, 24]$$and his range is
$$[0,\ 24{,}000]$$Aly's are \([0, 20]\) and \([0,\ 28{,}000]\).
The bracket carries the whole meaning.
Summary of bracket forms. The table below collects the four shapes you will meet.
| Written | Says | In words |
|---|---|---|
| \([0, 24]\) | \(0 \le x \le 24\) | from 0 to 24, both endpoints included |
| \((0, 24)\) | \(0 < x < 24\) | strictly between 0 and 24, neither endpoint included |
| \([0, 24)\) | \(0 \le x < 24\) | 0 is included, 24 is not |
| \([0, \infty)\) | \(x \ge 0\) | 0 and everything above it, forever |
The symbol \(\infty\) never gets a square bracket. A square bracket means "this endpoint is in the set," and infinity is not a number that could be in a set — it is a direction. So it always takes a round bracket.
Which bracket to use is a question about the situation, not about the notation. Is the pool's starting moment part of the story? Yes — at \(x = 0\) the pump switches on and the pool holds 24,000 gallons, which is a real reading. So the bracket is square. Is the finishing moment part of the story? Also yes: at \(x = 24\) the pool holds 0 gallons, which is a real reading too.
There is one thing interval notation cannot do, and §1.4.7 is about it. An interval is an unbroken stretch of the number line. If a domain has gaps in it — if some numbers inside the two endpoints are not allowed — then no interval describes it, and writing one down claims something false. Hold that thought.
Write each in interval notation, or explain why you cannot.
a) \(-2 \le x \le 7\)
b) \(x > 0\)
c) all real numbers except 4
Solution
a) Both endpoints are included, so both brackets are square: \([-2, 7]\).
b) Zero itself is excluded and there is no upper end, so a round bracket on both sides: \((0, \infty)\). Infinity always takes a round bracket.
c) You cannot write this as a single interval. An interval is one unbroken stretch, and this set is the number line with a hole punched in it at 4. It needs two intervals joined together, or the plain description "all real numbers except 4."
Answer: a) \([-2, 7]\). b) \((0, \infty)\). c) not a single interval — the set has a gap.
1.4.5 The Domain of the Rule and the Domain of the Situation
This is the part of the section that does the most work, and it is where domain stops being bookkeeping and starts being modeling.
Take Dayne's rule as bare arithmetic, with no pool attached:
$$d(x) = 24{,}000 - 1000x$$Which numbers does that accept? All of them. Feed it \(-3\) and it returns \(27{,}000\). Feed it \(30\) and it returns \(-6000\). Feed it \(1{,}000{,}000\) and it returns a very negative number. The arithmetic never breaks, so the domain of the rule is every real number: \((-\infty, \infty)\).
Now put the pool back. The domain of the situation is \([0, 24]\).
Look at what the rule was willing to say outside that interval. \(d(30) = -6000\) claims the pool contains negative six thousand gallons. There is no such thing. The rule is not wrong — it is answering a question nobody should have asked it. A model is a rule plus the set of inputs on which the rule means something, and the second half is supplied by the situation.
The reasonable domain is the set of input values that make sense in the situation being modeled, which is usually much smaller than the set the rule alone would accept.
The same split applies to Aly's function. The rule \(a(x) = 28{,}000 - 1400x\) accepts every real number; the situation permits \([0, 20]\). And it applies to the ranges: the rule's outputs run over every real number, while the situation's run over \([0,\ 28{,}000]\).
Whenever you are asked for "the domain," find out which of the two is being asked for. In a modeling problem it is almost always the situation's.
Definition 1.4.3 — Reasonable domain: the rule accepts an unbroken number line, and the situation clips it three times.
When the rule refuses too
Sometimes the arithmetic does break, and then both kinds of restriction are in play at once.
A rectangle has an area of \(24 \text{ cm}^2\), so its length is a function of its width: \(L(w) = \dfrac{24}{w}\). Which of \(3\), \(0.5\), \(48\), \(-6\), and \(0\) are possible inputs?
Solution
Work down the list, and for each one ask both questions: does the arithmetic work, and does the situation allow it?
| Candidate input \(w\) | Possible? | Why |
|---|---|---|
| \(3\) | yes | \(L(3) = 8\). A 3 cm by 8 cm rectangle has area 24. |
| \(0.5\) | yes | \(L(0.5) = 48\). A long thin rectangle, but a real one. |
| \(48\) | yes | \(L(48) = 0.5\). The same rectangle, turned on its side. |
| \(-6\) | no | A width cannot be negative. The arithmetic is fine — it would return \(-4\) — but the situation refuses. |
| \(0\) | no | The rule divides by \(w\), and \(\tfrac{24}{0}\) is undefined. Here the arithmetic refuses, and so does the situation: a rectangle with zero width is not a rectangle. |
Answer: \(3\), \(0.5\) and \(48\) are possible; \(-6\) and \(0\) are not.
The last two rows fail for different reasons that happen to agree. When a value is excluded, it is worth knowing which of the two reasons excluded it, because only one of them can be found by looking at the rule.
A rule that refuses, with no situation at all
Strip the situation away entirely and the arithmetic restriction is still there.
Build a table for \(f(x) = \dfrac{6}{x - 2}\) at \(x = -10, 0, 1, 2, 8\), and describe what the graph does near the excluded value.
Solution
Step 1 — the table.
| \(x\) | \(-10\) | \(0\) | \(1\) | \(2\) | \(8\) |
|---|---|---|---|---|---|
| \(f(x)\) | \(-0.5\) | \(-3\) | \(-6\) | ? | \(1\) |
Check two of them: \(f(-10) = \dfrac{6}{-10 - 2} = \dfrac{6}{-12} = -0.5\), and \(f(8) = \dfrac{6}{8 - 2} = \dfrac{6}{6} = 1\).
Step 2 — the fifth entry. When \(x = 2\) the denominator becomes \(2 - 2 = 0\), and the rule asks you to compute \(\dfrac{6}{0}\). There is no such number, and a calculator will not produce one either — it returns an error, because there is nothing to return. So the domain is all real numbers except 2, written \(x \neq 2\).
Step 3 — creep up on the forbidden input from both sides.
| \(x\) | \(1.8\) | \(1.9\) | \(1.95\) | \(1.999\) | \(2.001\) | \(2.05\) | \(2.1\) | \(2.2\) |
|---|---|---|---|---|---|---|---|---|
| \(f(x)\) | \(-30\) | \(-60\) | \(-120\) | \(-6000\) | \(6000\) | \(120\) | \(60\) | \(30\) |
As \(x\) creeps toward 2 from the left the outputs plunge; as it creeps toward 2 from the right they rocket upward instead.
Answer: The domain is \(x \neq 2\), and the graph is not one connected curve — it splits into two pieces at \(x = 2\). The gap in the graph is the gap in the domain, drawn.
This is also the first domain in the section that no interval can express. "All real numbers except 2" is a stretch of the number line with a hole punched in it, and an interval has no way to punch a hole.
A square's area is \(A(s) = s^2\), where \(s\) is the side length in centimeters.
a) What is the domain of the rule, ignoring the square?
b) What is the reasonable domain?
c) Name a number that the rule accepts and the situation refuses, and say what the rule would report for it.
