Relations and Functions · Lesson 5 of 5
Chapter Summary and Practice
“Connect the complete chapter through a shared language of pairs, rules, graphs and carefully justified restrictions.”
• Connect Cartesian products, relations and functions without confusing their defining conditions. • Compare domain, range and codomain in finite and infinite examples. • Recall the seven graph families and explain their domains and ranges. • Apply function operations with complete domain checks. • Solve mixed problems involving boundaries, output ranges and relation properties. • Use examples and counterexamples to justify conclusions about functions.
The chapter as one connected idea
This chapter begins with all possible combinations and gradually adds more structure. A Cartesian product supplies possible pairs; a relation selects pairs; a function makes that selection reliable for every declared input. Graphs and algebra then offer ways to see and use the same input–output information.
The language stays consistent throughout. The first entry is an input candidate and the second is an output candidate. A statement about membership checks whether a pair is permitted, while a statement about a function checks whether every input has a unique output. The formulas and diagrams are different representations of those underlying pairs.
| Chapter idea | What it records or requires | Useful check |
|---|---|---|
| Ordered pair | Two entries with specified positions | Equal pairs agree in both positions |
| Cartesian product | Every allowed first–second combination | Check membership in both sets |
| Empty and infinite products | Whether pairs can be formed at all | An empty factor wins; nonempty infinite factors give infinitely many pairs |
| Union, intersection and subsets | How restrictions on entries combine | Use “either”, “both” and inclusion conditions |
| Ordered triplets and coordinates | Three positions; locations in plane or space | ℝ² is the plane and ℝ³ is three-dimensional space |
| Relation | Any subset of a product | Collect actual first and second entries |
| Counting relations | Independent inclusion decisions for each pair | pq pairs give 2^(pq) relations |
| Function | Exactly one image for every declared input | No missing inputs and no conflicting outputs |
| Function notation | Evaluation, image and preimage | f(a) = b means the pair (a, b) is selected |
| Seven graph families | The shape and output behaviour of a rule | Distinguish a finite window from the full graph |
| Algebra of functions | Arithmetic on outputs at the same input | Use common domain; quotient excludes denominator zeros |
| Linear and piecewise rules | Fixed-rate changes or interval-specific expressions | Use stated form and inspect boundary values |
| Relation reasoning | Consequences of a stated membership condition | Prove universal claims; disprove with a counterexample |
Rebuilding the set foundations
The order in a pair is essential because each position has a different role. Counting a Cartesian product and counting its subsets also answer different questions. Revisit these foundations before interpreting more complicated numerical rules.
In general A × B and B × A are different sets despite equal size when finite. Products distribute over unions and intersections because membership in their second entries follows the same “either” or “both” condition. If A ⊆ B and C ⊆ D, every pair from A × C also belongs to B × D. These facts are all consequences of the definition, rather than separate tricks to memorise.
Problem
Let A = {1, 2} and B = {3, 4}. Form the product, count all relations, and decide whether R = {(1, 3), (2, 3)} is a function from A to B.
- 1.The product is {(1, 3), (1, 4), (2, 3), (2, 4)}. There are 2 × 2 = 4 available pairs.
- 2.Each pair may be included or excluded independently, so there are 2⁴ = 16 possible relations. R is one of them.
- 3.R uses each input in A once, so it is a function from A to B. Its domain is A, its range is {3}, and its codomain is B.
- 4.The shared output 3 is permitted. The unused codomain entry 4 is also permitted. If (1, 4) were added, input 1 would have two outputs and the relation would cease to be a function.
Domain, range and codomain answer different questions
Domain concerns actual inputs, range concerns actual outputs, and codomain records the declared destination. For a function from A to B, its domain must be all of A. For a general relation from A to B, the domain can be a smaller subset of A.
| Question | Set to inspect | Example: x² declared from ℝ to ℝ |
|---|---|---|
| Which inputs are used? | Domain | ℝ |
| Which outputs occur? | Range | [0, ∞) |
| Which outputs were allowed by the declaration? | Codomain | ℝ |
To establish a range, show two things: every output produced lies in your proposed set, and every element of that set is actually attained. Merely finding a lower bound is not enough. For example, x² is nonnegative, and for any y ≥ 0 the input √y produces y; together these facts prove the range [0, ∞).
Problem
Find domain and range of f(x) = √(x − 1) and g(x) = |x − 1| on their largest real domains.
- 1.For f, the square-root input must satisfy x − 1 ≥ 0, hence x ≥ 1. Its domain is [1, ∞). Outputs are nonnegative, and every y ≥ 0 occurs at x = y² + 1. Its range is [0, ∞).
- 2.For g, every real x is allowed. Modulus is nonnegative and reaches zero at x = 1. Every y ≥ 0 occurs at x = y + 1, so the range is [0, ∞).
