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Secondary Math I · Module 3 · Features of Functions

3.7 — To Function or Not to Function

Supplement for Mathematics Vision Project (MVP) curriculum · Secondary Math I, Module 3, Section 3.7
Independently created to support students using MVP Math. Not official publisher material and not affiliated with or endorsed by the Mathematics Vision Project.

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Decide whether a relationship is a function by identifying its input and checking whether that input determines exactly one output.

1. In class: A Practice Understanding Task

In your MVP packetSection 3.7 · Main task · pp. 34–35 · Problems 1–15

Identify the input and output, then decide whether the relationship is a function.

Start with the input

The independent variable is the input; the dependent variable is the output. A function assigns exactly one output to each input in its domain. Ask: Could the same input give two different outputs? If so, the relationship is not a function.

Main Problem 3 — Cost of gas versus amount pumped

In your MVP packetSection 3.7 · Main 3 · p. 34

In the MVP packet, you are asked about this relationship: “The cost of gas versus the amount of gas pumped.”

The amount pumped is the input; the cost is the output. Think of choosing an amount of gas and then finding what it costs. At a fixed price per gallon, each amount determines exactly one cost, so the relationship is a function.

For a refresher on checking graphs and reading their domain and range, see the function test and graph features in Section 3.3.

Repeated outputs are allowed

Different inputs can have the same output. In Main 8, the inputs 3 and 5 both lead to 5, while 7 leads to 9. Each input still has just one output, so this is a function. Likewise, several percentage scores can produce the same letter grade under a fixed grading scale.

Worked example — Check the table’s inputs

In your MVP packetSection 3.7 · Main 6 · p. 34

The left column gives the inputs (distance); the right column gives the outputs (days). Distance 6 appears twice, paired with both 2 days and 5 days. Because the same input has two different outputs, this relation is not a function.

Does the input give enough information?

For a circle, each radius determines one area, and each area determines one radius. For water in a fixed cylinder, each water height determines one volume. Each of these relationships is a function in the stated direction.

Worked example — The same area can need different amounts of fence

In your MVP packetSection 3.7 · Main 12 · p. 35

Original problem: “The length of fence needed with respect to the amount of rectangular area to be enclosed.”

Solution: Treat area as the input and fence length as the output. Both rectangles below enclose 12 square units:

Side lengthsAreaFence length (perimeter)
3 and 4\(3\cdot4=12\)\(2(3+4)=14\)
2 and 6\(2\cdot6=12\)\(2(2+6)=16\)

The same area gives two different fence lengths, so area alone does not determine the perimeter. This relationship is not a function of area unless another condition fixes the rectangle’s shape.

The function test also applies to rules

Main 13’s sequence gives one term for each term number: \(f(1)=3\) and \(f(n+1)=f(n)+4\). Its explicit rule is \(f(n)=3+4(n-1)\), for integers \(n\ge1\). The input is the term number and the output is the term’s value.

In Main 14, a rational number can be written as a ratio of two integers with a nonzero denominator; an irrational number cannot. The rule gives output 1 to rational inputs and 0 to irrational inputs. Every real number falls into exactly one category, so every input has exactly one output. It is a function even though its outputs do not form a continuous curve.

The same test works for the other situations in the packet: choose the input, identify the output, and justify whether that input determines one output.

2. Ready — Functions, domain, and range

In your MVP packetSection 3.7 · Ready · pp. 36–37 · Problems 1–8

For ordered pairs, the domain is the set of first coordinates and the range is the set of second coordinates. List repeated values only once. For a refresher on graphs and interval notation, see Section 3.2. Try the problems before checking the answers.

Check your work — Ready 1–8 answer key
  1. Function: yes

    Domain: \(\{-7,-6,-2,3,8\}\)

    Range: \(\{2,3,4,5\}\)

  2. Function: no; input 0 has outputs 4 and −3.

    Domain: \(\{0,2,3,4,5,9\}\)

    Range: \(\{-3,-1,0,2,3,4,7\}\)

  3. Function: yes

    Domain: \(\{1,2,3,4,5,6,7,8\}\)

    Range: \(\{2,3,4,5,6,7,8,9\}\)

  4. Domain: \([1,\infty)\)

    Range: \([0,\infty)\)

  5. Domain: \([-5,0]\cup[2,8]\)

    Range: \([-3,3]\)

    The graph skips inputs between 0 and 2; both displayed pieces are included.

  6. Domain and range: all real numbers, \((-\infty,\infty)\)

  7. Domain: \((-\infty,\infty)\)

    Range: \((0,\infty)\)

  8. Domain: \(\{1,2,3,4\}\)

    Range: \(\{9,98,987,9876\}\)

    The table defines the entire function, so do not add unlisted inputs or outputs.

