Where a model holds: domain, interpolation and extrapolation
A model that fits the data will give an answer for any input you hand it. For which inputs can that answer be believed?
Where you are 5 in Modeling with functions
Why this exists
A model is fitted to a handful of measurements and then asked about cases nobody measured: a pumpkin’s mass in a week it was not weighed, the drinks a kiosk will sell on a hotter day than any on record, next year’s winning time. The formula answers every such question in the same tone. It cannot tell you which of its answers rest on data and which are guesses about a world it never saw. That has to come from outside the formula: from where the data lie, and from what the situation allows.
Line of best fit, Interpreting the slope and intercept, Modeling with linear relationships, Fitting a quadratic to data and A reasonable domain for a real situation each met this in passing. This lesson gathers those warnings into one idea.
Definition
Measured data cover a stretch of inputs, from the smallest input measured to the largest. Using a model at an input inside that stretch, the two ends included, is interpolation. Using it at an input outside the stretch is extrapolation. (Latin inter, between, and extra, outside.)
When a question gives an output and asks for the input (“when will it reach 30 kg?”), the input the model lands on decides which it is.
The domain of a model is the set of inputs on which the model is trusted to describe the situation. It is stated with the model, as part of it:
The formula alone accepts every number, so a model stated without its domain is incomplete. The domain is never wider than the reasonable domain the situation allows, and when the model was fitted to data it is usually the data’s stretch, or a little beyond it if there is a reason to trust a short step further.
Intuition
Severin keeps a drinks kiosk on a beach. On six days he recorded the day’s highest temperature t in °C and the number of cans of cold drink sold, D (made-up data): (18, 64), (21, 79), (23, 94), (25, 106), (27, 115) and (30, 136). The line D = 6t - 45, drawn by eye as in Line of best fit, passes within 2 cans of every day.
Inside the shaded band, every prediction has measured days on both sides of it. At 26 °C the line says 111 cans, and the days at 25 °C and 27 °C, which sold 106 and 115, stand on either side. For the line to be badly wrong at 26 °C, sales would have to jump and then jump back between two days the line fits closely.
Beyond 30 °C there is no day to check against. The line keeps climbing because climbing is all a line can do, not because anything measured says sales keep rising at 6 cans a degree. One degree past the hottest day, the pattern would have to change abruptly to make the line badly wrong. Ten degrees past, there is room for any change at all.
Results
Interpolation is checked; extrapolation is assumed. Inside the data’s stretch the model has been compared with measurements on both sides of the input. Outside it, the answer assumes the pattern carries on, and nothing measured says it does.
Trust falls with distance. Just past an end of the data a prediction is usually close. Far past it, the prediction can be anything.
The data cannot choose between models outside their stretch. Two models that both fit the same data closely can give answers far apart outside it, and the further out, the further apart.
A model must stop before an impossible output. If a count would drop below 0, a time would reach 0, or sales would pass the stock, the model’s domain ends before the input where that happens.
Why it is true
The first two are what the words mean: outside, nothing was compared, and the further out, the longer the stretch over which the pattern could change unseen. The third is Example 3 below. Evander’s line and Jessamy’s exponential both land within 0.1 kg of all five weighings of a pumpkin, made at weeks 0 to 4. Yet at week 12 they give 23.7 kg and 31.4 kg, about 7.7 kg apart, and at week 30 more than 100 kg apart. The weighings agree with both models, so nothing in them can say which pattern carries on past week 4. Data pin a model down only where the data are; outside, the choice of model is doing all the work.
The fourth follows from how lines behave. A falling line T = a - bx with b > 0 reaches 0 at \displaystyle x = \frac{a}{b}, however small b is; a smaller b only pushes that point further out. So a falling line used for something that must stay above 0 has to stop somewhere before \displaystyle x = \frac{a}{b}.
Worked examples
Inside or outside the kiosk’s data
Use Severin’s model D = 6t - 45, fitted to days from 18 °C to 30 °C. Predict the sales at (a) 26 °C, (b) 31 °C, (c) 12 °C and (d) 40 °C. Say whether each is interpolation or extrapolation, and how far it can be trusted.
SolutionThe data’s stretch is 18 \le t \le 30.
(a) 26 lies inside, so this is interpolation. D = 6 \times 26 - 45 = 111. It sits between the 106 cans of the 25 °C day and the 115 of the 27 °C day, and the line is within 2 cans of every day, so about 111 cans, give or take a few.
