Function notation in a real situation
A statement such as C(12) = 40 packs a whole sentence into a few symbols. Which sentence is it, and how do you turn sentences about a real situation into notation and back again?
Where you are 3 in Function notation
When you reach for it
Evaluating a function turns an input into an output. In a real situation the harder half is the meaning. A table or a chart hands you statements such as P(10) = 5300 or U(3) = U(11), and the job is to say what they claim, in a sentence with units. Just as often a question asked in words has to be written in notation before anyone can answer it. This lesson is about that reading and writing, not about computing.
The procedure
Input, output and their units
A function in a real situation pairs two quantities, and its description says what each measures:
C(n) is the cost, in dollars, of n skeins of wool.
That fixes the name, C; the input n, a number of skeins; and the output C(n), an amount in dollars. The input is the quantity that is chosen and the output is the one that depends on it (Independent and dependent variables).
Now read C(12) = 40. The number inside the brackets is always the input, so it is 12 skeins; the number on the other side is the output, 40 dollars. The statement says
Twelve skeins of wool cost 40 dollars.
The units are the check. The swapped reading, “40 skeins cost 12 dollars”, is a different claim, written C(40) = 12.
When the input is a time, the description must also say where the clock starts. Let T(t) be the temperature, in degrees Celsius, of a pie t minutes after it comes out of the oven. Then T(0) = 100 says “the pie comes out of the oven at 100^\circ\text{C}”, and T(20) = 60 says “twenty minutes after it comes out, the pie is at 60^\circ\text{C}”. Without the starting point, T(20) would not say twenty minutes after what.
A quantity, a claim, a question
The same notation writes three different things.
| Written | What it is | In words |
|---|---|---|
| C(12) | a quantity | the cost of 12 skeins |
| C(12) = 40 | a claim, true or false | 12 skeins cost 40 dollars |
| C(n) = 100 | a question about n | how many skeins cost 100 dollars? |
On its own, C(12) is a noun: it names an amount of money and says nothing about it. An equals sign and a number make it a claim. When the input is unknown, the equation is a question, and its answer is an input: here, a number of skeins.
Comparing two outputs
Each f(\text{input}) is an output, in the output’s units, so comparisons read the same way.
- f(a) = f(b): the inputs a and b give the same output.
- f(a) > f(b): the output at a is larger than the output at b.
- f(a) - f(b) = d: the output at a is d more than at b, with d in the output’s units.
A bike-share scheme counts its bikes out on the streets through one weekday. Let R(h) be the number of bikes in use h hours after midnight, for 0 \le h \le 24, graphed below as a simplified model made of straight pieces. Then R(10) = R(14) says “as many bikes were in use at 10 in the morning as at 2 in the afternoon” (200 each), and R(17) > R(12) says “more bikes were in use at 5 in the afternoon than at noon”.
R(10) = R(14) does not say 10 = 14. Two inputs may share an output; what a function forbids is one input with two outputs. So f(a) \neq f(b) guarantees a \neq b, but equal outputs guarantee nothing about the inputs.
What solving f(x) = k asks
An equation f(x) = k fixes an output and asks for the input, so its answer is in the input’s units. “Solve R(h) = 200” asks: at what times were exactly 200 bikes in use? The dashed line at 200 meets the graph four times, at 7 and 10 in the morning, 2 in the afternoon and 7 in the evening. Finding such values on a picture is reading a value from a graph; here the point is what the question asks and what kind of thing answers it.
The situation tells you how many answers to expect. A quantity that only falls, like a cooling pie’s temperature, or only rises, passes each value at most once. A quantity that rises and falls, like the bikes in use, can return to a value several times. A value never reached gives no answer at all, and “never” is a complete answer.
Which inputs make sense
A formula accepts many numbers that the situation does not. The inputs that make sense form the function’s reasonable domain, and a value or a solution outside it does not describe the situation, even when the algebra is correct.
Worked examples
Reading statements aloud
P(y) is the number of people living in a town at the start of the year 2000 + y. Say in words what each means. (a) P(10) = 5300 (b) P(0) = 4100 (c) P(25) (d) P(y) = 6500
SolutionThe input y is a number of years after 2000; the output is a number of people.
(a) Input 10 is the start of 2010, output 5300 people: “At the start of 2010, 5300 people lived in the town.”
