Relations and Functions · Lesson 4 of 5
Mixed Applications and Edge Cases
“Use linear rules, relation reasoning, rational restrictions and piecewise graphs to resolve less obvious cases.”
• Interpret and sketch a linear rule f(x) = mx + c. • Recover a stated linear rule from input–output information and check other pairs. • Justify the three given properties of the integer-difference relation on rational numbers. • Factor a rational denominator to determine excluded inputs. • Evaluate and graph a piecewise function, including its boundary values. • Check whether overlapping pieces assign the same output to a shared input.
Linear rules and a shifted graph
The identity function gives output x. Adding a fixed number shifts every output by that same amount, while multiplying x by a fixed number changes how rapidly the output varies with the input. These ideas combine in the linear rule f(x) = mx + c.
In this chapter, a rule f(x) = mx + c with fixed real numbers m and c is called a linear function. When m = 0, it is a constant function.
The constant c is the output at input zero. The coefficient m measures the change in output for a one-unit increase in input: m(x + 1) + c − (mx + c) = m. A positive m gives a rising line, a negative m a falling line and m = 0 a horizontal line. These conclusions come directly from comparing outputs.
Problem
Sketch f(x) = x + 10 on ℝ and identify its domain and range.
- 1.At inputs −10, −1, 0, 1 and 10, the outputs are 0, 9, 10, 11 and 20. Two convenient plotting points are (−10, 0) and (0, 10).
- 2.Every real input is allowed. Compared with y = x, the output at each input is ten larger, so the line is shifted upward by ten units.
- 3.Every real y occurs at x = y − 10. Thus domain and range are both ℝ. The graph continues in both directions; the chosen table does not bound the function.
Recovering a rule from pairs
A set of input–output pairs does not normally determine a formula at every other input. Here the additional information that the rule is linear makes the task possible. Use that stated form first, then use pairs to determine its two constants.
Problem
A linear function from ℤ to ℤ is known to include the pairs (1, 1), (2, 3), (0, −1) and (−1, −3). Find its rule.
- 1.Write f(x) = mx + c because linearity is given. The pair (0, −1) gives f(0) = c = −1.
- 2.The pair (1, 1) gives m + c = 1. Substituting c = −1 gives m − 1 = 1, so m = 2.
- 3.Therefore f(x) = 2x − 1. Check the remaining data: f(2) = 4 − 1 = 3 and f(−1) = −2 − 1 = −3.
- 4.For every integer x, 2x − 1 is an integer, so the rule is compatible with the declared codomain. Its range is all odd integers, not just the four listed outputs.
The four displayed pairs are information about this function, not its complete graph on ℤ. If the graph really contained only four pairs, its domain would contain only four inputs. Once the linear rule on all integers is specified, it supplies the remaining pairs as well.
Many different functions can agree at a few inputs. Two distinct inputs determine m and c only after you are told the function has the form mx + c. Always distinguish supplied data, a stated form and the conclusions that follow from them.
Problem
A linear rule satisfies f(2) = 7 and f(5) = 16. Find its rule and check both pairs.
- 1.Let f(x) = mx + c. The two conditions give 2m + c = 7 and 5m + c = 16.
- 2.Subtract the first equation from the second: 3m = 9, hence m = 3.
- 3.Substitute into 2m + c = 7: 6 + c = 7, giving c = 1. Thus f(x) = 3x + 1.
- 4.Check f(2) = 6 + 1 = 7 and f(5) = 15 + 1 = 16. Subtracting the equations removed the constant term because it was identical in both.
Reasoning about an integer-difference relation
A relation can express a condition on a difference rather than a direct output formula. To establish a property that must hold for every input, use an arbitrary pair or triple and the arithmetic implied by the definition. A few numerical examples can illustrate a statement, but cannot prove its universal form.
Let ℚ be the rational numbers, meaning numbers expressible as fractions of integers with nonzero denominators. On ℚ, declare (a, b) to belong to R exactly when a − b is an integer. Not every pair of rational numbers meets this condition: 3/2 − 1/2 = 1 does, while 3/2 − 1/3 = 7/6 does not.
Problem
For the relation a − b ∈ ℤ on ℚ, show that (a, a) always belongs; reversing a selected pair preserves membership; and two successive selected pairs give a third.
- 1.For any rational a, a − a = 0. Since 0 is an integer, (a, a) belongs to R. This works for every a, not just integer inputs.
- 2.If (a, b) belongs, then a − b is an integer. Its negative b − a is also an integer, so (b, a) belongs.
- 3.If (a, b) and (b, c) belong, both a − b and b − c are integers. Their sum is an integer: (a − b) + (b − c) = a − c. Hence (a, c) belongs.
- 4.For an illustration, a = 7/2, b = 3/2 and c = −1/2 give differences 2, 2 and 4. The proof, however, used arbitrary a, b and c.
This relation is not a function from ℚ to ℚ. For example, input a = 1/2 is related to both b = 1/2 and b = 3/2, because the differences are 0 and −1. Satisfying the three properties above does not establish the unique-output condition for a function. We need only the three stated properties here; no broader classification of relations is required.
Rational denominators: factor before excluding
A rational expression can be undefined at several inputs, even if its numerator has a value everywhere. Factor the denominator to identify every input that makes it zero. Keep the exclusion attached to the function, rather than treating it as an optional detail after the algebra.
