Functions and their properties · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

05 Functions and their properties · g10-05-01 · 5–10 min · TEACHER · ANSWERS

Repair one mapping arrow

Can two inputs share an output? Which single input breaks the function rule?

Predict: Before changing the diagram, mark each input with 0, 1 or 2 distinct outputs.

Compare	Computed observation
Many-to-one	A = {1, 2, 3}; B = {5, 7}; attained range = {5, 7}.
Missing input 3	A = {1, 2, 3}; B = {5, 7}; attained range = {5}.
Input 2 has two outputs	A = {1, 2, 3}; B = {5, 7}; attained range = {5, 7}.
Constant output 8	A = {1, 2, 3}; B = {6, 8, 10}; attained range = {8}.

Conclusion: A function gives every allowed input exactly one output. Many-to-one is permitted; a missing arrow or two different outputs from one input violates the rule.

Conditions: The input set is {1,2,3}. Count distinct outputs from each input, not arrows arriving at an output.

Justification: Check the defining statement for each of the three inputs. This exhausts this finite mapping; it does not prove a claim about every possible mapping.

Ask: Ask a student to repair the missing-arrow case, then ask another to justify allowing shared output 5.

Watch for: Different inputs must have different outputs.

Extension: Make every input map to 8. Explain why unused elements of the output set are harmless.

Exit: All three inputs map to 8. Function or not, and why?

Yes: each input has exactly one output.

Before: Read function inputs, outputs and the stated domain; write your first prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

05 Functions and their properties · g10-05-02 · 5–10 min · TEACHER · ANSWERS

Same formula: the same function?

Does x = 0.5 distinguish a finite domain from the interval [−2,2]?

Predict: Predict whether 0.5 is allowed before dropping its chip onto each domain.

Compare	Computed observation
0.5 in the finite domain	Same fixed rule x² + 1. First range: {1, 2, 5}. Second range: [1, 5].
0.5 in the interval	Same fixed rule x² + 1. First range: {1, 2, 5}. Second range: [1, 5].
−1 in the first paper domain	Same fixed rule x². First range: {0, 1}. Second range: {0, 1}.
−1 in the second paper domain	Same fixed rule x². First range: {0, 1}. Second range: {0, 1}.

Conclusion: Formula arithmetic and defined function output are different questions. For x²+1, 0.5 gives arithmetic 1.25, but this is a function value only when 0.5 belongs to the domain.

Conditions: Compare both the rule and the complete input domain. The range contains attained outputs only.

Justification: Exhibit one input contained in exactly one domain to refute equality. For a finite domain, evaluate every input to obtain its range.

Ask: Require two sentences: “the arithmetic is …” and “the input is allowed/rejected because …”.

Watch for: The same expression defines the same function on any domain.

Extension: For the two finite x² domains, explain how equal ranges {0,1} can coexist with different functions.

Exit: Find the range of x²+1 on {−2,−1,0,1,2}.

{1,2,5}; repeated outputs are listed once.

05 Functions and their properties · g10-05-03 · 5–10 min · TEACHER · ANSWERS

Which endpoint belongs?

Why is 3 allowed in √(x−3), while 2 is forbidden in 1/(x−2)?

Predict: Draw the boundary point as open or closed before selecting each formula.

Compare	Computed observation
Included endpoint 0	Painted set: [0, 4). Infinity is a direction and always uses an open bracket.
Excluded endpoint 4	Painted set: [0, 4). Infinity is a direction and always uses an open bracket.
Reciprocal at 2	Painted set: (−∞, 2) ∪ (2, ∞). Infinity is a direction and always uses an open bracket.
Square root at 3	Painted set: [3, ∞). Infinity is a direction and always uses an open bracket.

Conclusion: A real square root permits zero radicand, giving x ≥ 3. A denominator cannot be zero, giving x ≠ 2. In [0,4), 0 is included and 4 excluded.

Conditions: Work over real inputs. Infinity is a direction, never an included endpoint; integers and real intervals are different domains.

Justification: Translate the defining restrictions: x−3 ≥ 0 and x−2 ≠ 0. These algebraic conditions justify all inputs; plotted examples only illustrate them.

Ask: Ask “what operation fails here?” before naming an interval bracket.

Watch for: Every domain boundary must be excluded.

Extension: Toggle integer inputs: identify a noninteger that leaves the domain even though it lies between the endpoints.

Exit: Write the domain of 1/√(x−3).

(3,∞): the radicand must be positive because the root is also a denominator.

05 Functions and their properties · g10-05-04 · 5–10 min · TEACHER · ANSWERS

Keep every restriction after simplification

Does cancelling x−1 restore the missing input x=1?