(b) Input 0 is where the counting starts: “At the start of 2000 the town had 4100 people.”
(c) There is no equals sign, so this is a quantity, not a claim: “the number of people in the town at the start of 2025”.
(d) The input is unknown, so this is a question: “In which year did (or will) the town have 6500 people at the start of the year?” The answer is a value of y, and the year is 2000 + y.
From words to notation
Use the bike-share function R(h) and its graph. Write each sentence in function notation. (a) At 8 in the morning, 350 bikes were in use. (b) More bikes were in use at 5 in the afternoon than at noon. (c) How many bikes were in use at half past seven in the morning? (d) At what times were 100 bikes in use? Answer (d) from the graph.
SolutionIn each sentence the time in hours after midnight is the input and the number of bikes the output.
(a) R(8) = 350, not R(350) = 8, which would claim 8 bikes in use 350 hours after midnight.
(b) Five in the afternoon is 17 hours after midnight and noon is 12, and “more bikes” compares two outputs: R(17) > R(12). The graph gives 400 and 250, so it is true.
(c) Half past seven is 7.5 hours after midnight. The sentence asks for an output, so its notation is a quantity with no equals sign: R(7.5).
(d) Now the count is known and the time is wanted, so this is an equation to solve for h: R(h) = 100. The graph reaches 100 once on the morning climb, at h = 6, and once on the evening fall, halfway from 150 at h = 20 to 50 at h = 22, which is h = 21: 6 in the morning and 9 in the evening.
Comparisons from a table
A school records its electricity use each month for one year. U(m) is the energy used in month m, in kilowatt-hours, with m = 1 for January up to m = 12 for December.
| m | Month | U(m) (kWh) |
|---|---|---|
| 1 | January | 42\,000 |
| 2 | February | 41\,000 |
| 3 | March | 35\,000 |
| 4 | April | 28\,000 |
| 5 | May | 24\,000 |
| 6 | June | 22\,000 |
| 7 | July | 9\,000 |
| 8 | August | 11\,000 |
| 9 | September | 24\,000 |
| 10 | October | 30\,000 |
| 11 | November | 35\,000 |
| 12 | December | 39\,000 |
Say each statement in words and decide whether it is true. (a) U(1) > U(6) (b) U(3) = U(11) (c) U(10) - U(9) = 6000 (d) U(4) < U(8) (e) What does U(m) = 30\,000 ask, and what is its answer?
SolutionEach U(\text{month}) is an amount of energy, so every comparison is between amounts of energy.
(a) “The school used more energy in January than in June.” 42\,000 > 22\,000, so it is true.
(b) “The school used the same energy in March as in November.” Both are 35\,000, so it is true: two different months, one output.
(c) “In October the school used 6000 kilowatt-hours more than in September.” 30\,000 - 24\,000 = 6000, so it is true.
(d) “The school used less energy in April than in August.” April used 28\,000 and August 11\,000, so the claim is false.
(e) It asks: “In which month did the school use exactly 30\,000 kilowatt-hours?” Only one row has that output, so m = 10: October. The answer is a month, not an amount of energy.
What solving asks, on a graph
The pie’s temperature T(t), in degrees Celsius t minutes after it leaves the oven, is read to the nearest degree below and drawn in the second figure.
| t (minutes) | T(t) (^\circC) |
|---|---|
| 0 | 100 |
| 10 | 77 |
| 20 | 60 |
| 30 | 48 |
| 40 | 40 |
| 50 | 34 |
| 60 | 30 |
(a) Say what T(20) = 60 claims, and check it. (b) What does T(t) = 48 ask, and what is the answer? (c) Solve T(t) = 70 from the graph and explain why there is exactly one answer. (d) What does T(t) = 15 ask, and what is the answer?
Solution(a) “Twenty minutes after it came out of the oven, the pie was at 60^\circ\text{C}.” The row for t = 20 shows 60, so it is true, to the nearest degree.
(b) It asks: “How many minutes after leaving the oven was the pie at 48^\circ\text{C}?” The output 48 is in the row for t = 30, so the answer is about 30 minutes, in the input’s unit. “About”, because the table is rounded.
(c) Go across from 70 to the curve, then down: about t = 14, between the table’s 77^\circ at 10 minutes and 60^\circ at 20. There is only one answer because the pie only cools, so it passes 70 once and never comes back.