(b) 31 lies 1 degree past the hottest day: extrapolation. D = 6 \times 31 - 45 = 141. So short a step is a fair estimate, held less firmly than (a).
(c) 12 lies 6 degrees below the coolest day: extrapolation. D = 6 \times 12 - 45 = 27. On a cool day the beach may be nearly empty, and nothing measured says sales follow the same line there.
(d) 40 lies 10 degrees past the hottest day: extrapolation, far out. D = 6 \times 40 - 45 = 195. Not to be trusted; Example 2 finds a reason it cannot be right.
Where the kiosk’s line must stop
Severin’s kiosk can stock at most 180 cans a day. For which temperatures does D = 6t - 45 give an impossible number? State the model with its domain.
SolutionTwo edges come from the situation, as in A reasonable domain for a real situation.
Sales cannot be negative. 6t - 45 = 0 gives t = 7.5, and below 7.5 °C the line gives fewer than 0 cans.
Sales cannot pass the stock. 6t - 45 = 180 gives 6t = 225, so t = 37.5, and above 37.5 °C the line sells more cans than the kiosk holds.
So outside 7.5 \le t \le 37.5 the model is certainly wrong. That is an outer limit, not the stretch where it is trusted: inside it, only 18 to 30 °C is backed by measured days. Stated with its domain, the model is
where t is the day’s highest temperature in °C and D the cans sold. Someone could argue for a degree more at either end; stating the domain makes that judgment visible.
Two models through the same five weighings
Faustina weighed her pumpkin once a week (made-up data). With w the weeks since the first weighing and M the mass in kilograms: (0, 10.0), (1, 11.1), (2, 12.1), (3, 13.3) and (4, 14.6). Evander fits the line M = 9.9 + 1.15w, a steady 1.15 kg a week. Jessamy fits the exponential M = 10 \times 1.1^w, a steady 10\% a week (see Interpreting the numbers in a model). Compare the two inside the data, then predict the mass at week 12 and at week 30.
Inside the data, put each week into each model:
| w | measured | line | exponential |
|---|---|---|---|
| 0 | 10.0 | 9.9 | 10 |
| 1 | 11.1 | 11.05 | 11 |
| 2 | 12.1 | 12.2 | 12.1 |
| 3 | 13.3 | 13.35 | 13.31 |
| 4 | 14.6 | 14.5 | 14.641 |
Both models land within 0.1 kg of every weighing, a single step on a scale that reads to 0.1 kg, and in the figure they lie almost on top of each other across the band.
At week 12 the line gives 9.9 + 1.15 \times 12 = 23.7 kg and the exponential gives 10 \times 1.1^{12} \approx 31.4 kg. They differ by about 7.7 kg, more than the 4.6 kg the pumpkin gained over all five weighings.
At week 30 the line gives 9.9 + 1.15 \times 30 = 44.4 kg and the exponential 10 \times 1.1^{30} \approx 174.5 kg.
Which is right? The data cannot say. The line adds the same mass each week and the exponential the same fraction, and over four weeks those two growth patterns look alike. Choosing between them needs knowledge of how pumpkins grow (Choosing a function family for data), or later weighings; Exercise 14 shows that neither model holds.
A winning time that must stop falling
A made-up annual 1500 m lake swim was first held in 1990. With x the years since 1990 and T the winning time in minutes, the results every fifth year were (0, 24.0), (5, 22.9), (10, 22.1), (15, 21.0), (20, 20.1), (25, 19.0) and (30, 18.1). The line T = 24 - 0.2x passes within 0.1 minutes of each. (a) What does it predict for 2040? (b) In what year does the line reach 0? (c) What domain should go with it?
(a) 2040 is x = 50, and T = 24 - 0.2 \times 50 = 14 minutes. This is extrapolation, 20 years past the last result.
(b) 24 - 0.2x = 0 gives 0.2x = 24, so x = 120, the year 2110. After that the line gives negative times.
(c) A winning time cannot be 0: somebody has to swim the 1500 m. So the line must fail before 2110, and the data cannot say when. On the other side, the race began in 1990: the line gives 24 - 0.2 \times (-10) = 26 minutes for 1980, the winner of a race that never took place. The trusted domain is the results’ stretch:
Working backwards from an output
Use the pumpkin models of Example 3. (a) When does Evander’s line give 13 kg? (b) When does it give 30 kg? (c) When does Jessamy’s exponential give 30 kg?