Solution
a) Squaring never breaks, so the rule accepts every real number: \((-\infty, \infty)\).
b) A side length must be positive — a square with side 0 or a negative side is not a square. So the reasonable domain is \(s > 0\), or \((0, \infty)\).
c) Take \(s = -5\). The rule reports \(A(-5) = 25\), a perfectly ordinary number. But there is no square with a side of \(-5\) cm, so the situation refuses it. The arithmetic gave an answer to a question that should not have been asked.
Answer: a) \((-\infty, \infty)\). b) \((0, \infty)\). c) e.g. \(s = -5\), for which the rule reports 25.
1.4.6 Domain and Range from a Table or a List of Pairs
When a function arrives as a table or as a set of ordered pairs, finding its domain and range takes no reasoning at all. You just read.
Recall from §1.3 that a coordinate pair is written \((\text{input}, \text{output})\) — the input first, the output second. So in a list of ordered pairs:
- the domain is the collection of all the first coordinates,
- the range is the collection of all the second coordinates.
State the domain and range of \((1, 3)\), \((2, 6)\), \((3, 12)\), \((4, 24)\).
Solution
Step 1 — first coordinates give the domain. They are \(1, 2, 3, 4\), so the domain is \(\{1, 2, 3, 4\}\).
Step 2 — second coordinates give the range. They are \(3, 6, 12, 24\), so the range is \(\{3, 6, 12, 24\}\).
Answer: domain \(\{1, 2, 3, 4\}\), range \(\{3, 6, 12, 24\}\).
Those outputs should look familiar. Each one is twice the one before it, which is the doubling of §1.2's Growing, Growing Dots, started at 3 instead of 1. Here that pattern is again with the step numbers attached, written as pairs instead of as a sequence.
The same information in a table:
| Input \(x\) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Output \(y\) | 3 | 6 | 12 | 24 |
Reading across the top row gives the domain. Reading across the bottom row gives the range.
Two conventions apply when you write these out.
List each value only once. If a function pairs \((1, 5)\), \((2, 5)\), and \((3, 5)\), the range is \(\{5\}\) — a single element — even though the output 5 occurs three times. A set records which values appear, not how many times.
Order does not matter, but writing them in order helps. \(\{4, 1, 3, 2\}\) and \(\{1, 2, 3, 4\}\) are the same set. Sorting them makes the set easier to read and much harder to make a mistake with.
Five convenient rows out of infinitely many moments is a snapshot, not a domain. Read the surrounding words before you report what you see in a table.
There is one caution, and it matters. A table usually shows a sample of a function, not the whole of it. Here is a table for Dayne's pool:
| \(x\) (minutes) | 0 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| \(d(x)\) (gallons) | 24,000 | 19,000 | 14,000 | 9000 | 4000 |
Reading the top row gives \(\{0, 5, 10, 15, 20\}\), and that is emphatically not the domain of the function. The pump ran continuously for 24 minutes; the table shows five convenient moments out of infinitely many, and it does not even reach the end of the story.
A function is given by \((-2, 7)\), \((0, 7)\), \((3, 1)\), \((5, 9)\).
a) State the domain and the range.
b) Explain why the range has three elements and not four.
Solution
a) First coordinates: \(-2, 0, 3, 5\), so the domain is \(\{-2, 0, 3, 5\}\). Second coordinates: \(7, 7, 1, 9\), so the range is \(\{1, 7, 9\}\), written in order.
b) The output 7 occurs twice — once from \(-2\) and once from \(0\) — but a set records which values appear, not how many times each appears. So 7 is listed once. This is §1.3's point that repeated outputs are perfectly allowed, seen from the range's side.
Answer: a) domain \(\{-2, 0, 3, 5\}\), range \(\{1, 7, 9\}\). b) because 7 appears twice as an output and a set lists it once.
1.4.7 Discrete and Continuous
Every domain so far in this section has been an unbroken stretch: every time between 0 and 24 minutes, every width greater than zero. Not every domain looks like that, and the difference has a name.
The Library of Congress in Washington, D.C. is the largest library in the world, and it makes a good place to see both kinds at once, because everything about it gets counted somehow.
Books arriving in boxes. Books are heavy, so they ship in small boxes — on average about 8 books to a box. The number of books received is a function of the number of boxes:
$$B(n) = 8n$$Six boxes bring \(B(6) = 48\) books; ten boxes bring \(B(10) = 80\).
What is the domain? Whole numbers of boxes, starting at zero: \(\{0, 1, 2, 3, \ldots\}\). And that is a domain with gaps in it. There is no such thing as \(6.5\) boxes arriving at a loading dock, so \(6.5\) is not an allowed input, even though it sits between two inputs that are allowed. Graph this function and you do not get a line; you get a row of separated dots, one above each whole number, with empty space between them.
The range has gaps for the same reason: \(\{0, 8, 16, 24, \ldots\}\). Ask whether 32.5 books could have been received and the answer is no, twice over — 32.5 is not a multiple of 8, and half a book is not a book.
Medicine leaving a bloodstream. A dog is given a 60 milligram dose of an anti-parasite medicine, and 20% of what remains becomes ineffective every hour. So each hour, 80% of the previous amount is still active:
$$M(t) = 60(0.8)^t$$After 3 hours, \(M(3) = 60(0.8)^3 = 60(0.512) = 30.72\) milligrams — a little over half the dose is still working.
What is the domain here? Time in hours, and time can be split as finely as you like. Asking for the amount at \(4.25\) hours is a perfectly sensible question: \(M(4.25) \approx 23.2\) milligrams. There are no gaps. The domain is the unbroken interval \([0, \infty)\) — or however long the dog is observed for.
Those are the two kinds.
A discrete function is a function whose domain consists of separated values with gaps between them, rather than an unbroken stretch. Its graph is a set of separate points.
A continuous function is a function whose domain is an unbroken stretch of values and whose graph has no breaks in it. Its domain and range can be described by intervals.
Definition 1.4.5 — Continuous function: separated dots against one unbroken curve, and the description has to match.
The test: can the input quantity be split?
That one question decides it nearly every time. Boxes, cuts, students, tickets and buses cannot be split, so functions of those quantities are discrete. Hours, minutes, lengths, weights and temperatures can be split as finely as you like, so functions of those are continuous.
Two more from the library, to practise on.
A giant sheet of paper is cut into three equal pieces; each of those is cut into three; and so on. After \(n\) rounds of cutting there are
$$P(n) = 3^n$$pieces — after 10 rounds, \(3^{10} = 59{,}049\) of them. Discrete. The input counts rounds of cutting, and there is no such thing as \(5.2\) rounds. Asking how many pieces there are after \(5.2\) cuts is asking about a moment that does not exist. And running it backwards fails too: asking when there are \(53.6\) papers is asking about a count that cannot happen, since every output is a whole power of 3.