Problem
Find the largest real domain of f(x) = (x² + 3x + 5)/(x² − 5x + 4).
- 1.The numerator is a polynomial, so it places no real-input restriction. Division requires x² − 5x + 4 ≠ 0.
- 2.To factor the denominator, find two numbers with product 4 and sum −5: −1 and −4. Thus x² − 5x + 4 = (x − 1)(x − 4).
- 3.A product is zero exactly when at least one factor is zero. The excluded inputs are x = 1 and x = 4.
- 4.The domain is ℝ ∖ {1, 4}. At other real inputs the denominator is nonzero, so the rule is defined. At 1 and 4, the numerators are 9 and 33, but neither can be divided by zero.
Problem
Compare r(x) = (x² − 9)/(x − 3) with s(x) = x + 3.
- 1.Factor x² − 9 = (x − 3)(x + 3). For x ≠ 3, cancel x − 3 to get r(x) = x + 3.
- 2.The original rule r has domain ℝ ∖ {3}; s on its largest real domain has domain ℝ.
- 3.At x = 3, r is undefined but s(3) = 6. The rules agree on the common domain, but their domains differ.
- 4.This explains why simplifying an expression is not enough to specify a function completely.
Piecewise rules and boundary inputs
A piecewise function uses different expressions for different parts of its domain. The inequalities tell you which expression applies. Boundary values deserve an explicit check because they may be handled by a separate clause or lie in more than one listed interval.
Problem
Evaluate the displayed three-part rule and draw its graph.
- 1.For x < 0, use 1 − x. Inputs −4, −3, −2 and −1 give 5, 4, 3 and 2 respectively. These points lie on the left-hand straight branch.
- 2.At x = 0, the middle clause explicitly gives f(0) = 1. Plot a filled point at (0, 1).
- 3.For x > 0, use x + 1. Inputs 1, 2, 3 and 4 give 2, 3, 4 and 5. These points lie on the right-hand branch.
- 4.The branches meet the height 1 at the boundary, and the middle clause includes that point. Every real input has one output. The rule is also |x| + 1, so its range is [1, ∞).
| Input | Clause used | Output |
|---|---|---|
| −4 | 1 − x because x < 0 | 5 |
| −1 | 1 − x because x < 0 | 2 |
| 0 | Middle clause | 1 |
| 1 | x + 1 because x > 0 | 2 |
| 4 | x + 1 because x > 0 | 5 |
A hollow circle excludes an endpoint from a branch; a filled circle includes the corresponding point in the graph. When a separate boundary clause fills exactly the endpoint approached by a branch, the final graph has a filled point there. The final graph records function values, rather than each clause separately.
Problem
A relation uses x² on 0 ≤ x ≤ 3 and 3x on 3 ≤ x ≤ 10. A second relation uses x² on 0 ≤ x ≤ 2 and 3x on 2 ≤ x ≤ 10. Which is a function on [0, 10]?
- 1.For the first relation, both clauses apply at x = 3. They give 3² = 9 and 3(3) = 9. These are the same output; the resulting pair (3, 9) is included only once. Every other input lies in one interval, so this is a function.
- 2.For the second relation, both clauses apply at x = 2. They give 2² = 4 and 3(2) = 6. Both (2, 4) and (2, 6) are selected.
- 3.The second relation is not a function because one input has two distinct outputs. The problem is not overlap alone; it is disagreement on the overlap.
Do not decide that every multi-part rule is automatically a function. Read ≤ and < carefully. A missing boundary can leave a declared input without an output; overlapping clauses can give conflicting outputs. Both problems violate the definition.
Quiz
For f(x) = −2x + 5 on ℝ, what does the constant 5 represent?
A linear function satisfies f(0) = −1 and f(1) = 1. What is its rule?
What is the domain of (x² + 3x + 5)/(x² − 5x + 4)?
For the displayed three-part rule 1 − x, 1, x + 1, what is f(−3)?
If (a, b) and (b, c) satisfy the integer-difference condition, why does (a, c) satisfy it?
Two clauses apply at x = 3 and both give 9. Does that overlap alone prevent a function?
Practice Problems
- Sketch f(x) = −2x + 4 on ℝ. Use two axis intersections and explain the roles of −2 and 4.
- A linear function satisfies f(−1) = 5 and f(2) = −1. Recover m and c and check both values.
- For the integer-difference relation on ℚ, decide whether (5/3, 2/3) and (5/3, 1/2) belong. Then write the general argument for reversing a selected pair.
- Find the largest real domain of (x² + 2x + 1)/(x² − 8x + 12), showing the factorisation.
- Graph the rule 2 − x for x < 0, 2 for x = 0 and x + 2 for x > 0. State its range and explain its vertex.
- A relation uses x² on [0, 4] and 2x on [4, 8]. Is it a function? Change one boundary so that it becomes a function without changing either expression.
- Explain why specifying f(1) = 2 and f(2) = 3 does not alone determine f(3), then explain what changes if f is stated to be linear.
Key Takeaways
• For f(x) = mx + c, c gives f(0) and m gives the output change per unit input change. • Recovering a linear rule uses both the stated form and the supplied pairs. • General relation properties require reasoning with arbitrary entries, not only numerical checks. • Factor a rational denominator and retain every excluded input. • For piecewise rules, check each interval and every boundary; overlap is harmless only when the outputs agree.