Predict: Predict whether the original and simplified formulas both have value 2 at x=1.

Compare	Computed observation
Intersect two restrictions	Overlaid masks: 2/2. Current intersection accepts x = 0.
Root in the denominator at 1	Overlaid masks: 2/2. Current intersection rejects x = 1.
Cancel at the hole x=1	Overlaid masks: 1/1. Current intersection rejects x = 1. Simplified formula: x + 1, retaining original exclusion x ≠ 1.
Finite range	Learner output set: {0, 1, 4}. x = 2; output = —.
Interval range	Learner output set: [0, 4]. x = 2; output = —.

Conclusion: For √(x+1)/(x−2), take x ≥ −1 and x ≠ 2 together. For (x²−1)/(x−1), cancellation gives x+1 only on x ≠ 1. The original graph retains its hole.

Conditions: The original formula fixes the domain. Finite output sets are not filled in between their elements.

Justification: Factor x²−1=(x−1)(x+1); division by x−1 is equivalent only if x≠1. Separately, x² on [−2,1] attains every y∈[0,4] via x=−√y.

Ask: Let students overlay both restriction masks, then compare arithmetic and domain membership at the hole.

Watch for: A simpler expression automatically enlarges the original function domain.

Extension: Compare the ranges of x² on {−2,−1,0,1} and on [−2,1]. Why is only one a full interval?

Exit: Is (1,2) on the original graph of (x²−1)/(x−1)?

No. At x=1 the original denominator is zero.

05 Functions and their properties · g10-05-05 · 5–10 min · TEACHER · ANSWERS

Should the points be joined?

The same cost formula y=2x+4 describes time or item count. Which graph permits x=2.5?

Predict: Draw five table points. Predict whether connecting them preserves the intended inputs.

Compare	Computed observation
Continuous hours, joined	Fixed domain: [0, 4]. Learner placements: 5/5. Joining: on. The straight-line model is given.
Whole items, joining is misleading	Fixed domain: {0, 1, 2, 3, 4}. Learner placements: 5/5. Joining: on. The straight-line model is given.
Whole items, separate points	Fixed domain: {0, 1, 2, 3, 4}. Learner placements: 5/5. Joining: off. The straight-line model is given.

Conclusion: On the real interval [0,4], the affine model gives a line segment. On {0,1,2,3,4}, it gives five separate points. The point (2.5,9) belongs only to the time model.

Conditions: Linearity is a stated model assumption. A finite table alone does not establish a unique formula.

Justification: For each permitted x the formula gives one point. The domain decides which x occur. The rule 2x+4+x(x−1) matches two given points but is not the same rule.

Ask: Ask what half an item means before toggling the joining segments.

Watch for: All plotted data points should be connected.

Extension: Give another formula agreeing with y=2x+4 at x=0 and 1 but differing elsewhere.

Exit: For 0–4 whole tickets costing 2x+4, is (2.5,9) on the graph?

No: 2.5 is outside the ticket-count domain.

05 Functions and their properties · g10-05-06 · 5–10 min · TEACHER · ANSWERS

A missing boundary or two outputs?

If both pieces include the boundary, is the relation always invalid?

Predict: At t=1, count distinct outputs for the open, single-closed and double-closed versions.

Compare	Computed observation
Source boundary	Fixed total domain: 0 < t ≤ 4. First branch 0 < t ≤ 1; second branch t > 1, t ≤ 4. At 1: 1 applicable branch(es), 1 distinct output(s).
Both ends open	Fixed total domain: 0 < t ≤ 4. First branch 0 < t < 1; second branch t > 1, t ≤ 4. At 1: 0 applicable branch(es), 0 distinct output(s).
Both ends closed, same output	Fixed total domain: 0 < t ≤ 4. First branch 0 < t ≤ 1; second branch t ≥ 1, t ≤ 4. At 1: 2 applicable branch(es), 1 distinct output(s).
Overlap with distinct outputs	Fixed total domain: 0 < t ≤ 4. First branch 0 < t ≤ 1.5; second branch t ≥ 1, t ≤ 4. At 1.25: 2 applicable branch(es), 2 distinct output(s).

Conclusion: Both open endpoints omit the boundary. Both closed endpoints are harmless if both formulas give the same value there. An overlap giving two distinct values violates uniqueness.

Conditions: For the rental model, the values 6 and 6+4(t−1) agree at t=1. Equality at one point does not justify an interval of overlap.

Justification: At a shared input, form the set of distinct outputs. A function requires cardinality exactly one. The four constructions exhibit zero, one and two outputs.