A librarian shelving books sorts and shelves 3 books a minute, which is 180 books an hour. The number of books shelved is a function of the hours worked:
$$S(h) = 180h$$This one is worth arguing about, and the argument is the point. Hours can be split, so the input behaves continuously — a librarian can work \(3.5\) hours. But the output counts books, and \(S(3.5) = 630\) books is fine while \(S(3.502) = 630.36\) books is not. The honest description is that shelving is a discrete process that we model with a continuous function because the counts are large and the approximation is harmless. That is a modeling decision, not a fact about the world, and being able to say which is which is most of what this subsection is for.
Why interval notation matters here
An interval is a solid stretch. A count is a row of separated dots. Writing an interval for a count is like giving a shoe size as "somewhere between 8 and 9" — the units do not work that way.
Go back to the boxes. It is tempting to write the domain of \(B(n) = 8n\) for a delivery of at most 12 boxes as
$$0 \le n \le 12$$and that is wrong. Not "informal" — wrong. That inequality claims that \(7.5\) and \(11.9\) are allowed inputs, and they are not. The same objection kills the interval \([0, 12]\): an interval is by definition an unbroken stretch, so writing one claims there are no gaps.
The correct description lists the values, or says in words what they are: \(\{0, 1, 2, \ldots, 12\}\), or "any whole number of boxes from 0 to 12." A discrete domain needs a description with gaps in it, because the thing being described has gaps in it.
Put the two library functions side by side.
| Books in boxes | Medicine in the bloodstream | |
|---|---|---|
| Rule | \(B(n) = 8n\) | \(M(t) = 60(0.8)^t\) |
| Input quantity | number of boxes | hours elapsed |
| Can the input be split? | no | yes |
| Type | discrete | continuous |
| Graph | separated dots | one unbroken curve |
| Domain | \(\{0, 1, 2, 3, \ldots\}\) | \([0, \infty)\) |
| Range | \(\{0, 8, 16, 24, \ldots\}\) | \((0, 60]\) |
That last cell repays a second look. The medicine's range is written with a round bracket at zero, and the reason is that the amount of active medicine never actually reaches zero. Each hour it drops to 80% of what it was: 60, then 48, then \(38.4\), then \(30.72\) — always smaller, never nothing. The dose gets arbitrarily close to zero without ever arriving, so 0 is not in the range and does not get a square bracket.
Which also settles a question about the graph before you draw it: this curve has no horizontal intercept. It never crosses the time axis, because crossing it would mean the medicine had completely vanished. When a feature is missing from a graph, the missingness itself usually means something about the situation.
Decide whether each is discrete or continuous, and give the domain in the appropriate form.
a) The cost of buying \(n\) concert tickets at $45 each, for at most 6 tickets.
b) The distance a car has travelled after \(t\) hours at a steady 60 miles an hour, on a 3-hour trip.
Solution
a) Tickets cannot be split — there is no such thing as 2.5 tickets. So the function is discrete, and the domain must be written as a list: \(\{0, 1, 2, 3, 4, 5, 6\}\). Writing \(0 \le n \le 6\) would be wrong, because it claims 2.5 is allowed.
b) Time can be split as finely as you like; the car is somewhere definite at 1.37 hours. So the function is continuous, and the domain is the interval \([0, 3]\).
Answer: a) discrete, \(\{0, 1, 2, 3, 4, 5, 6\}\). b) continuous, \([0, 3]\).
1.4.8 Domain and Range from a Graph
A graph is the representation where domain and range become almost physical. There are two motions, and they are the whole technique:
- For the domain, sweep your eyes left to right. The domain runs from the leftmost input the graph reaches to the rightmost.
- For the range, sweep your eyes bottom to top. The range runs from the lowest output the graph reaches to the highest.
Domain is about how wide the graph is. Range is about how tall it is.
Sweep the water park graph and both answers fall out. Left to right, Dayne's segment runs from 0 to 24, so his domain is \([0, 24]\); Aly's stops at 20, so hers is \([0, 20]\). Bottom to top, Dayne's segment spans from 0 up to 24,000, so his range is \([0,\ 24{,}000]\); Aly's spans \([0,\ 28{,}000]\).
A graph that is not a straight line
Straight segments make the sweep easy, and easy can be misleading. Here is a graph where the technique earns its keep — and it is the graph the two sweeping figures above are drawn on.
In a bungee jump, the height of the jumper is a function of the time since the jump began. The graph starts high at the platform, drops fast to the low point where the cord snaps taut, springs back up to a peak lower than the platform, drops again to a second low point higher than the first, and keeps oscillating with smaller and smaller swings until the jumper hangs nearly still and is lowered down. The jump ends 35 seconds after it began.
Read the domain and range of the bungee jump from its graph, and say what each one means about the jump.
Solution
Step 1 — domain, sweeping left to right. The leftmost time on the graph is \(t = 0\), the instant of the jump; the rightmost is \(t = 35\), when it ends. So the domain is \([0, 35]\), or \(0 \le t \le 35\).
Step 2 — range, sweeping bottom to top. The lowest height the graph ever reaches is 10 feet — the bottom of the first plunge, the closest the jumper comes to the river. The highest is about 75 feet, at the platform. So the range is \([10, 75]\).
Step 3 — read them back into the situation. The domain says the jump lasted 35 seconds. The range says the jumper was never higher than 75 feet and never lower than 10 feet above the water — which is also the answer to "did they touch the river?" No; the range stops 10 feet short.
Answer: domain \([0, 35]\), range \([10, 75]\).
Now notice what the sweep ignores. The jumper passes through a height of 40 feet many times — on the way down, on the way up, on the way down again. The range does not care. It records which heights occurred, not how often. Every height between 10 and 75 occurred at least once, so all of them are in the range, and each is listed once. On the water park graph, where each line passes each height exactly once, that distinction never came up. Here it does.
Two cautions
Check whether the endpoints are really endpoints. A graph can simply run off the edge of the picture. If a curve is still climbing when it leaves the top of the frame, you cannot read a maximum off it, and you cannot read the right-hand end of the domain either. A partial graph shows part of a function.
Two graphs on one pair of axes are still two functions. The water park picture holds two of them, with different domains and different ranges. Sweep each one separately.
| Question | Direction to sweep | What you record |
|---|---|---|
| Domain | left to right | the smallest and largest inputs the graph reaches |
| Range | bottom to top | the smallest and largest outputs the graph reaches |
A graph is a straight segment running from \((2, 1)\) up to \((10, 9)\), including both endpoints.
a) State the domain and the range.
b) A second graph is a curve from \((0, 5)\) that dips to \((4, 1)\) and rises again to \((8, 5)\). State its domain and range, and say which output occurs twice.
Solution
a) Sweeping left to right, the inputs run from 2 to 10: domain \([2, 10]\). Sweeping bottom to top, the outputs run from 1 to 9: range \([1, 9]\).
b) Left to right, the inputs run from 0 to 8: domain \([0, 8]\). Bottom to top, the lowest output is 1 at the dip and the highest is 5 at both ends: range \([1, 5]\).
The output 5 occurs twice, at \(x = 0\) and at \(x = 8\). The range lists it once, exactly as in the bungee jump — a range records which outputs occurred, not how many times.
Answer: a) domain \([2, 10]\), range \([1, 9]\). b) domain \([0, 8]\), range \([1, 5]\); the output 5 occurs twice.