(d) It asks: “When was the pie at 15^\circ\text{C}?” The pie cools toward the room’s 20^\circ\text{C} and never gets down to it, so no time works. “Never” answers the question.
Which inputs make sense
For each function, say which inputs make sense and whether the value named means anything. (a) The wool cost C(n), and C(2.5). (b) The pie’s temperature T(t), and T(-10). (c) The bikes in use R(h), and R(7.5). (d) The school’s energy U(m), and U(13).
Solution(a) n counts skeins sold whole, so the sensible inputs are the whole numbers 0, 1, 2, 3 and so on, up to the shop’s stock. C(2.5) means nothing, even if a pricing rule prints a number for it.
(b) Any time t \ge 0 makes sense, fractions included. T(-10) would be ten minutes before the pie came out, when it was still in the oven, and the function describes only the cooling.
(c) h is a time of day, 0 \le h \le 24, and any time is allowed, so R(7.5) is the number of bikes in use at half past seven.
(d) m labels a month, so only 1, 2, \dots, 12 make sense. U(13) means nothing, since the function covers one year, and neither does U(2.5): a month number names a month rather than measuring time. A fraction is fine for a time and meaningless for a count or a label.
Two towns, one question
The rule P(y) = 4100 + 120y fits the town of Example 1: P(0) = 4100 and P(10) = 5300. A neighboring hill town has Q(y) = 5900 - 30y people at the start of the year 2000 + y. (a) What does P(y) = Q(y) ask? Solve it. (b) Say what P(y) > Q(y) means and find when it holds. (c) Solve Q(y) = 0 and say whether to trust the answer.
Solution(a) Both functions get the same input, so it asks: “In which year did the two towns have the same population?”
adding 30y to both sides and subtracting 4100, then dividing by 150. At the start of 2012 the towns were level: P(12) = 4100 + 1440 = 5540 and Q(12) = 5900 - 360 = 5540.
(b) It says “at the start of the year 2000 + y, the first town has more people than the hill town.” The same moves give 150y > 1800, so y > 12; dividing by the positive 150 keeps the direction. Since y counts whole years, the first town is larger from 2013 on, while the rules hold.
(c) It asks: “When would the hill town have nobody left?” From 5900 - 30y = 0,
which points to late in 2196. The algebra is right, but a straight-line fit to a few decades of counts says nothing reliable two centuries on.
Boundaries
The description comes first. P(10) = 5300 means nothing until it is said what the input measures, in what unit and from what starting point, and what the output measures.
Equal outputs say nothing about the inputs. f(a) = f(b) can hold with a \neq b.
Only inputs in the reasonable domain carry meaning, even when the arithmetic works.
Rounded and modeled values are approximate. T(30) = 48 from a table read to the nearest degree means “about 48^\circ\text{C}”, and a fitted rule is only as close as its fit.
Where this goes next
Reading the 4100 and the 120 in P(y) = 4100 + 120y is linear functions in context and interpreting the numbers in a model. A difference such as U(10) - U(9), divided by the change in input, is an average rate of change. P(y) = Q(y) is solved with graphs and tables in solving f(x) = g(x), and the peaks of the bike graph are turning points. An expression in the brackets belongs to evaluating at an expression, and choosing a rule for data is modeling with functions.
Exercises
Saying it in words
- Using the school’s table in Example 3, say U(2) = 41\,000 in words, and check it.
- For the cooling pie, say T(15) = 68 in words, and say whether it fits the table in Example 4.
- For the bike-share graph, say R(17) - R(8) = 50 in words, and check it against the graph.
- For the town of Example 1, say P(30) - P(20) = 1200 in words.
- For the same town, what does the equation P(y) = 5000 ask? What kind of number is its answer, and in what unit?
Into notation
- Write in notation: “Forty minutes after it came out of the oven, the pie had cooled to 40^\circ\text{C}.” Say which 40 is which.
- Write in notation: “The school used less energy in April than in March.” Is it true?
- Write in notation: “When were 150 bikes in use?” Then answer it from the graph.
- Write in notation: “At the start of 2015 the town of Example 1 had 1200 more people than at the start of 2005.”
Reading the bike-share graph
- Find R(8) and say what it means.
