Choosing a function for a relationship

A model includes its meaning and conditions

A function model assigns one predicted output to each permitted input. Start by naming both quantities, their units and the question the prediction should answer. Then state the domain and what remains fixed. In the measurement formulas below, variables mean numerical values in the named units, and coefficients carry the units needed by each equation. For a cost depending on item count, the input may be an integer count; for an area depending on a length, it may be a positive real measurement. A formula that can be evaluated at a negative input does not make that input meaningful in the problem.

There are two different reasons to propose a rule. A specified pricing scheme or a geometric identity can determine it from stated assumptions. A table of measurements can suggest a candidate that needs further checking. An exact deduction within an ideal model remains conditional on its assumptions; an approximate fit to records does not become an exact law merely because its graph looks convincing. State which kind of evidence you have.

Write a model statement such as: C(n) is the total price in euros for an integer order count n from 0 to 20, with a fixed setup charge and an unchanged price per item. The domain, units and conditions are part of the statement. If another factor changes the price at the same count, a rule using count alone may no longer describe the actual prices; it can still be an explicitly chosen prediction rule, rather than a claim that the raw records are a single exact function of count.

Constant additions suggest an affine rule; proportionality is more specific

An affine rule has the form f(x) = mx + b, with real constants m and b. It satisfies f(x + h) − f(x) = mh whenever both inputs are permitted. Thus equal input increments produce equal output increments, and every pair of distinct permitted inputs has the same average rate of change m. That rate has units of output per input. The constant b has the output's units; it is the actual output at zero only when zero is in the domain.

A proportional rule has the form f(x) = kx. Its ratio f(x)/x is the same k at every permitted nonzero input. It is the affine case with b = 0. Under an unchanged proportional relationship, doubling the input doubles the output. In a fixed fee plus a per-item charge, doubling the count doubles only the variable part: the fixed fee is paid once. A nonzero fixed fee therefore prevents the total from being proportional to count.

Some textbooks call both mx + b and kx linear functions. Here affine makes the possible offset explicit, while proportional means the offset is zero. A constant rule is also affine, with m = 0. These family names overlap. When the domain excludes zero, use the full relationship or a constant-ratio argument to establish proportionality; do not claim to have observed a zero-input output that the model never permits.

Scaling a length can produce a power rather than a constant addition

For a square of positive side length s, its area is A(s) = s². Doubling the side multiplies the area by four. For a cube, V(s) = s³, and doubling the side multiplies the volume by eight. Those conclusions come from multiplying the side lengths, with the shape unchanged. A model may explicitly add the degenerate zero-size case with output zero; the geometry of a nondegenerate square or cube uses s > 0.

More generally, a stated direct-power relationship y = kx^p, for p = 1, 2 or 3 here and nonzero k, has constant ratio y/x^p at permitted nonzero inputs. Scaling x by a factor r scales the output by r^p, provided both inputs are in the domain. Find k from a permitted nonzero input, then check the rule's dimensions. For area in square centimetres and length in centimetres, a coefficient of s² is dimensionless; a coefficient of a lone s in an area formula needs units of centimetres.

A general quadratic ax² + bx + c, with a ≠ 0, need not be the pure power kx². A rectangle with fixed perimeter 20 centimetres has sides s and 10 − s, so its area is s(10 − s) = −s² + 10s square centimetres for 0 < s < 10. The second side changes when the first changes; doubling s therefore need not quadruple this area. A pure square power is quadratic, polynomial and rational through denominator 1, while a quadratic with additional terms can have a different scaling pattern. Choose a relationship within a family rather than treating the family names as mutually exclusive boxes.

A fixed product suggests inverse variation

If two positive quantities satisfy a fixed product xy = k, then y = k/x on a positive domain. Doubling x halves y when both inputs are allowed. For example, an ideal job requiring 18 worker-hours would take T(n) = 18/n hours with n identical workers contributing independently at unchanged rates. If only integer worker counts 1 through 6 are allowed, retain that finite domain. Team coordination or a different rate would change the assumptions and could change the rule.