1.4.9 What a Graph Tells You: Intercepts, Maximum, and Minimum
Reading a domain and a range means finding where a graph starts, stops, bottoms out, and tops out. Those places have names, and so does the place where the graph crosses each axis. This subsection collects the vocabulary and, more importantly, what each feature means about the situation.
The vertical intercept is the point where the graph crosses the vertical axis. It is the output when the input is 0, so for a function \(f\) it is the point \((0, f(0))\). It is also called the \(y\)-intercept.
Definition 1.4.6 — Vertical intercept: on f(x) = 2x + 6 each intercept carries a zero, and the zero sits in a different slot.
A horizontal intercept is a point where the graph crosses the horizontal axis. Its output is 0, so it is the point \((a, 0)\) for an input \(a\) solving \(f(a) = 0\). It is also called the \(x\)-intercept. That input \(a\) on its own — the number, not the point — is called a zero of the function.
The maximum is the greatest output value the function reaches.
Definition 1.4.8 — Maximum: the greatest output the function reaches, and with the least output it bounds the range.
The minimum is the least output value the function reaches.
Both intercepts are points, written as ordered pairs. The word zero is the one that names a bare input, and keeping those two apart is worth the effort: "the zero is 24" and "the horizontal intercept is \((24, 0)\)" are the same fact said about two different kinds of object.
Two of these connect straight back to the last subsection: the maximum and minimum of a function are the top and bottom of its range. When you swept a graph bottom to top, the two values you wrote down were the minimum and the maximum. They are not new information — they are the range's endpoints, given names.
§1.2 met intercepts once already, informally, while reading the graph of a growing pattern. These are the definitions that section was pointing at, and they are stated here once for the whole book.
The features of the water park graph
Every one of the four is on the picture you have been reading all section.
| Feature | Dayne's pool | What it means |
|---|---|---|
| Vertical intercept | \((0,\ 24{,}000)\) | the pool held 24,000 gallons when the pump started |
| Horizontal intercept | \((24, 0)\) | 24 minutes in, the pool is empty — the drain is finished |
| Maximum | \(24{,}000\) gallons | it never held more water than it started with |
| Minimum | \(0\) gallons | it does get completely empty |
Aly's four are \((0,\ 28{,}000)\), \((20, 0)\), \(28{,}000\) and \(0\). Comparing the two columns is how you answer questions nobody phrased in terms of axes: whose pool was fuller to begin with? — compare the vertical intercepts. Whose finished first? — compare the horizontal intercepts.
Both functions are also decreasing across their whole domains: as the input grows, the output falls. Aly's decreases faster, which on the graph is the steeper segment and in the story is the stronger pump.
Find the vertical intercept, the horizontal intercept, the maximum, and the minimum of \(f(x) = 2x + 6\).
Solution
Step 1 — vertical intercept. Evaluate at 0:
$$f(0) = 2(0) + 6 = 6$$so the graph crosses the vertical axis at \((0, 6)\).
Step 2 — horizontal intercept. Solve \(f(x) = 0\):
$$2x + 6 = 0 \quad\Longrightarrow\quad 2x = -6 \quad\Longrightarrow\quad x = -3$$Check it: \(f(-3) = 2(-3) + 6 = 0\). The graph crosses the horizontal axis at \((-3, 0)\).
Step 3 — maximum and minimum. It has neither. The line climbs forever to the right and falls forever to the left, so there is no greatest output and no least one.
Answer: vertical intercept \((0, 6)\), horizontal intercept \((-3, 0)\), no maximum and no minimum.
A function is not required to have a maximum or a minimum, and a straight line with a nonzero slope never has either.
Take care not to swap the two intercepts. The vertical intercept is \((0, 6)\) — input 0, output 6. The horizontal intercept is \((-3, 0)\) — input \(-3\), output 0. Each one has a zero in it, but in a different slot, and mixing them up turns every answer inside out.
A maximum is an output
Its input is a second number that tells you when. Reporting them the wrong way round is the single most common error with maxima, and nothing in the notation stops you.
A kickball is kicked from ground level, and its height in feet is a function of the time in seconds since the kick. Its maximum is \(h(2.5) = 31\).
That single statement carries two numbers with two different jobs. The output, 31, is the maximum height: about 31 feet. The input, \(2.5\), is when that happened: about \(2.5\) seconds after the kick. Getting these backwards — saying the maximum height is \(2.5\) feet and it happened 31 seconds in — is worth guarding against out loud every time: the maximum is an output, and its input tells you when.
The water park graph makes the same point more quietly. Dayne's maximum is 24,000 gallons, and it occurs at \(x = 0\). The maximum is the 24,000, not the 0.
Flat stretches
One more feature comes up often enough to name informally. Suppose Dayne's pump had jammed and sat idle from minute 8 to minute 10. The graph would run flat across that stretch, and
$$d(8) = d(10)$$Two different inputs, the same output. That is not a violation of anything — §1.3's rule is that each input has exactly one output, and there is no rule against two inputs sharing an output. A flat stretch means the quantity did not change, which here would mean no water left the pool for those two minutes.
Whenever you see a horizontal section, ask what "not changing" means for that quantity. On a distance-from-home graph it means standing still. On a temperature graph it means holding steady. On a draining-pool graph it means the pump stopped.
Summary of graph features. The table below collects how to find each one and what it usually means.
| Feature | How to find it | Typical meaning |
|---|---|---|
| Vertical intercept | evaluate \(f(0)\) | the starting value, before anything happens |
| Horizontal intercept | solve \(f(x) = 0\) | when the quantity runs out, arrives, lands, or hits the ground |
| Maximum | the highest point on the graph | the largest value reached, and the input tells you when |
| Minimum | the lowest point on the graph | the smallest value reached, and the input tells you when |
| Flat stretch | a horizontal section | a period where the quantity did not change |
A candle's height in centimeters after \(t\) hours is \(h(t) = 20 - 2.5t\).
a) Find the vertical intercept and say what it means.
b) Find the horizontal intercept and say what it means.
c) State the maximum and the minimum over the candle's reasonable domain.
Solution
a) \(h(0) = 20 - 0 = 20\), so the vertical intercept is \((0, 20)\). The candle was 20 cm tall when it was lit.
b) Solve \(20 - 2.5t = 0\):
$$2.5t = 20 \quad\Longrightarrow\quad t = 8$$The horizontal intercept is \((8, 0)\). After 8 hours the candle has burned all the way down.
c) The reasonable domain is \([0, 8]\) — the candle does not exist before it is lit or after it is gone. Over that stretch the outputs run from 20 down to 0, so the maximum is 20 cm (at \(t = 0\)) and the minimum is 0 cm (at \(t = 8\)).
Answer: a) \((0, 20)\), the starting height. b) \((8, 0)\), when it burns out. c) maximum 20 cm, minimum 0 cm.
Notice that the maximum is 20, not 0 — the maximum is the output, and \(t = 0\) is only when it happened.
1.4.10 Two Functions on One Graph
Everything so far has read one function at a time. The water park picture was drawn with two on it for a reason, and function notation is what lets you ask questions about both at once. Section 1.3 built the notation; here is what it was for.