Ask: Have students state both formulas at the boundary before judging the brackets.

Watch for: A function must never have overlapping piece conditions, even with identical outputs.

Extension: Use the paper model x+2 and x² at x=0 to explain why closing both ends now fails.

Exit: Both formulas give 6 at t=1 and both include it. How many function values occur?

One distinct value, 6.

05 Functions and their properties · g10-05-07 · 5–10 min · TEACHER · ANSWERS

Swap two function machines

Does doubling then adding 3 give the same result as adding 3 then doubling?

Predict: For input 2, predict both intermediate values and final outputs before sending the token.

Compare	Computed observation
Double, then add	Input 2; order: f(u) = 2u → g(u) = u + 3.
Add, then double	Input 2; order: g(u) = u + 3 → f(u) = 2u.
Square-root pair at 0	Input 0; order: g(u) = u − 1 → f(u) = √u.
Square-root pair at 1	Input 1; order: g(u) = u − 1 → f(u) = √u.

Conclusion: The orders give 2x+3 and 2(x+3)=2x+6, differing by 3. With square roots, order can also change which inputs are allowed.

Conditions: Every intermediate result must belong to the next machine’s domain. Stop when a stage is undefined.

Justification: Algebra gives (2x+6)−(2x+3)=3 for every real x. For the root pair, the domains are x≥1 and x≥0, so they cannot be equal functions.

Ask: Read the pipeline left to right; require the intermediate value as well as the final answer.

Watch for: The same two rules always commute.

Extension: Compare √(x−1) and √x−1 at x=0 and 1; explain why one equal sample does not prove equality.

Exit: At x=0, which is defined: √(x−1) or √x−1?

Only √x−1, whose value is −1.

05 Functions and their properties · g10-05-08 · 5–10 min · TEACHER · ANSWERS

Can the largest value be reached?

Does y=x on (0,2) have a maximum just because the graph approaches 2?

Predict: Write a candidate maximum; predict what happens when the endpoint is closed.

Compare	Computed observation
Water: local versus global	Domain [0, 6]. Cursor input 2: output 5.
Open interval: build a larger value	Domain (0, 2). Cursor input 1.75: output 1.75.
Close the right endpoint	Domain (0, 2]. Cursor input 1: output 1.

Conclusion: On (0,2), every allowed x has a larger allowed input (x+2)/2. Thus 2 is a bound but no maximum is attained. On (0,2], the maximum is 2.

Conditions: Extrema depend on the complete stated domain; local peaks or a finite graph window do not establish global extrema.

Justification: For 0<x<2, x<(x+2)/2<2. This defeats every proposed maximum, not merely the sampled candidates.

Ask: Ask a student to propose a largest value and another to construct a strictly larger allowed value.

Watch for: A curve reaching the top of the visible window has a maximum.

Extension: Trim the water graph to [3,6]. Explain why its earlier peak no longer counts.

Exit: State maximum and minimum of y=x on (0,2].

Maximum 2; no minimum, because any allowed x has smaller allowed x/2.

05 Functions and their properties · g10-05-09 · 5–10 min · TEACHER · ANSWERS

Decreasing on each piece or the whole domain?

Is 1/x decreasing on all nonzero real numbers?

Predict: Predict the order of f(−1) and f(1), then compare it with the order of the inputs.

Compare	Computed observation
Both inputs positive	x₁ = 1, x₂ = 2. f(x₂) − f(x₁) = -0.5.
Both inputs negative	x₁ = -2, x₂ = -1. f(x₂) − f(x₁) = -0.5.
Cross zero	x₁ = -1, x₂ = 1. f(x₂) − f(x₁) = 2.
Forbidden input zero	x₁ = 0, x₂ = 2. One or both inputs are outside the domain.

Conclusion: 1/x decreases separately on (−∞,0) and (0,∞). Across zero, −1<1 but f(−1)=−1<f(1)=1, which refutes decrease on the whole domain.

Conditions: For strict decrease, every allowed pair x₁<x₂ must satisfy f(x₁)>f(x₂). x=0 is not allowed.

Justification: For same-sign x₁<x₂, 1/x₁−1/x₂=(x₂−x₁)/(x₁x₂)>0. A crossing-zero pair has negative denominator and reverses this sign.

Ask: Keep x₁<x₂ visible and ask which word in “for every pair” the counterexample defeats.

Watch for: Two decreasing branches imply decrease on their union.

Extension: For 2x²+1 on (0,∞), use a difference of squares to prove increase.