A decreasing table alone does not establish inverse variation. Check the fixed product xy, or derive it from the stated mechanism. An affine decreasing rule can also decrease, but it preserves a constant rate of change rather than a constant product. Likewise, an average cost b + k/x with a nonzero offset is not pure inverse variation: subtract b before looking for the fixed product.

Inverse variation is not the same as an inverse function. The rule k/x describes a relationship with a constant product; the notation f⁻¹ refers to reversing a one-to-one function on its actual range. Choose words and notation that keep those two ideas distinct. Zero is excluded from k/x when k ≠ 0, and negative inputs allowed by an algebraic extension need not be allowed by the physical model.

Differences and ratios test candidates at the inputs you have

For equally spaced inputs x, x + h, x + 2h with h ≠ 0, an affine rule has equal first differences mh. A quadratic ax² + bx + c has constant second difference 2ah²: subtract one successive output difference from the next. For unitless inputs 0, 1, 2, 3 and outputs 1, 4, 9, 16, the first differences are 3, 5, 7 and the second differences are 2, 2, consistent with the quadratic (x + 1)². They are not constant first differences, and the output at zero shows this is not the pure power kx².

Equal spacing matters. For x = 0, 1, 3 and y = 2, 5, 11, the output increments 3 and 6 differ because the input increments differ. The pairwise rates are 3/1 = 3 and 6/2 = 3, consistent with y = 3x + 2. Do not compare raw differences as if unequal input steps were equal. Similarly, ratios for proportional or power candidates require nonzero input, and a reciprocal candidate requires the original nonzero denominator.

These checks describe a finite table. Exact agreement can rule out a conflicting candidate, and, under a prior choice of family, can determine parameters. It does not establish that all future or intermediate outputs follow that family. Recorded measurements may also contain rounding or variation; approximate differences suggest a candidate rather than prove an exact identity. Keep every observation's meaning and precision visible, and use the structural conditions as well as the numerical pattern.

A changed rule needs a precise boundary

Suppose a stated service charges 4 euros per kilogram for the first 3 kilograms and 2 euros per kilogram for any additional mass, with allowed mass 0 ≤ w ≤ 10 kilograms. Its charge in euros is C(w) = 4w for 0 ≤ w ≤ 3 and C(w) = 12 + 2(w − 3) for 3 < w ≤ 10. The second expression retains the 12-euro charge for the first three kilograms; applying 2w to the whole mass would describe a different pricing scheme.

The boundary w = 3 belongs to the first piece and gives C(3) = 12. The second formula approaches the same charge there, so this example has no jump. At w = 4 the charge is 14 euros. Specify which inputs belong to each piece, including zero, the threshold and any upper bound. On the full stated domain each input must receive one output. Overlapping pieces are consistent only if their formulas agree at every shared input; otherwise they do not define a function there.

A different stated scheme could discount the whole order after a threshold, producing a jump, or use integer counts rather than continuously measured mass. Do not assume every piecewise model is continuous or replace a discrete count by all real inputs. Draw any open or closed endpoint according to the actual inequalities, and let the words of the scheme determine whether the cheaper rate applies only to an extra portion or to the entire purchase.

Parameters can be unique within a family without the family being established

Once an affine family is specified, two observations at distinct inputs determine its parameters: m = (y₂ − y₁)/(x₂ − x₁) and b = y₁ − mx₁. The nonzero input difference is essential. For a specified pure power y = kx^p, one observation at a nonzero input determines k = y/x^p. For specified inverse variation y = k/x, a permitted observation gives k = xy. A zero-input zero-output observation alone cannot determine the coefficient of kx or kx².

After calculating parameters, check other observations and the entire context. A candidate giving negative area or a fractional number of people can contradict the output meaning even when its algebra is valid. Different output records at the same input cannot all be exact values of a single function of that input alone. They may indicate changing conditions or measurement variation; a chosen prediction rule is separate from that raw collection of records.

The phrase two points determine a line means a unique affine rule through those points, not a unique function of every kind. Three distinct inputs similarly determine at most one polynomial of degree at most two: the difference between two such polynomials would have three distinct zeros, so it would have to be the zero polynomial. This is conditional uniqueness in a stated family. It does not eliminate cubic, piecewise or other rules unless the model's assumptions do so.