Solution(a) 9.9 + 1.15w = 13 gives 1.15w = 3.1, so \displaystyle w = \frac{3.1}{1.15} \approx 2.7 weeks. That input lies inside 0 \le w \le 4, so this is interpolation.
(b) 9.9 + 1.15w = 30 gives 1.15w = 20.1, so w \approx 17.5 weeks. That is 13.5 weeks past the last weighing, more than three times as long as the weighings lasted: extrapolation, far out.
(c) 10 \times 1.1^w = 30 means 1.1^w = 3. There is no step yet in this course that undoes a power, so try whole weeks: 1.1^{11} \approx 2.85 gives 28.5 kg, and 1.1^{12} \approx 3.14 gives 31.4 kg. The exponential reaches 30 kg between weeks 11 and 12.
Two models that agreed to 0.1 kg on every weighing put “30 kg” about six weeks apart. Stating each with its domain, 0 \le w \le 4, flags both answers before anyone acts on them.
Boundaries
Interpolation assumes nothing sudden happens between measurements. A frost, a sale, a rule change between two measured inputs breaks it as surely as extrapolation. It is safer, not safe, and the wider the gaps between measurements, the less it is worth.
Inside the stretch the model still has scatter. Interpolation predicts the trend. Single days at the kiosk sat up to 2 cans off the line, and a new day can too.
The other conditions must match as well. A 25 °C day at a different beach, or in winter, is inside the stretch of temperatures and outside the data all the same: the kiosk was only measured on these summer days.
The edge of the domain is a judgment. The data stop sharply; trust fades gradually. That is why the domain is stated, not left to the reader.
Where this goes next
Judging and comparing models compares a model’s outputs with fresh measurements to see where it starts to drift, which is how the edge of a domain is found in practice. The vertical gaps between data and model get a name in Residuals, and Predicting with a regression line, and extrapolation returns to interpolation and extrapolation with the line a calculator fits. Choosing a family from what is known about the situation, rather than from five dots alone, is Choosing a function family for data. Later, statistics attaches to a prediction a margin that says how far off it may be.
Exercises
Inside or outside the data
Odile runs a drinks kiosk on another beach. On days whose highest temperature ran from 16 °C to 28 °C she fitted the model D = 5t - 30, with t in °C and D the cans sold (made-up data).
- Would a prediction from Odile’s model be interpolation or extrapolation at 20 °C, 28 °C, 31 °C and 9 °C?
- Predict her sales at 20 °C and at 31 °C. Which prediction deserves more trust?
- At what temperature does her model predict 0 cans? What does it give at 2 °C?
- Her kiosk can stock at most 150 cans a day. Above what temperature does the model predict more than she can sell?
- State Odile’s model with its domain, and give the temperatures outside which it is certainly wrong.
- Ulrich uses Severin’s model from Examples 1 and 2 to predict sales at 33 °C and at 45 °C. Find both, and say what each is worth.
Two models, one set of data
Pascoe weighed a smaller pumpkin weekly (made-up data), with w the weeks since the first weighing: (0, 4.0), (1, 4.8), (2, 5.8), (3, 6.9) and (4, 8.3), in kilograms. One model is the line M = 3.8 + 1.1w; the other is the exponential M = 4 \times 1.2^w.
- Find each model’s value at weeks 0 to 4, and the largest miss of each from the data.
- Predict the mass at week 10 with both models. How far apart are they?
- Use each model to say when the pumpkin reaches 12 kg. Is each answer interpolation or extrapolation?
- Use Evander’s line M = 9.9 + 1.15w from Example 3 to find the mass at week -9, and the week at which the line gives 0 kg. What do these say?
Trends that must stop
A made-up annual 800 m river swim was first held in 2000. The results from 2000 to 2024 are fitted with the line T = 12.5 - 0.08x, with x the years since 2000 and T the winning time in minutes.
- Predict the winning time in 2012 and in 2030, and say which is interpolation.
- In what year does the line reach a winning time of 0? Must the line fail before then?
- What does the line give for 1990, and why does the question have no answer?
Reasoning
- Faustina kept weighing her pumpkin from Example 3. At week 8 it weighed 17.5 kg and at week 12 it weighed 19.0 kg. Compare these with both models’ predictions. What happened, and what does it say about choosing a model by its fit inside the data?
- In Exercise 7 the exponential misses Pascoe’s data by less than the line does. Does that make its week 10 prediction the right one?