Exit: Give one pair refuting whole-domain decrease of 1/x.

x₁=−1, x₂=1 gives −1<1 and f(x₁)<f(x₂).

05 Functions and their properties · g10-05-10 · 5–10 min · TEACHER · ANSWERS

Symmetry needs the matching input

Is x² still even when its domain is restricted to [0,3]?

Predict: For x=2, first ask whether −2 is allowed; only then compare f(−x) and f(x).

Compare	Computed observation
Square on symmetric domain	Domain ℝ. f(2) = 4; actual f(-2) = 4.
Square on [0,3]	Domain [0, 3]. f(2) = 4; actual f(-2) = Undefined.
Shifted line	Domain ℝ. f(2) = 3; actual f(-2) = -1.
Zero function	Domain ℝ. f(2) = 0; actual f(-2) = 0.

Conclusion: Odd/even definitions require a domain symmetric about zero. x² on [0,3] fails that requirement. On a symmetric domain, x² is even, x³ odd, and the zero function both.

Conditions: Check domain symmetry before the identities f(−x)=f(x) or f(−x)=−f(x).

Justification: On a symmetric domain, (−x)²=x², (−x)³=−x³, and 0=−0. The restricted interval lacks −2 while containing 2.

Ask: Ask for the domain test and identity test as two separate decisions.

Watch for: The formula alone decides parity; odd and even are mutually exclusive.

Extension: Compare the mirrored point with the actual value for x+1. Identify a numerical failure of evenness.

Exit: Can a function be both odd and even on a symmetric nonempty domain?

Yes: f(x)=0. If both identities hold, f(x)=−f(x), so every value is zero.

05 Functions and their properties · g10-05-11 · 5–10 min · TEACHER · ANSWERS

Five powers, different forbidden inputs

Which of x, x², x³, √x and 1/x are defined at −1 and at 0?

Predict: On paper, mark allowed/rejected before matching each curve to its formula.

Compare	Computed observation
Negative versus positive input	Probe a = -1; b = 2. 0/5 cards placed. The window shows only part of each graph.
Zero versus negative input	Probe a = 0; b = -1. 0/5 cards placed. The window shows only part of each graph.
Positive fractional input	Probe a = 0.25; b = 2. 0/5 cards placed. The window shows only part of each graph.

Conclusion: Real √x requires x≥0; 1/x requires x≠0. The integer positive powers x, x² and x³ accept every real input.

Conditions: These are the five specified real-valued rules, not a claim that all real powers have the same domain.

Justification: Apply each operation’s real-domain condition. Polynomial integer powers require no divisions or roots; √x and 1/x do.

Ask: Use the zero case to distinguish “output zero” from “undefined”.

Watch for: Every power function is defined on all real numbers.

Extension: Explain why x² and x³ agree at 0 and 1 but disagree at −1.

Exit: List the five outputs at x=0, using “undefined” where needed.

0, 0, 0, 0, undefined, in the given order.

05 Functions and their properties · g10-05-12 · 5–10 min · TEACHER · ANSWERS

When does the larger exponent give less?

For positive x, does x³ always exceed x²?

Predict: Predict the ordering for x=0.5, 1 and 2 before moving the probe.

Compare	Computed observation
Base 0.5	Positive inputs only. x² = 0.25; x³ = 0.125; difference = 0.125.
Boundary base 1	Positive inputs only. x² = 1; x³ = 1; difference = 0.
Base 2	Positive inputs only. x² = 4; x³ = 8; difference = -4.

Conclusion: x³−x²=x²(x−1). Since x²>0 for x>0, the sign is determined by x−1: less below 1, equal at 1, greater above 1.

Conditions: The comparison here assumes x>0 and compares powers of the same base. Comparing two bases is a different question.

Justification: Factor the difference rather than extrapolating three samples. For positive a<b, b²−a²=(b−a)(b+a)>0.

Ask: Keep the equality case x=1 between the two opposite inequalities.

Watch for: A larger exponent always produces a larger positive value.

Extension: In same-power mode, compare a² and b² for positive a<b using factorisation.

Exit: Compare 0.8² and 0.8³ without decimal multiplication.

0.8²>0.8³ because multiplication by 0.8 reduces a positive value.

05 Functions and their properties · g10-05-13 · 5–10 min · TEACHER · ANSWERS

Undo the input rule in the right order

If f(2x+1)=6x+5, how do you obtain f(u) rather than f((u−1)/2)?

Predict: For u=3, place “subtract 1” and “halve” in the order that recovers x.