A finite fit can leave infinitely many possible continuations

For unitless exact records (0, 1), (1, 3), (2, 5), the rule f(x) = 2x + 1 fits every record. So does g(x) = 2x + 1 + x(x − 1)(x − 2), because its added product is zero at each recorded input. At the unrecorded input 1/2, f gives 2 while g gives 19/8. At input 3 they give 7 and 13. Agreement at all three samples therefore leaves both an intermediate and a later value undetermined if those inputs are in the intended domain.

More generally, suppose f fits records at finitely many distinct inputs a₁ through aₙ. The rules gλ(x) = f(x) + λ(x − a₁)⋯(x − aₙ) preserve every recorded value. On a domain containing at least one allowed unrecorded input, the product is nonzero at that input, so different real λ produce different functions there. For dimensional quantities the coefficient λ must have the units needed to make the added term an output; the unitless construction already proves the mathematical point. Additional physical or output restrictions may rule out some of these alternatives, so the construction by itself does not validate them as real-world models.

There is an important finite-domain exception. If the entire domain is {0, 1, 2} and all three outputs have been recorded exactly, those records determine the complete function on that domain. The formulas above then represent the same function there, despite differing on extensions. To choose a model beyond a finite set of records, use stated mechanisms, further permitted observations and explicit assumptions. A simplest-looking formula is a choice that needs an explanation, not a proof supplied by the finite table.

A residual is observed minus predicted

At an input with an observed output, define the residual as r = observed output − predicted output. It has the output's units. A positive residual means the observation is above the prediction, so the model underpredicts; a negative residual means it overpredicts. A zero residual means agreement at that input. At an unobserved input there is a prediction, but no observed residual to calculate.

For a candidate water-level prediction M(t) = 10 + 2t centimetres, with numerical t in minutes, a record of 15.5 centimetres at t = 3 gives prediction 16 centimetres and residual −0.5 centimetres. The model overpredicts by 0.5 centimetres. The same numerical difference should be expressed as −5 millimetres if the outputs are converted to millimetres. Compare models on the same observations and in the same units rather than comparing unconverted numbers.

A smaller absolute residual at one input is useful local evidence, not proof that a model is correct everywhere. Several positive or negative residuals can signal a systematic mismatch worth investigating; that pattern still does not identify its cause on its own. Check whether assumptions, units, omitted factors or recorded precision need revision. We can compare explicit candidates here without introducing an automatic regression procedure or declaring an unspecified rule best.

Interpolation and extrapolation concern the observed inputs

For forward prediction from input to output, compare the selected input with the smallest and largest observed inputs. A prediction at an unobserved permitted input between them is interpolation. A prediction at a permitted input outside their span is extrapolation. Evaluating a model at a recorded input is a comparison with that record. This positional classification uses the input span; it does not require the predicted output to lie between the recorded output heights.

Check admissibility independently. If the input is an integer worker count, 2.5 workers may lie between observed counts 1 and 4 but remains outside the model's domain. A formula value there is a mathematical extension, not an allowed worker prediction. Likewise, a count beyond a stated maximum can lie beyond the data but is not authorized by that finite model. Even at allowed inputs, interpolation is not guaranteed correct: two rules fitting the same records can disagree between them, as the product construction shows.

Extrapolation needs particular attention to whether the same conditions continue beyond the observed span. A capacity limit, changed price or altered process can make a formerly useful rule fail. Exact structural models can justify predictions throughout their stated domain by their assumptions, while empirical predictions need supporting evidence. In either case, explain the output and units, retain the domain and identify what would make the rule need revision.

Compare candidates, evidence and permitted predictions separately

The explorer has seven contexts: distance at a stated constant speed, a fixed fee plus distance charge, square area, a fixed job shared by integer worker counts, a delivery tariff with a mass threshold, unknown temperature records and cube volume. Each offers three candidate rules. Read the stated assumptions and units before comparing the formulas. Proportional rules also belong to the affine family, square powers are quadratic, and a constant rule is affine; these labels can overlap without making every candidate suitable for the context.

