Equipment rental has a hypothetical setup charge of 40 currency units plus 30 per hour. Two hours cost 100. How can one statement answer not only this question but every permitted duration? A function assigns exactly one output to each allowed input.
You only need variables and expressions. This invented tariff allows any real duration from 0 to 4 hours inclusive, charges fractional hours proportionally and has no other fees.
A rule together with its domain
Let t be numerical duration in hours and C(t) numerical cost in currency units. Then
C(t) = 40 + 30t, 0 ≤ t ≤ 4.
t is the argument, or input. C names the function; C(t) is its value at t, not multiplication of C by t. For example, C(2) = 40 + 30 ⋅ 2 = 100. Zero duration here means a rental booked without use: the setup fee remains.
The domain contains all allowed inputs, [0,4]. The range contains all outputs actually attained on those inputs, [40,160]. It is not merely a list of a few calculated prices. Every price between 40 and 160 occurs at t = (C − 40)/30, which lies between 0 and 4. Function definition, OpenStax §3.1.
Three representations of the same relationship
The formula computes any allowed input. A table conveniently lists selected pairs. The graph places them in a plane: first coordinate horizontally, second vertically.
Table and graph of C(t)=40+30t
| t, h | C, c.u. |
|---|---|
| 0 | 40 |
| 1 | 70 |
| 2 | 100 |
| 4 | 160 |
The point (2,100) means “two hours, one hundred currency units,” not two currency units or one hundred hours. Horizontal ticks represent 1 hour, vertical ticks 40 currency units. Equal screen lengths do not make different units equal.
A solid segment is justified by the model’s allowance of any real duration and its given formula. A four-row table alone does not specify intermediate values. If the service charges only for whole started hours, the rule, domain or graph would differ; the graph might be stepped.
A unique output does not mean different outputs
For the same allowed t = 2, this tariff cannot simultaneously give 100 and 120. Such a relationship would not be a function of duration alone: another input, such as equipment category, is missing. Different inputs, however, may share one output. With a flat fee F(t) = 40, all permitted durations have the same cost, 40.
Geometrically this gives the vertical-line test: a vertical line at fixed t meets a function graph at most once. Outside the domain it does not meet it at all. A horizontal line may intersect many times without violating the definition.
A formula does not override the context
Substituting t = − 1 into 40 + 30t is arithmetically possible, but negative duration is outside our function. Likewise t = 6 exceeds the agreed 4 hours. The same formula on a wider domain would define a different function and would not guarantee a real price. Domain and range, OpenStax §3.2.
For an unfamiliar relationship, first find the input, its units and its allowed domain; then identify the rule and attained outputs. Formula, table and graph become three ways of reading one story.