Using \(a(x) = 28{,}000 - 1400x\) and \(d(x) = 24{,}000 - 1000x\): find \(a(5)\); solve \(d(x) = 2000\); find when \(a(x) = d(x)\); and find when \(a(x) > d(x)\).
Solution
Part 1 — find \(a(5)\).
$$a(5) = 28{,}000 - 1400(5) = 28{,}000 - 7000 = 21{,}000$$Five minutes after the pumps started, Aly's pool still held 21,000 gallons. On the graph it is the height of Aly's segment above \(x = 5\). Notice which number is which: the 5 is a time and the 21,000 is an amount of water.
Part 2 — solve \(d(x) = 2000\). This is the backwards question, and §1.4.3 already answered it: \(x = 22\). Dayne's pool is down to its last 2000 gallons at 22 minutes. Part 1 gave a time and asked for an amount; this one gives an amount and asks for a time. Same graph, opposite directions.
Part 3 — when is \(a(x) = d(x)\)? Two functions are equal when their outputs match, which on a graph is exactly where the two segments cross.
$$28{,}000 - 1400x = 24{,}000 - 1000x$$ $$4000 = 400x$$ $$x = 10$$Check both sides: \(d(10) = 24{,}000 - 10{,}000 = 14{,}000\), and \(a(10) = 28{,}000 - 14{,}000 = 14{,}000\) as well. Ten minutes in, both pools hold exactly 14,000 gallons.
Part 4 — when is \(a(x) > d(x)\)? Aly's pool holds more whenever Aly's segment is above Dayne's. It starts above at \(x = 0\) — 28,000 against 24,000 — and stays above until the crossing at \(x = 10\):
$$0 \le x < 10$$After that Aly's is below: the stronger pump has caught up and overtaken. And the answer stops making sense past \(x = 20\), where Aly's function runs out of domain altogether.
Answer: \(a(5) = 21{,}000\) gallons; \(d(x) = 2000\) at \(x = 22\) minutes; the pools are equal at \(x = 10\) minutes with 14,000 gallons each; and \(a(x) > d(x)\) on \(0 \le x < 10\).
That crossing is the point you spotted in the very first list of observations, before there were any numbers on the axes: at some point in time, both pools will have the same amount of water. It is worth dwelling on, because it is a fact about graphs you will use for the rest of the course: the input at which two graphs cross is the solution of the equation that sets them equal. Reading the crossing off the picture and solving the equation on paper are the same question asked two ways, and they had better give the same answer.
Notice too that the answer to a question about two functions is an interval of inputs, not a single number. "Which is bigger?" has different answers at different times, and naming when the answer changes is the whole content of the question.
Two candles are lit at the same time. One is \(h(t) = 20 - 2.5t\) and the other is \(g(t) = 30 - 5t\), both in centimeters after \(t\) hours.
a) Which is taller at the moment they are lit?
b) When are they the same height?
c) Which burns out first?
Solution
a) Evaluate both at \(t = 0\): \(h(0) = 20\) and \(g(0) = 30\). The second candle starts taller.
b) Set them equal:
$$20 - 2.5t = 30 - 5t$$ $$2.5t = 10$$ $$t = 4$$Check: \(h(4) = 20 - 10 = 10\) and \(g(4) = 30 - 20 = 10\). Both are 10 cm tall after 4 hours.
c) Solve each for zero. \(h(t) = 0\) gives \(t = 8\); \(g(t) = 0\) gives \(t = 6\). The second candle burns out first, at 6 hours, even though it started 10 cm taller — it burns twice as fast.
Answer: a) the second, at 30 cm. b) at 4 hours, both 10 cm. c) the second, at 6 hours.
This is the water park again with different numbers: the one that starts higher and falls faster is overtaken partway through and finishes first.
Both words in this section's title are just new names for things §1.3 already gave you. Domain is the collection of inputs; range is the collection of outputs. What the names bought you is the ability to ask a sharper question than "what does this function do?" — namely, what is this function even willing to accept, and what can it possibly produce?
That question turned out to have two kinds of answer, and separating them is the habit to carry forward. Sometimes an input is excluded because the arithmetic breaks on it: \(\dfrac{6}{x - 2}\) has nothing to hand back when \(x = 2\). Sometimes an input is excluded because the situation says so: Dayne's rule will happily report \(-6000\) gallons at 30 minutes, and a pool cannot hold that. The first exclusion you find by looking at the rule. The second you find only by reading the situation and thinking about what could really happen — and that thinking is what turns a formula into a model.
The vocabulary from the last two subsections — intercepts, maximum and minimum, increasing and decreasing, and the crossing of two graphs — is not finished with. Chapter 2 takes the straight lines that drained these two pools and studies them on their own terms, and every one of those features turns out to be readable straight off the equation.
Section 1.5 changes subject to exponents, and builds the rules for combining and simplifying powers. Two threads from this chapter run through it. The doubling sequence — \(3, 6, 12, 24\), and Growing, Growing Dots before it — is an exponent rule wearing a story, and the medicine leaving the dog's bloodstream, \(60(0.8)^t\), is the same shape pointed downward. Keep the domain question in your pocket while you are there, because a negative exponent puts a quantity underneath a fraction bar, and that reopens exactly the question you just spent this section learning to ask: is there any input this rule refuses?
Problem Set 1.4
Review — reaching back to §1.2 and §1.3.
Problem 1. Section 1.3 said the water depth in Sylvia's pool "never goes below zero and never goes above the rim." Rewrite that sentence using the word range.
Solution
Step 1 — identify what the original sentence is describing: "Never goes below zero and never goes above the rim" is a statement about which outputs (depths) the function actually produces — that is exactly what the word range names.
Step 2 — restate it using the vocabulary word: The range of the depth function runs from 0 up to the height of the rim; every depth in between occurs, and none outside it does.
Answer: The range of Sylvia's pool-depth function is from 0 up to the rim height, inclusive.
Problem 2. For §1.2's Growing Dots, \(D(t) = 4t + 1\), state the domain of the situation and say why it is discrete.
Solution
Step 1 — identify what \(t\) counts: In \(D(t) = 4t + 1\), \(t\) is the step (round) number of the Growing Dots pattern, not a length of time that can be split.
Step 2 — write the domain of the situation: The pattern begins at step 0 and continues through step 1, step 2, step 3, and so on, so the domain is \(\{0, 1, 2, 3, \ldots\}\).
Step 3 — say why it is discrete: There is no such thing as step \(2.5\) — a step number cannot be split into a fraction of a step, so the domain is a row of separated whole numbers rather than an unbroken stretch.
Answer: Domain \(\{0, 1, 2, 3, \ldots\}\); discrete because the input counts a step number, and step numbers cannot be split.
Problem 3. Explain how the vertical line test from §1.3 and the left-to-right sweep in this section are looking at the same graph for two different purposes.
Practice — this section's own work.
Solution
Step 1 — what the vertical line test checks: It sweeps a vertical line across the graph and asks, at each fixed input, whether more than one output is stacked above it. It is a check that the picture actually represents a function — that each input has exactly one output.