Compare every candidate with every supplied record using exact observed-minus-predicted residuals. Then choose a test input independently. Check whether that input is permitted, whether its position is within or outside the observed span, and what each candidate predicts. The extra polynomial term in the distance case can fit all supplied records while conflicting with the unchanged-speed assumption. Two temperature candidates can both fit all supplied records while disagreeing at unrecorded inputs; the records alone do not establish a temperature mechanism. Celsius temperatures may be negative, unlike a physical square area.

Inspect the delivery threshold using the stated piecewise conditions and the worker case using integer counts 1 through 6. The square and cube cases explicitly allow the ideal zero-size value as well as positive sizes. A table of selected controls does not prove a full relationship, and a matching record does not prove causation. The next chapter develops exponential and logarithmic relationships; a repeated multiplicative change over equal input steps will need a different candidate from an affine rule's repeated additive change.

Worked examples

Try it yourself

1. A stated service has an unchanged per-item price and a fixed charge paid even for zero items. Its total is 13 euros for two items and 25 euros for six items. It permits integer counts 0 through 10. Find the cost rule and the cost of five items; state the coefficient units and decide whether the total is proportional to count.

   Show the answer and reasoning

   C(n) = 3n + 7 euros on integer counts 0 through 10; C(5) = 22 euros. The rate is 3 euros per item and the fixed charge is 7 euros. The total is affine and is not proportional to count.

   The distinct counts give rate (25 − 13)/(6 − 2) = 3 euros per item. Substituting the two-item total gives fixed charge 13 − 3·2 = 7 euros. The stated constant pricing structure justifies the affine family; a finite pair alone would not establish that structure. The nonzero offset prevents a constant total-cost-to-count ratio.

2. An ideal family of similarly shaped containers has capacity V(s) = k s³, where s > 0 is a numerical length in centimetres and V is in cubic centimetres. At s = 2 the capacity is 12 cubic centimetres. Find k, predict capacity at s = 6 and give the factor by which tripling s changes capacity. Does a constant addition per centimetre explain this scaling?

   Show the answer and reasoning

   k = 3/2, dimensionless under the stated units; V(s) = (3/2)s³ cubic centimetres. V(6) = 324 cubic centimetres. Tripling s multiplies capacity by 27, rather than adding a fixed amount per centimetre.

   The stated power relationship gives k = 12/2³ = 3/2. Then V(6) = (3/2)·216 = 324. In general V(3s)/V(s) = 3³ = 27 for permitted positive s. Because capacity and s³ both have cubic-length units, k is dimensionless; the derivation relies on the supplied shape-scaling assumption.

3. A stipulated ideal job requires 18 worker-hours. Each identical worker contributes independently at the same rate, and allowed team sizes are integer counts n from 1 to 6. Choose a time rule, find times for three and six workers, and explain why the fact that a candidate decreases would not by itself establish inverse variation. Is a prediction for 2.5 workers permitted?

   Show the answer and reasoning

   T(n) = 18/n hours for integer n = 1, …, 6. T(3) = 6 hours and T(6) = 3 hours. The constant product nT(n) = 18 worker-hours justifies inverse variation; a decreasing pattern alone does not. A count of 2.5 workers is outside the domain.

   The unchanged independent contributions give the fixed work product nT = 18. Divide by the permitted positive nonzero n to obtain the rule. Doubling the team from 3 to 6 halves the predicted time. An affine rule could also decrease without preserving that product, and a fractional numerical input does not become an allowed team size merely because 18/2.5 can be calculated.

4. Unitless inputs 0, 2, 4, 6 have outputs 1, 9, 33, 73. Assuming a polynomial of degree at most two, find its formula using the equal-step differences. Does this table alone prove that the same quadratic applies at unrecorded inputs?

   Show the answer and reasoning

   The conditional quadratic is f(x) = 2x² + 1. Its first differences are 8, 24, 40 and second differences are 16, 16 at input step h = 2. The finite table alone does not prove that rule at unrecorded inputs.

   For ax² + bx + c, the second difference is 2ah² = 8a, so a = 2. At x = 0, c = 1; at x = 2, 9 = 8 + 2b + 1 gives b = 0. The value at 6 checks as 73. Adding λx(x − 2)(x − 4)(x − 6) leaves all four records unchanged and can change an allowed unsampled value, showing that the chosen family assumption remains necessary.