- 3.The ranges match but the domains differ. A graph’s shape and a rule’s restrictions must both be checked; a nonnegative output does not imply a nonnegative input.
Recognising the seven graph families
The seven families give a small set of useful visual patterns. Recalling the reason behind each shape is more reliable than memorising a picture. Domain and range must refer to the whole specified function, including endpoints and restrictions.
| Family and representative rule | Domain | Range | Visual or conceptual feature |
|---|---|---|---|
| Identity: x | ℝ | ℝ | Straight line through origin; output equals input |
| Constant: c | ℝ | {c} | Horizontal line; one fixed height |
| Polynomial: x² | ℝ | [0, ∞) | U-shaped square graph; opposite inputs share outputs |
| Polynomial: x³ | ℝ | ℝ | Cube graph includes positive and negative heights |
| Rational: 1/x | ℝ ∖ {0} | ℝ ∖ {0} | Two separate branches; zero excluded twice |
| Modulus: |x| | ℝ | [0, ∞) | V-shape; distance from zero |
| Signum: sgn(x) | ℝ | {−1, 0, 1} | Only the sign is retained; special origin value |
| Greatest integer: [x] | ℝ | ℤ | Steps include left endpoint and exclude right endpoint |
The two polynomial rows are representatives of one family, not two extra families. Other polynomials can have different ranges. Likewise, a rational function does not always have the reciprocal’s range; its particular rule determines it. Use the family to understand the form, then reason about the actual expression and domain.
x² and |x| have the same domain and range on ℝ, but they give different outputs at many inputs: at 2 they give 4 and 2. A function includes its actual assignment of outputs, not just the sets it uses.
Problem
Find the range of r(x) = x²/(1 + x²) on ℝ.
- 1.The denominator 1 + x² is always positive, so the domain is ℝ. Write r(x) = 1 − 1/(1 + x²).
- 2.Since x² ≥ 0, the original quotient is nonnegative. Since 1/(1 + x²) is strictly positive, the rewritten rule is strictly less than 1. The value 0 occurs at x = 0.
- 3.For any proposed output y with 0 ≤ y < 1, solve y = x²/(1 + x²): y + yx² = x², so y = (1 − y)x² and x² = y/(1 − y).
- 4.This right-hand side is nonnegative and defined. Choosing x = √(y/(1 − y)) produces the proposed y. Thus the range is exactly [0, 1), with 1 excluded.
Operating on functions while retaining restrictions
Function operations use ordinary arithmetic, but their domains are part of the answer. A legal input must be accepted by both original functions. For division, the denominator’s output must also be nonzero.
Problem
For f(x) = x + 1 and g(x) = 2x − 3 on ℝ, find f + g, f − g and f/g.
- 1.The sum is (x + 1) + (2x − 3) = 3x − 2, defined on ℝ.
- 2.The difference is (x + 1) − (2x − 3) = x + 1 − 2x + 3 = 4 − x, also defined on ℝ.
- 3.The quotient is (x + 1)/(2x − 3). Solving 2x − 3 = 0 gives x = 3/2, so its domain is ℝ ∖ {3/2}.
- 4.The zero of the numerator at x = −1 is allowed: the quotient equals 0/(−5) = 0 there. A numerator zero is not the same problem as a denominator zero.
Problem
For f(x) = x², evaluate (f(1.1) − f(1))/(1.1 − 1).
- 1.First evaluate the two outputs: f(1.1) = (1.1)² = 1.21 and f(1) = 1.
- 2.Their difference is 0.21. The input difference is 0.1, which is nonzero.
- 3.Dividing gives 0.21/0.1 = 2.1. The quotient compares the output change over these two specified inputs with their input change; no new calculus concept is needed.
Boundary checks and counterexamples
Some claims ask whether a relation is a function or whether a stated property always holds. A universal claim requires a general argument. A single carefully chosen failure, however, is enough to show that a universal claim is false.
Problem
Let R = {(ab, a + b) : a, b ∈ ℤ}. Is this a function from ℤ to ℤ?
- 1.Every integer input n can occur as a product, using a = n and b = 1. Thus missing integer inputs are not the obstruction.
- 2.Choose a = 1, b = 6. This gives the pair (6, 7). Choose a = 2, b = 3. This gives (6, 5).
- 3.The same first entry 6 has two distinct second entries, so R is not a function. A formula involving a and b does not guarantee a unique output for their product.
Problem
On positive natural numbers, let R contain (a, b) when a = b². Decide whether each of the three properties from the previous lesson holds.
- 1.The statement that (a, a) belongs for every a is false. At a = 2 it would require 2 = 2², which is false.
- 2.The statement that reversing any selected pair preserves membership is false. (4, 2) belongs because 4 = 2², but (2, 4) does not because 2 ≠ 4².