Step 2 — what the left-to-right sweep checks: It sweeps horizontally and asks which inputs the graph reaches at all, from the leftmost to the rightmost. It is a check of how wide the graph is — the domain.
Step 3 — say how they relate: Both scan the same picture, but the vertical line test moves a line left-to-right while asking a vertical question at each stop (how many points are directly above this \(x\)?), while the domain sweep asks a horizontal question over the whole picture (which \(x\)'s are used at all?). One verifies the graph is a function; the other measures the function's domain.
Answer: The vertical line test uses vertical checks at each input to confirm the graph is a function; the left-to-right sweep uses the same horizontal direction to find which inputs occur — one is about validity, the other about extent.
Problem 4. In your own words, state what the domain of a function is and what its range is.
Solution
Step 1 — state the domain in your own words: The domain of a function is the collection of all the input values the function is willing to accept and produce an output for.
Step 2 — state the range in your own words: The range of a function is the collection of all the output values the function actually produces when every allowed input is run through it.
Answer: Domain = the set of allowed inputs; range = the set of resulting outputs.
Problem 5. A function is given by the ordered pairs \((0, 4)\), \((1, 9)\), \((2, 16)\), and \((3, 25)\). State the domain and the range.
Solution
Step 1 — read the first coordinates for the domain: The pairs are \((0, 4)\), \((1, 9)\), \((2, 16)\), \((3, 25)\), so the first coordinates are \(0, 1, 2, 3\).
Step 2 — read the second coordinates for the range: The second coordinates are \(4, 9, 16, 25\).
Answer: Domain \(\{0, 1, 2, 3\}\); range \(\{4, 9, 16, 25\}\).
Problem 6. A function is given by the ordered pairs \((-3, 8)\), \((-1, 8)\), \((2, 5)\), and \((6, 1)\). State the domain and the range, listing each value only once.
Solution
Step 1 — read the first coordinates for the domain: The pairs are \((-3, 8)\), \((-1, 8)\), \((2, 5)\), \((6, 1)\), so the first coordinates are \(-3, -1, 2, 6\), all distinct.
Step 2 — read the second coordinates for the range, listing repeats once: The second coordinates are \(8, 8, 5, 1\). The value 8 appears twice (from inputs \(-3\) and \(-1\)), but a set records which values occur, not how many times, so 8 is listed only once.
Answer: Domain \(\{-3, -1, 2, 6\}\); range \(\{1, 5, 8\}\).
Problem 7. Find the domain of \(\;f(x) = \dfrac{4}{x - 7}\), and explain what goes wrong at the excluded value.
Solution
Step 1 — find where the rule breaks: \(f(x) = \dfrac{4}{x - 7}\) has a denominator that becomes 0 when \(x = 7\).
Step 2 — explain what goes wrong there: At \(x = 7\), the rule asks for \(\dfrac{4}{0}\), and division by zero has no value — there is no number the function can hand back.
Answer: The domain is \(x \neq 7\); at \(x = 7\) the rule would require dividing by zero, which is undefined.
Problem 8. Find the domain of \(\;g(x) = \dfrac{x + 1}{x + 5}\).
Solution
Step 1 — find where the denominator is zero: \(g(x) = \dfrac{x + 1}{x + 5}\) breaks when \(x + 5 = 0\), i.e., \(x = -5\).
Step 2 — state the domain: Every other real number is accepted, since the numerator being zero (at \(x = -1\)) causes no trouble — only the denominator matters.
Answer: The domain is \(x \neq -5\).
Problem 9. The area of a circle is \(A(r) = \pi r^2\), where \(r\) is the radius in centimeters. Give the reasonable domain and explain where the restriction comes from.
Solution
Step 1 — check what the bare rule accepts: \(A(r) = \pi r^2\) squares \(r\) and multiplies by \(\pi\); arithmetically this never breaks, so the rule alone would accept any real number, including negative ones.
Step 2 — bring in the situation: \(r\) is a radius, a physical length, and a length cannot be negative or (meaningfully) zero — a "circle" with radius 0 isn't a circle at all.
Step 3 — state the reasonable domain: \(r > 0\), or \((0, \infty)\).
Answer: Reasonable domain \((0, \infty)\); the restriction comes from the situation (a radius must be a positive length), not from the arithmetic.
Problem 10. A school club sells popcorn for $4 a bag and has 50 bags. The money collected is \(M(b) = 4b\).
a) State the reasonable domain.
b) State the range.
c) Is $18 a possible total? Explain.
Solution
Step 1 — reasonable domain: \(M(b) = 4b\) counts bags sold, and there are only 50 bags on hand. Bags cannot be split — there is no such thing as 2.5 bags sold — so the domain is discrete and must be listed: \(\{0, 1, 2, \ldots, 50\}\).
Step 2 — range: Running every allowed input through the rule gives multiples of 4 from 0 up to \(4(50) = 200\): \(\{0, 4, 8, \ldots, 200\}\).
Step 3 — is $18 possible? Solve \(4b = 18\): \(b = 4.5\). This is not a whole number of bags, so it is not an allowed input — and 18 is not a multiple of 4, confirming it is not in the range.
Answer: a) \(\{0, 1, 2, \ldots, 50\}\). b) \(\{0, 4, 8, \ldots, 200\}\). c) No — \(b = 4.5\) is not a whole number of bags, so $18 is never collected.
Problem 11. For \(C(n) = 15n\), the cost of a field trip at $15 per student, decide whether each of 45, 60, and 70 is a possible output.
Solution
Step 1 — work backwards for each candidate: For \(C(n) = 15n\), test whether each output corresponds to a whole-number input, since \(n\) counts students.
Step 2 — check 45: \(15n = 45 \Rightarrow n = 3\), a whole number. Possible.
Step 3 — check 60: \(15n = 60 \Rightarrow n = 4\), a whole number. Possible.
Step 4 — check 70: \(15n = 70 \Rightarrow n = 4.\overline{6}\), not a whole number of students. Not possible.
Answer: 45 and 60 are possible outputs (at \(n=3\) and \(n=4\)); 70 is not.
Problem 12. A rectangle has area \(36 \text{ cm}^2\), so its length is \(L(w) = \dfrac{36}{w}\). Decide whether each of \(4\), \(0\), \(-9\), and \(72\) is a possible input, and say why.
Solution
Step 1 — check \(w = 4\): \(L(4) = \dfrac{36}{4} = 9\). Positive width, positive length — a real \(4\text{ cm} \times 9\text{ cm}\) rectangle. Possible.
Step 2 — check \(w = 0\): \(\dfrac{36}{0}\) is undefined, so the arithmetic itself refuses this input. It also isn't a rectangle with zero width. Not possible — refused by both the arithmetic and the situation.
Step 3 — check \(w = -9\): The arithmetic computes \(L(-9) = -4\) without complaint, but a width cannot be negative. Not possible — refused by the situation only.
Step 4 — check \(w = 72\): \(L(72) = \dfrac{36}{72} = 0.5\). A thin but real rectangle. Possible.
Answer: \(4\) and \(72\) are possible inputs; \(0\) is refused by both the arithmetic (division by zero) and the situation; \(-9\) is refused only by the situation (a negative width).