5. A service charges 3 euros per kilogram for the first 2 kilograms and 1 euro per kilogram only for the extra mass. Its measured mass domain is 0 ≤ w ≤ 5 kilograms. Write a piecewise total charge, find the charges at 2 and 2.5 kilograms, and prove its complete range. Explain why w euros for all w > 2 would describe a different scheme.

   Show the answer and reasoning

   C(w) = 3w euros for 0 ≤ w ≤ 2, and C(w) = 6 + (w − 2) = w + 4 euros for 2 < w ≤ 5. C(2) = 6 euros and C(2.5) = 6.5 euros. The complete range is 0 ≤ C ≤ 9 euros.

   The first portion costs 6 euros at the threshold and remains included above it. The first piece attains every charge from 0 to 6 via w = C/3; the second attains every charge above 6 through 9 via w = C − 4, so their union proves the whole range. Both formulas agree at 2, which the stated definition assigns to the first piece. Using w above 2 would apply the cheaper rate to the whole mass and discard the first portion's charge.

6. Exact synthetic cost records at integer counts 1, 3, 5 are 4, 8, 12 euros. A candidate M(n) = 2n + 2 euros permits integer counts 1 through 8. Before a new record, classify a prediction at n = 4 and at n = 7, and find both values. A new total at n = 4 is 11 euros: calculate its residual and direction. Is n = 2.5 an allowed prediction merely because it lies inside the observed input span?

   Show the answer and reasoning

   M(4) = 10 euros is an interpolated prediction; M(7) = 16 euros is an extrapolated prediction, both at permitted counts. The new residual is 11 − 10 = +1 euro, so M underpredicts. Input 2.5 is not permitted.

   The original observed span is 1 through 5: 4 is inside it and 7 outside it, while both belong to the stated integer domain. The residual uses observed minus predicted, not the reverse. Adding the record at 4 does not change the span endpoints. A fractional count remains inadmissible even inside that span; its formula value would concern an extension rather than this count model.

7. Unitless records at x = 0, 1, 2 have outputs 2, 5, 8. Compare f(x) = 3x + 2 with g(x) = 3x + 2 + 2x(x − 1)(x − 2). Check every record and find both values at x = 1/2 and x = 3. How does the conclusion differ when the domain is all real numbers versus exactly {0, 1, 2}?

   Show the answer and reasoning

   Both rules give 2, 5, 8 at the records. At 1/2, f = 7/2 and g = 17/4; at 3, f = 11 and g = 23. They are different functions on all real inputs, but the same complete function on domain {0, 1, 2}.

   The added product vanishes at each recorded input. At 1/2 it equals 2·(1/2)·(−1/2)·(−3/2) = 3/4, and at 3 it equals 2·3·2·1 = 12. Thus a finite fit does not choose a unique continuation on a domain with those unsampled inputs. If the domain contains only the three sampled inputs, no additional allowed values remain to distinguish the rules.

What to carry forward

• A useful model names its quantities, units, permitted inputs and fixed conditions. Distinguish an exact deduction under stated assumptions from a candidate fitted to observations.
• Proportional means f(x) = kx; affine means f(x) = mx + b and permits an offset. Equal changes and constant ratios answer different questions, and function-family names can overlap.
• Derive a power relationship from a specified scaling rule and inverse variation from a fixed product. Decreasing outputs alone do not establish k/x, and inverse variation is different from an inverse function.
• First and second differences need equally spaced inputs. Parameter uniqueness within a chosen family and agreement at a finite table do not prove that family describes every allowed unrecorded input.
• A piecewise model needs complete, consistent boundary conditions. Preserve the charge or relationship for an earlier portion when a new rate applies only to an extra portion.
• Residual means observed minus predicted, in output units: positive indicates underprediction and negative indicates overprediction. Local agreement or smaller errors alone do not establish a complete law or its cause.
• Forward interpolation and extrapolation classify a prediction by its input's position relative to the observed span. Check domain admissibility separately, and retain the model's assumptions before extending any prediction.