Compare	Computed observation
Subtract 1, then halve	u = 3. Your cards propose how to obtain x from the whole input 2x + 1.
Wrong inverse order	u = 3. Your cards propose how to obtain x from the whole input 2x + 1.
Discount then subtract	Original price = 100. Original-price eligibility met.
Reverse discount order	Original price = 100. Original-price eligibility met.

Conclusion: Set u=2x+1, so x=(u−1)/2. Substitution yields f(u)=6(u−1)/2+5=3u+2. Undoing the input is not inverting f.

Conditions: The affine input rule is invertible over real numbers. In the discount context, eligibility is checked using the original price.

Justification: Substitute x=(u−1)/2 into the stated identity; then 2x+1=u exactly. For prices, 0.9p−20 differs from 0.9(p−20) by −2.

Ask: After each rearrangement, run the recovered x forward through 2x+1 and check whether it returns u.

Watch for: Halve then subtract 1 is the inverse of doubling then adding 1.

Extension: Compare “10% off then subtract 20” with the reverse order at an eligible original price of 100.

Exit: From f(2x+1)=6x+5, find f(−3).

x=−2, hence f(−3)=−12+5=−7.

05 Functions and their properties · g10-05-14 · 5–10 min · TEACHER · ANSWERS

Cheapest at this quantity or most within budget?

Can the best plan change when the decision question changes?

Predict: At n=10, 30 and 50, predict the cheaper plan; then use budget 160 to predict the greatest affordable count.

Compare	Computed observation
Tie at 10 booklets	Integer quantities only. Budget 200: P permits 60 booklets; Q permits 70. P = 50; Q = 50; P − Q = 0
30 booklets	Integer quantities only. Budget 200: P permits 60 booklets; Q permits 70. P = 110; Q = 120; P − Q = -10
50 booklets	Integer quantities only. Budget 200: P permits 60 booklets; Q permits 70. P = 170; Q = 160; P − Q = 10
Budget 160	Integer quantities only. Budget 160: P permits 46 booklets; Q permits 50. P = 50; Q = 50; P − Q = 0
Rental at excluded time zero	Time is continuous on 0 < t ≤ 40. t = 0 is outside the stated domain; both costs are undefined.

Conclusion: P(n)=20+3n. Q(n)=5n for n≤20 and 2n+60 for n>20. Ties occur at integer counts 10 and 40. With budget 160, P allows 46 booklets and Q allows 50.

Conditions: Printing counts are integers 1–80. Rental time uses 0<t≤40 and different cost rules; t=0 is excluded.

Justification: On n≤20, P−Q=20−2n, giving n=10; on n>20, P−Q=n−40, giving n=40. Budget counts follow integer inequalities, not rounded real solutions.

Ask: State the objective and domain aloud before students choose a quote token.

Watch for: The plan cheaper at one quantity must always be the best choice.

Extension: Explain the two printing tie points using the two Q branches instead of just spotting crossings.

Exit: With budget 160, why is P limited to 46 rather than 47?

P(46)=158≤160, but P(47)=161>160.

05 Functions and their properties · g10-05-15 · 5–10 min · TEACHER · ANSWERS

Squaring can create a false root

Why does squaring √(x+2)=x produce two candidates but only one solution?

Predict: Predict whether x=−1 and x=2 satisfy the original equation before checking either graph.

Compare	Computed observation
Accepted root 2 Test input x = 2	At x=2: √(x+2)=2, x=2; original equation holds, squared equation holds.
Extraneous candidate −1 Test input x = -1	At x=-1: √(x+2)=1, x=-1; original equation fails, squared equation holds.
Outside radical domain −3 Test input x = -3	At x=-3, x+2<0: the real radical is undefined.
Radicand boundary −2 Test input x = -2	At x=-2: √(x+2)=0, x=-2; original equation fails, squared equation fails.

Conclusion: Squaring gives x²=x+2, with candidates −1 and 2. The original square root is nonnegative, so its right side must satisfy x≥0. Substitution rejects −1 and accepts 2.

Conditions: Real radicand requires x≥−2; equality with the principal square root additionally requires x≥0.

Justification: For real A≥0, √A=B is equivalent to A=B² together with B≥0. Here x²−x−2=(x−2)(x+1)=0; retaining x≥0 leaves x=2 only.

Ask: Write an implication arrow for squaring, then ask what added condition makes it equivalence.

Watch for: Squaring an equation always preserves exactly its solutions.

Extension: Compare the intersections of y=√(x+2), y=x, and y=x² versus y=x+2.

Exit: Solve √(x+2)=x and identify the extraneous candidate.

Solution x=2; candidate x=−1 is extraneous because 1≠−1.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.