- True or false: R(12) > R(16). Say the statement in words.
- Solve R(h) = 200 and give each answer as a time of day.
- Solve R(h) = 450.
- Find every input h with R(h) = 20. How many are there?
The school, the pie and the town
- Solve U(m) = 24\,000 and U(m) = 25\,000, and say what each equation asks.
- For which months m is U(m) > U(12)? Say in words what you found.
- Esmond extends the pie’s curve to the left and reads off a value for T(-10). Does it say anything about the pie? Then explain why T(t) = 110 has no solution.
- For the town with P(y) = 4100 + 120y: (a) solve P(y) = 4700 and give the year; (b) solve P(y) = 3500 and say what the answer would mean and whether to trust it; (c) solve P(y) = 4750, and say what the answer tells you about the counts taken each year.
Reasoning
- Delphine-Mae reads T(30) = 48 as “the pie was at 30^\circ\text{C} after 48 minutes”. Explain which reading is right and how the units settle it.
- Explain why T(t) = 70 has exactly one solution but R(h) = 200 has four. Then Colm says: “U(3) = U(11), so 3 = 11, which is impossible, so the table must be wrong.” What is wrong with his reasoning?
Answers
- “The school used 41\,000 kilowatt-hours in February.” True: the row for m = 2 shows 41\,000.
- “Fifteen minutes after it came out of the oven, the pie was at 68^\circ\text{C}.” It fits: 68 lies between 77 at 10 minutes and 60 at 20.
- “At 5 in the afternoon, 50 more bikes were in use than at 8 in the morning.” True: R(17) = 400 and R(8) = 350.
- “The town had 1200 more people at the start of 2030 than at the start of 2020.” If the counts stop before 2030, it is a forecast.
- It asks in which year the town had (or will have) 5000 people at the start of the year. The answer is a value of y, a number of years after 2000.
- T(40) = 40. The 40 in the brackets is the input, 40 minutes; the 40 on the right is the output, 40^\circ\text{C}.
- U(4) < U(3). True: 28\,000 < 35\,000.
- Solve R(h) = 150. The answers are h = 6.5 (halfway from 100 at 6 to 200 at 7) and h = 20: half past six in the morning and 8 in the evening.
- P(15) - P(5) = 1200, since 2015 is 15 years after 2000 and 2005 is 5.
- R(8) = 350: at 8 in the morning, 350 bikes were in use, the morning peak.
- False. It says “more bikes were in use at noon than at 4 in the afternoon”, but R(12) = 250 and R(16) = 300.
- h = 7, 10, 14 and 19: 7 and 10 in the morning, 2 in the afternoon and 7 in the evening.
- No solution. The most bikes in use at any time is 400, at 5 in the afternoon.
- Every h with 0 \le h \le 5, and also h = 24: infinitely many inputs. In this simplified model the count stays at 20 from midnight to 5 in the morning, and is back to 20 at the next midnight.
- Each asks in which months the school used exactly that many kilowatt-hours. U(m) = 24\,000: m = 5 and m = 9, May and September. U(m) = 25\,000: no month has that output.
- U(12) = 39\,000, and only m = 1 (42\,000) and m = 2 (41\,000) are larger: only January and February used more energy than December.
- No. Ten minutes before it left the oven the pie was in the oven, outside the cooling the function describes. T(t) = 110 asks when the pie was at 110^\circ\text{C}; it starts at 100 and only cools, so never.
- (a) 120y = 600, so y = 5: the start of 2005. (b) 120y = -600, so y = -5: the start of 1995, before the years the rule describes, so a guess rather than a fact. (c) 120y = 650, so \displaystyle y = \frac{65}{12} \approx 5.42, not whole. The counts were P(5) = 4700 and P(6) = 4820, so no yearly count was exactly 4750; by the rule the town passed it during 2005.
- The number in the brackets is the input, here minutes, so T(30) = 48 says “after 30 minutes the pie was at 48^\circ\text{C}”. Her reading would be written T(48) = 30.
- The pie only cools, so its temperature passes 70^\circ\text{C} once; the bikes in use rise and fall, passing 200 four times. Colm assumes equal outputs force equal inputs, but a function only forbids one input having two outputs, so March and November can both use 35\,000 kilowatt-hours.