Problem 13. Explain the difference between what a rule would accept and what a situation permits, using an example of your own.
Solution
Step 1 — state the general distinction: A rule is bare arithmetic and will accept any input that doesn't make it undefined; a situation additionally rules out inputs that don't make sense for the real thing being modeled, even when the arithmetic is perfectly happy to compute an answer.
Step 2 — give an original example: A laundromat charges $1.75 per wash cycle, so the cost for \(n\) cycles is \(C(n) = 1.75n\). The rule accepts any real number of cycles — feed it \(n = 2.5\) and it reports \(C(2.5) = \$4.375\) without complaint. But a washing machine only runs whole cycles; there is no such thing as running a machine for half a cycle. The situation restricts \(n\) to whole numbers, even though the arithmetic never objects to \(2.5\).
Answer: The rule accepts anything that doesn't break the arithmetic (like \(n = 2.5\) cycles, giving $4.375); the situation only permits whole numbers of cycles, since a machine can't run a fractional cycle.
Problem 14. The graph of a function is a straight segment from \((2, 1)\) to \((10, 9)\). State the domain and the range.
Solution
Step 1 — domain, sweeping left to right: The segment runs from input 2 to input 10, so the domain is \([2, 10]\).
Step 2 — range, sweeping bottom to top: The segment runs from output 1 to output 9, so the range is \([1, 9]\).
Answer: Domain \([2, 10]\); range \([1, 9]\).
Problem 15. A ball is thrown and its graph runs from \((0, 0)\) up to a highest point at \((3, 45)\) and back down to \((6, 0)\). State the domain and the range.
Solution
Step 1 — domain, sweeping left to right: The graph runs from \(t = 0\) (thrown) to \(t = 6\) (lands), so the domain is \([0, 6]\).
Step 2 — range, sweeping bottom to top: The lowest height reached is 0 (at the start and the end), and the highest is 45 (the peak), so the range is \([0, 45]\).
Answer: Domain \([0, 6]\); range \([0, 45]\).
Problem 16. A partial graph of a function leaves the top of the picture while still climbing. Explain why you cannot report a maximum for it.
Solution
Step 1 — recall what a maximum requires: A maximum is the greatest output the function actually reaches, which means the graph must show its true top.
Step 2 — explain the problem with a partial graph: If the curve is still climbing when it exits the top of the frame, the picture has not shown where — or whether — the graph stops climbing. Any "highest point visible" is only the highest point that happened to fit in the frame, not necessarily the function's true maximum.
Answer: You cannot report a maximum because the graph is cut off before reaching its top; a partial graph shows only part of the function, and the true peak may lie beyond the edge of the picture (or may not exist at all).
Problem 17. Decide whether each is discrete or continuous, and say why.
a) the number of chairs in a classroom as a function of the number of tables
b) the weight of water in a bucket as a function of the time it has been filling
c) the number of buses needed as a function of the number of passengers
Solution
Step 1 — part a), chairs as a function of tables: The number of tables cannot be split — there is no such thing as \(2.5\) tables — so this is discrete.
Step 2 — part b), weight of water as a function of time filling: Time can be split as finely as you like, and the weight increases smoothly with no gaps, so this is continuous.
Step 3 — part c), number of buses as a function of number of passengers: Passengers cannot be split into fractions, and buses are counted in whole units, so this is discrete.
Answer: a) discrete (tables can't be split). b) continuous (time can be split arbitrarily finely). c) discrete (passengers can't be split).
Problem 18. Explain why writing the domain of \(B(n) = 8n\), for a delivery of at most 12 boxes, as \(0 \le n \le 12\) is wrong. Then write it correctly.
Solution
Step 1 — explain why the inequality is wrong: \(0 \le n \le 12\) is an interval, and an interval is an unbroken stretch of the number line — it claims every real number between 0 and 12, including \(6.5\) boxes, is an allowed input. But boxes cannot be split; there is no such thing as \(6.5\) boxes arriving. The domain has gaps in it, so an interval misdescribes it — this is wrong, not merely informal.
Step 2 — write it correctly: List the whole numbers only: \(\{0, 1, 2, \ldots, 12\}\).
Answer: \(0 \le n \le 12\) falsely claims fractional numbers of boxes are allowed. The correct domain is the discrete list \(\{0, 1, 2, \ldots, 12\}\).
Problem 19. A garage charges a $6 entry fee plus $3 an hour and closes 8 hours after opening, so \(P(h) = 3h + 6\). Find the domain and the range, and say whether the function is discrete or continuous.
Solution
Step 1 — domain: The garage is open for parking durations from 0 up to 8 hours, and hours can be split as finely as you like (a fraction of an hour is a sensible parking time), so the domain is the interval \([0, 8]\).
Step 2 — range: \(P(h) = 3h + 6\) is increasing, so evaluate the endpoints: \(P(0) = 6\) and \(P(8) = 3(8) + 6 = 30\). Since the domain is an unbroken interval and the function is continuous, every value between is reached too, giving range \([6, 30]\).
Step 3 — discrete or continuous: Time (hours parked) can be split arbitrarily finely, so the function is continuous.
Answer: Domain \([0, 8]\); range \([6, 30]\); continuous.
Problem 20. For \(f(x) = 3x - 12\), find the vertical intercept and the horizontal intercept.
Solution
Step 1 — vertical intercept: Evaluate at 0: \(f(0) = 3(0) - 12 = -12\), so the vertical intercept is \((0, -12)\).
Step 2 — horizontal intercept: Solve \(3x - 12 = 0\): \(3x = 12 \Rightarrow x = 4\), so the horizontal intercept is \((4, 0)\).
Answer: Vertical intercept \((0, -12)\); horizontal intercept \((4, 0)\).
Problem 21. For \(f(x) = -5x + 20\), find the vertical intercept and the horizontal intercept.
Solution
Step 1 — vertical intercept: Evaluate at 0: \(f(0) = -5(0) + 20 = 20\), so the vertical intercept is \((0, 20)\).
Step 2 — horizontal intercept: Solve \(-5x + 20 = 0\): \(-5x = -20 \Rightarrow x = 4\), so the horizontal intercept is \((4, 0)\).
Answer: Vertical intercept \((0, 20)\); horizontal intercept \((4, 0)\).
Problem 22. Explain why a straight line with a nonzero slope has no maximum and no minimum.
Solution
Step 1 — describe the shape of such a line: A straight line with a nonzero slope climbs forever in one direction and falls forever in the other — it never levels off.
Step 2 — connect that to maximum and minimum: A maximum requires a greatest output the graph actually reaches, but if the line climbs without bound as \(x \to \infty\) (for positive slope) or as \(x \to -\infty\) (for negative slope), no output is ever the greatest — there's always a bigger one further along. The same reasoning rules out a least output on the falling side.
Answer: A nonzero-slope line has no maximum and no minimum because it increases without bound in one direction and decreases without bound in the other, so no output is ever the largest or smallest.
Problem 23. The height of a rocket in meters is \(H(t)\), and its maximum is \(H(9) = 400\). State the greatest height reached and when it happened, and explain which number is the output.