3. Set — Functions in context

In your MVP packetSection 3.7 · Set · p. 37 · Problems 9–13

State the input and output before describing domain and range. These situations do not supply enough measurements to give exact numerical endpoints; describe the relevant times, heights, values, names, or outcomes instead. Try each problem yourself, then check your reasoning below.

Check your work — Set 9–13 answer key
  1. Input: time during one Ferris wheel ride. Output: the rider’s height above the ground.

    Domain: the times during the ride. Range: heights from the lowest to the highest position reached, including both.

    Function: yes. At each time, the rider has one height; returning to the same height at different times is allowed.

  2. Input: date during a specified calendar year at a fixed location. Output: that day’s number of daylight hours.

    Domain: the dates in that year. Range: the daylight durations for those dates; numerical extremes depend on the location.

    Function: yes. Each date has one daylight duration at that location.

  3. Input: time from the car’s purchase in 1978 to the chosen present date. Output: its value.

    Domain: that time interval. Range: the car values over that interval, which are not provided.

    Function: yes, when using one value for this car at each time. Its value need not always decrease; a collectible car can appreciate.

  4. Input: a name. Output: a corresponding phone number.

    Domain: the names considered. Range: the corresponding phone numbers.

    Function: not in general. The same name can correspond to several phone numbers.

  5. Taking stadium as the input and game outcome as the output:

    Domain: stadiums where the player plays. Range: the observed outcomes, such as wins, losses, and ties if permitted.

    Function: not in general. Games played at the same stadium can have different outcomes.

4. Go — Graph features and function values

In your MVP packetSection 3.7 · Go · pp. 37–38 · Problems 14–16

For a refresher, see graph features in Section 3.3, distance-function notation in Section 3.6, and comparing and adding function outputs in Section 3.5. Try the problems before checking the answers.

Check your work — Go 14–16 answer key

14

The graph’s lowest point is approximately \((4,-3)\).

Decreasing: approximately \((-5,4)\)

Increasing: approximately \((4,15)\)

Minimum: approximately −3 near \(x=4\)

Maximum: 8 at \(x=-5\)

Domain: \([-5,15]\)

Range: approximately \([-3,8]\)

15

  1. The distance traveled in 4 hours: \(d(4)=78(4)=312\) miles.
  2. \(d(t)=450\) means the distance traveled after \(t\) hours is 450 miles.
  3. The distance traveled in 3.5 hours: \(d(3.5)=78(3.5)=273\) miles.
  4. Solve \(78t=800\): \(t=\frac{800}{78}=\frac{400}{39}\approx10.26\) hours.

16

  1. \(f(x)=g(x)\) at \(x=-8\) and \(x=0\), at the points \((-8,5)\) and \((0,-4)\)
  2. \(g(0)+f(0)=-4+(-4)=-8\)
  3. \(g(x)>f(x)\) when \(x<-8\) or \(x>0\); in interval notation, \((-\infty,-8)\cup(0,\infty)\)
  4. \(g(-8)+f(-8)=5+5=10\)

Additional learning resources

ResourceWhat it supportsScope
Khan Academy — Recognizing functions from verbal description word problemApplying the function test to a situation described in words.Close match to the in-class task and Set.
Khan Academy — Recognize functions from verbal description word problemsAdditional practice deciding whether a situation defines a function.Close match; optional practice.
The Organic Chemistry Tutor — Relations and Functions | AlgebraDecide whether a relation is a function using ordered pairs, tables, mapping diagrams, and the vertical line test.Close match to Ready’s function tests; reinforces the one-output-per-input rule used in class and Set.

Teacher’s review of the source packet

No substantial mathematical corrections are needed for this section. The supplement explains the “versus” convention and describes contextual domains and ranges without inventing numerical bounds.

Coverage — Section 3.7

Packet locationSkills supportedWhere supported
Main 1–3, 5–7, 9–12, 15 · pp. 34–35Input/output choices and function tests in contextIn class: input, repeated-output, and context explanations
Main 4, 6, 8 · p. 34Pairs, tables, and mappings; repeated inputs/outputsFunction definition; table example; mapping discussion
Main 13–14 · p. 35Sequences and rules with different input categoriesIn class: rules explanation
Ready 1–8 · pp. 36–37Function tests, domain and range across representationsReminder, earlier-section links, and answer key
Set 9–13 · p. 37Contextual domains/ranges and function testsContext reminder and explained answer key
Go 14–16 · pp. 37–38Graph descriptions, distance notation, comparisons and sumsEarlier-section links and answer key

Source: Mathematics Vision Project, Secondary Mathematics One, Module 3: Features of Functions (May 2016), Section 3.7, printed pp. 34–38; teacher notes for 3.7. Packet excerpts are attributed under CC BY 4.0. Equations use MathJax and require an internet connection.

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