- A prediction one degree past the data and one twenty degrees past are both extrapolation. Are they equally untrustworthy?
- Explain why any falling line, however slowly it falls, must give an impossible winning time if it is carried far enough.
Answers
- 20 °C and 28 °C are interpolation (28 is an end of the stretch, which counts as inside); 31 °C and 9 °C are extrapolation.
- 70 cans at 20 °C (5 \times 20 - 30) and 125 cans at 31 °C (5 \times 31 - 30). The 20 °C prediction has measured days on both sides; the 31 °C one is 3 degrees past them.
- t = 6 °C, from 5t - 30 = 0. At 2 °C it gives 5 \times 2 - 30 = -20 cans, which is impossible: extrapolation gone wrong.
- Above 36 °C: 5t - 30 = 150 gives 5t = 180, so t = 36.
- D = 5t - 30, 16 \le t \le 28. Below 6 °C it gives negative sales and above 36 °C more than 150 cans, so outside 6 \le t \le 36 it is certainly wrong.
- 153 cans at 33 °C (6 \times 33 - 45): extrapolation 3 degrees past the data, a rough estimate at best. 225 cans at 45 °C (6 \times 45 - 45): more than the 180 the kiosk can stock, so impossible; 45 °C is past even the outer limit of 37.5 °C.
- Line: 3.8, 4.9, 6, 7.1, 8.2, missing by 0.2, 0.1, 0.2, 0.2, 0.1, largest 0.2 kg. Exponential: 4, 4.8, 5.76, 6.912, about 8.29, missing by 0, 0, 0.04, 0.012, about 0.006, largest 0.04 kg.
- Line: 3.8 + 1.1 \times 10 = 14.8 kg. Exponential: 4 \times 1.2^{10} \approx 24.8 kg. About 10 kg apart, more than the pumpkin’s whole mass at week 4.
- Line: 3.8 + 1.1w = 12 gives \displaystyle w = \frac{8.2}{1.1} \approx 7.5 weeks. Exponential: 4 \times 1.2^6 \approx 11.94 and 4 \times 1.2^7 \approx 14.33, so just after week 6. Both are extrapolation: 12 kg is past the heaviest weighing, and both answers lie beyond week 4.
- At w = -9: 9.9 + 1.15 \times (-9) = -0.45 kg, an impossible mass. The line gives 0 kg at \displaystyle w = -\frac{9.9}{1.15} \approx -8.6, about 8.6 weeks before the first weighing. Both are extrapolation backwards; the line was built on weeks 0 to 4 and says nothing about when or how the pumpkin started.
- 2012 (x = 12): 12.5 - 0.08 \times 12 = 11.54 minutes, interpolation. 2030 (x = 30): 12.5 - 0.08 \times 30 = 10.1 minutes, extrapolation, 6 years past the last result.
- 12.5 - 0.08x = 0 gives x = 156.25: in 2156 the line gives 0.02 minutes (about one second) and from 2157 negative times. Yes, it must fail before then: an 800 m swim takes time, so the winning time must stay well above 0.
- x = -10 gives 12.5 + 0.8 = 13.3 minutes. The race was first held in 2000, so there was no 1990 winner; the input is outside the situation itself, not only outside the data.
- Week 8: line 19.1 kg, exponential about 21.4 kg, measured 17.5. Week 12: line 23.7 kg, exponential about 31.4 kg, measured 19.0. Both were too high and the exponential more so: the pumpkin’s growth slowed, a turn five early weighings could not show. Fitting the data closely says nothing about which pattern continues, and the truth may be neither model.
- No. The closer fit (largest miss 0.04 kg against 0.2 kg) is a fact about weeks 0 to 4. Week 10 is outside, and as Example 3 shows, two models that both fit those weeks can be far apart there. Only knowledge of how the pumpkin grows, or a later weighing, can decide.
- No. Both are extrapolation, but trust falls with distance. One degree past the data, the pattern would have to change abruptly to make the prediction badly wrong; twenty degrees past, there is room for any change at all.
- A falling line is T = a - bx with b > 0. It gives T = 0 at \displaystyle x = \frac{a}{b} and negative times after it; a slower fall (smaller b) only moves that point further out, never removes it. A winning time must stay above 0, so the line’s domain must end before \displaystyle x = \frac{a}{b}. A time that keeps improving but never reaches 0 needs a curve that levels off, not a line.