Solution
Step 1 — identify the output and the input in \(H(9) = 400\): The maximum is the output — 400 — and the input 9 tells you when that output occurred.
Step 2 — state the answer in words: The rocket reached its greatest height of 400 meters at \(t = 9\) seconds.
Answer: Greatest height reached: 400 meters, occurring at \(t = 9\) seconds. The 400 is the output (the maximum); the 9 is only the input telling you when it happened.
Problem 24. A candle's height in centimeters is \(h(t)\), with a horizontal intercept at \(t = 6\) hours. Say what that intercept means about the candle.
Solution
Step 1 — recall what a horizontal intercept means: It's the point where the output is 0 — here, where the candle's height is 0 centimeters.
Step 2 — apply it to the candle: At \(t = 6\) hours, \(h(6) = 0\), so the candle has burned all the way down.
Answer: The intercept \((6, 0)\) means the candle has completely burned out by 6 hours.
Problem 25. The temperature of a cooling drink flattens out near room temperature and never reaches \(0^{\circ}\text{F}\). Explain why its graph has no horizontal intercept, and what that missing feature tells you.
Solution
Step 1 — recall what a horizontal intercept requires: It requires some input at which the output is exactly 0 — here, some time at which the temperature is exactly \(0^{\circ}\text{F}\).
Step 2 — explain why that never happens: The temperature only flattens out near room temperature and never actually reaches \(0^{\circ}\text{F}\); it gets arbitrarily close to its limiting value without arriving at 0. Since no input ever produces an output of exactly 0, there is no horizontal intercept.
Step 3 — say what the missing feature tells you: The graph never crosses the time axis, which tells you the drink never actually cools all the way down to \(0^{\circ}\text{F}\) — it only approaches some warmer limiting temperature.
Answer: No horizontal intercept exists because the temperature never reaches \(0^{\circ}\text{F}\); the missing intercept signals that the drink approaches, but never attains, that temperature.
Problem 26. A graph of the total number of books a library has received over a year is flat from week 12 to week 15. Explain what that flat stretch means.
Solution
Step 1 — recall what a flat stretch means: A horizontal section of a graph means the output did not change over that stretch of inputs.
Step 2 — apply it here: The output is the total number of books received so far (a running total), and it stays the same from week 12 to week 15.
Answer: The flat stretch means the library received no new books between week 12 and week 15 — the running total held steady because nothing new arrived during that period.
Problem 27. A swing seat hangs from chains 8 feet long. Let \(d\) give the distance from the seat to the top beam, in feet, as a function of time. State the range and explain why it holds only one number.
Solution
Step 1 — think about what stays fixed as the swing moves: The chains are a fixed length, 8 feet, and (assumed taut) don't stretch or go slack as the seat swings back and forth.
Step 2 — state the range: No matter what time \(t\) you pick, the straight-line distance from the seat to the top beam along the taut chain is always 8 feet — the seat's position along the swing's arc changes, but its distance from the beam does not.
Answer: The range is \(\{8\}\) — it holds only one number because the chain length is fixed at 8 feet, so the seat-to-beam distance never changes as the swing moves.
Problem 28. Using \(a(x) = 28{,}000 - 1400x\), find \(a(12)\) and say what it means about Aly's pool.
Solution
Step 1 — evaluate \(a(12)\):
$$a(12) = 28{,}000 - 1400(12) = 28{,}000 - 16{,}800 = 11{,}200$$Step 2 — say what it means: \(x = 12\) is a time in minutes since the pump started, and \(11{,}200\) is the resulting number of gallons remaining.
Answer: \(a(12) = 11{,}200\); twelve minutes after the pump started, Aly's pool holds 11,200 gallons.
Problem 29. Two tanks drain according to \(u(x) = 18{,}000 - 900x\) and \(v(x) = 12{,}000 - 400x\). Find when they hold the same amount, and say how much that is.
Preview — pointing at §1.5.
Solution
Step 1 — set the two rules equal:
$$18{,}000 - 900x = 12{,}000 - 400x$$Step 2 — solve for \(x\):
$$6000 = 500x \quad\Longrightarrow\quad x = 12$$Step 3 — find the shared amount and check both sides:
$$u(12) = 18{,}000 - 900(12) = 18{,}000 - 10{,}800 = 7200$$ $$v(12) = 12{,}000 - 400(12) = 12{,}000 - 4800 = 7200$$Both agree.
Answer: The tanks hold the same amount at \(x = 12\) minutes, when each holds 7200 gallons.
Problem 30. The medicine function \(M(t) = 60(0.8)^t\) has a base smaller than 1. Compute \(M(0)\), \(M(1)\) and \(M(2)\), and describe in words what the base being less than 1 does to the outputs.
Solution
Step 1 — compute \(M(0)\): \(M(0) = 60(0.8)^0 = 60(1) = 60\).
Step 2 — compute \(M(1)\): \(M(1) = 60(0.8)^1 = 48\).
Step 3 — compute \(M(2)\): \(M(2) = 60(0.8)^2 = 60(0.64) = 38.4\).
Step 4 — describe the effect of a base less than 1: Each output is 80% of the one before it — a base smaller than 1 makes the outputs shrink every time the input increases by 1. The values get smaller and smaller, always positive, getting closer to (but never reaching) zero.
Answer: \(M(0) = 60\), \(M(1) = 48\), \(M(2) = 38.4\). A base less than 1 makes the function decay — each output is a fixed fraction (here, 80%) of the previous one, so the outputs shrink steadily toward zero without ever reaching it.
Problem 31. The paper-cutting function is \(P(n) = 3^n\). What would \(P(0)\) have to be for the pattern of the other values to continue, and what does that answer say about raising a number to the power zero?
Solution
Step 1 — see how each output relates to the one before it: \(P(n) = 3^n\) triples with each round: \(P(1) = 3\), and in general \(P(n) = 3 \cdot P(n-1)\).
Step 2 — apply that relationship backwards to \(n = 0\): For the pattern to continue, \(P(1)\) should equal \(3 \cdot P(0)\). Since \(P(1) = 3\), this requires \(3 \cdot P(0) = 3\), so \(P(0) = 1\).
Step 3 — say what this means about raising to the power zero: For the pattern to hold, any (nonzero) base raised to the power 0 must equal 1 — not 0 and not the base itself.
Answer: \(P(0) = 1\); this shows that raising a number to the power zero gives 1, keeping the tripling pattern consistent one round earlier than \(n=1\).
Key Terms
domain — the set of all input values a function accepts.
range — the set of all output values a function produces.
interval notation — a way of writing a set of numbers by its two endpoints, square bracket for included and round for excluded.
reasonable domain — the set of inputs that make sense in the situation being modeled.
discrete function — a function whose domain is separated values with gaps between them.
continuous function — a function whose domain is an unbroken stretch and whose graph has no breaks.
vertical intercept — the point where a graph crosses the vertical axis, \((0, f(0))\).
horizontal intercept — a point where a graph crosses the horizontal axis, \((a, 0)\) with \(f(a) = 0\).
zero of a function — an input whose output is 0.
maximum — the greatest output a function reaches.
minimum — the least output a function reaches.