- 3.The statement that (a, b) and (b, c) imply (a, c) is also false. Take a = 16, b = 4, c = 2: 16 = 4² and 4 = 2², but 16 ≠ 2².
- 4.Each counterexample addresses one universal claim. We do not need to list the whole infinite relation to disprove it.
Piecewise definitions need the same unique-output check. When two intervals overlap, calculate both outputs there. Equal outputs describe one pair and cause no conflict; unequal outputs create two pairs with the same first entry. Also check for a gap: an input in a declared domain cannot be left without an output.
Problem
Let A = {9, 10, 11, 12, 13}, and let f(n) be the highest prime factor of n. Find the range.
- 1.Factor each input: 9 = 3², 10 = 2 × 5, 11 is prime, 12 = 2² × 3, and 13 is prime.
- 2.Their highest prime factors are 3, 5, 11, 3 and 13 respectively. Each input has a unique largest prime factor, so this is a function.
- 3.The range is {3, 5, 11, 13}; repeated output 3 is written only once. If the codomain is ℕ, it contains many unused values, which is allowed.
How the language of functions developed
The modern input–output definition is the result of a long development in mathematical language. Early uses of the word function were linked to curves and variable expressions. This history helps explain why the same concept can be represented by a graph, a formula or a collection of pairs.
Leibniz used the word in a Latin manuscript in 1673 in connection with curves. A 1698 exchange with Johann Bernoulli helped establish a more specialised analytical use. The chapter records an English use in Chambers’ Cyclopaedia in 1779. The present definition focuses on unique correspondence, which also handles functions without a single algebraic formula.
Quiz
Which description gives the correct progression?
For f(x) = √(x − 1), what is the largest real domain?
What is the range of x²/(1 + x²) on ℝ?
For f(x) = x + 1 and g(x) = 2x − 3, which input is excluded from f/g?
Which observation disproves that {(ab, a + b) : a, b ∈ ℤ} is a function?
A relation from {1, 2, 3, 4} contains (1, 5), (2, 9), (3, 1), (4, 5), (2, 11). Why is it not a function?
For f(x) = x², what is (f(1.1) − f(1))/(1.1 − 1)?
Which pair of sets represents the domain and range of greatest integer on ℝ?
Practice Problems
- Construct {a, b} × {1, 2, 3}, then draw a relation using every first entry but only two second entries. Explain whether your chosen relation is a function.
- For (2x + 1, y − 4) = (7, −2), find x and y. Then explain how ordered-pair equality differs from equality of unordered sets.
- For sets with 3 and 4 elements, compare the number of pairs with the number of relations. Explain the two counting arguments.
- On {1, 2, ..., 12}, select (x, y) satisfying y = 2x. Find the domain, range and codomain, and decide whether this is a function from the whole set to itself.
- Sketch representative graphs of all seven named families. Label domains, ranges and important endpoints; give both x² and x³ as polynomial examples.
- Find the largest real domains of √(x + 2), 1/(x² − 9) and (x² − 4)/(x − 2). Explain why simplifying the third expression does not restore input 2.
- Find the ranges of |x − 1|, √(x − 1), x² + 3 on ℝ where defined, and 4 − 2x on x > 0. Justify which endpoints occur.
- Prove that the range of x²/(1 + x²) is [0, 1) by showing both containment and attainability.
- Let f(x) = √x and g(x) = x − 4 on [0, ∞). Find f + g, f − g, 3f, fg and f/g, retaining all restrictions.
- A linear function has f(−2) = −5 and f(1) = 4. Recover its rule and check the two given outputs.
- A relation uses x² for 0 ≤ x ≤ 3 and 3x for 3 ≤ x ≤ 10. Compare its boundary behaviour with the version whose dividing point is 2. Explain when it is a function.
- For f(x) = x², compute (f(2.2) − f(2))/(2.2 − 2), showing the separate output and input differences.
- For the integer-difference relation on ℚ, prove the three stated properties using arbitrary a, b and c. For a = b² on ℕ, give a counterexample to each property.
- For A = {14, 15, 16, 17, 18}, define f(n) as its highest prime factor. Find its range and explain why two inputs may share an output.
- Decide whether {(ab, a + b) : a, b ∈ ℤ} is a function. Give a second conflicting pair of outputs for one input, different from the example in this lesson.
Key Takeaways
• Cartesian products supply all pairs, relations select pairs, and functions require one output for every declared input. • Domain, range and codomain record different information; range is always contained in codomain. • Use graph shapes together with exact domain and range reasoning, including endpoint checks. • Function operations act on outputs at the same input and must retain restrictions from the original functions. • Linear assumptions, piecewise boundaries and rational denominators all require explicit checks. • General proofs establish universal claims; one valid counterexample disproves a universal claim.
Previous · Lesson 4
Mixed Applications and Edge Cases
Next
End of chapter