Mathematics Advanced • Year 11 • Module 1 • Lesson 1
Functions & Relations
Apply the function/relation distinction and function notation to realistic settings — pricing, security, geometry.
Problem 1 — Face ID and the one-output rule
A smartphone Face ID system takes a captured face image x and produces a decision F(x) of either "unlock" (1) or "do not unlock" (0). For the system to be reliable, the decision rule must be a function of the image.
Set up: What are we solving for?
(i) Explain in 1-2 sentences why F must satisfy the function definition (one input, one output) for the device to be usable. 2 marks
(ii) Suppose a buggy version of the system returns F(image_A) = 1 on Monday and F(image_A) = 0 on Tuesday for the exact same image. Using the formal definition of a function, state which property is violated and what the consequence is for the user. 2 marks
(iii) A different version of Face ID also accepts faces by Bluetooth from a paired watch, so the system has two inputs (image x, watch ID w). State whether the new rule F(x, w) is still a function in the Year 11 sense, and what the input set would need to look like. 2 marks
Stuck? Revisit lesson § What Is a Function? — the "machine" analogy.Problem 2 — Taxi fare as a function
A taxi company charges a $5 flag-fall plus $2 per kilometre. Let C(d) be the total cost ($) for a journey of d kilometres.
Set up: What are we solving for?
(i) Write the rule for C(d) and state the natural domain in this context. 2 marks
(ii) Find C(0), C(7) and C(20). Interpret C(0) in context. 3 marks
(iii) A passenger paid $33. By solving an equation in C(d), find the distance travelled, and justify that the function approach guarantees only one possible distance. 3 marks
Problem 3 — Geometric function: area of a square
The area A of a square as a function of its side length s is given by A(s) = s².
Set up: What are we solving for?
(i) State the independent and dependent variables, and the natural domain of A(s) in the geometric context. 2 marks
(ii) Evaluate A(3) and A(4.5), showing brackets. 2 marks
(iii) If the same equation A = s² is plotted on x = s, y = A axes without the geometric constraint s ≥ 0, then explained as "the side length is a function of the area", explain whether the reverse rule s(A) is a function of A on its natural domain. 3 marks
Stuck on (iii)? Solve A = s² for s and count the answers. Then apply the geometric restriction.Problem 4 — A circular bowl at the skate park
A cross-section of a skateboard bowl is the lower half of the circle x² + y² = 36, where x is horizontal distance from the centre (in metres) and y is the height (in metres, with y ≤ 0 below ground level).
Set up: What are we solving for?
(i) Show that the full circle x² + y² = 36 is not a function of x by giving a single counterexample with the vertical line test. 2 marks
(ii) Show that restricting to the lower half (y ≤ 0) gives a function y = b(x). Write the explicit rule for b(x) and state its domain. 3 marks
(iii) Find the depth of the bowl directly under a skater standing at x = 4 m from the centre. 2 marks
Problem 5 — Parking-meter table
A council parking meter accepts integer numbers of hours from 1 to 4 and shows the following price:
| Hours (t) | Cost C(t) ($) |
|---|---|
| 1 | 2.50 |
| 2 | 4.50 |
| 3 | 6.00 |
| 4 | 7.00 |
Set up: What are we solving for?
(i) State the domain of C and explain in one line why this table defines a function. 2 marks
(ii) Compute the average cost per hour for each duration and tabulate the results. Is the function linear? Justify in one sentence. 3 marks
(iii) A driver claims: "Since two different hours can give the same average rate, the table is not a function." Decide whether the driver is right or wrong, and quote the precise definition of a function in your answer. 2 marks
Stuck on (iii)? Re-read the definition: it's about one output per input, not one input per output.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Face ID
Set up. We are testing whether the decision rule of a security system satisfies the one-output-per-input definition of a function, and what happens when it doesn't.
(i) Each captured image (input) must produce exactly one decision (output) — either "unlock" or "do not unlock". If the same image could yield both, the user could never trust the system: sometimes their face would unlock the phone, sometimes not. Reliability requires the function property.
(ii) The same input (image_A) gives two different outputs (1, then 0). This violates the function definition "each input has exactly one output". Consequence: the device is non-deterministic; a single owner image could be accepted or rejected at random.
(iii) Still a function — but on a different input set. The input is now an ordered pair (x, w), one element of the set of (image, watch-ID) pairs. F is a function on that paired set provided each (image, watch) pair gives exactly one decision. We've changed the domain, not the definition.
Problem 2 — Taxi fare
Set up. We are modelling cost with function notation and using it to evaluate, interpret a fixed value, and invert.
(i) C(d) = 5 + 2d. Natural (contextual) domain: d ≥ 0 (distance cannot be negative).
(ii) C(0) = 5 + 2(0) = $5. C(7) = 5 + 2(7) = $19. C(20) = 5 + 2(20) = $45. C(0) is the flag-fall: the cost when the taxi has been hired but no distance has yet been travelled.
(iii) 5 + 2d = 33 ⇒ 2d = 28 ⇒ d = 14 km. Because C is a function, every cost value comes from at most one distance (here, C is linear and one-to-one), so $33 unambiguously corresponds to 14 km — there is no ambiguity in the back-calculated distance.
Problem 3 — Area of a square
Set up. We are using function notation for a geometric rule and considering whether the inverse rule is also a function.
(i) Independent: s (side length). Dependent: A (area). Natural domain in context: s ≥ 0 (side length is non-negative; some texts use s > 0 to exclude the degenerate square).
(ii) A(3) = (3)² = 9. A(4.5) = (4.5)² = 20.25.
(iii) Solving A = s² gives s = ±√A. With no geometric restriction, an input A > 0 produces two outputs (positive and negative), so s is not a function of A. With the geometric restriction s ≥ 0, we keep only the principal root and get s(A) = +√A, which is a function. So the reverse rule is a function only under the contextual restriction.
Problem 4 — Skate-park bowl
Set up. We use the vertical line test to decide whether a cross-section can be expressed as y = function of x, and we restrict to make it work.
(i) At x = 0 the full circle gives y² = 36 ⇒ y = ±6, so the vertical line x = 0 meets the circle at (0, 6) and (0, −6). Two outputs for one input → not a function.
(ii) Restricting to y ≤ 0 and solving y² = 36 − x² gives b(x) = −√(36 − x²). Domain: 36 − x² ≥ 0 ⇒ −6 ≤ x ≤ 6, i.e. [−6, 6].
(iii) Depth at x = 4: b(4) = −√(36 − 16) = −√20 = −2√5 ≈ −4.47 m. So the bowl is about 4.47 m deep at that point.
Problem 5 — Parking-meter table
Set up. We test the definition against a small table and explore what one-to-one means.
(i) Domain: {1, 2, 3, 4}. Each hour appears once and has exactly one cost, so the rule is a function.
(ii)
| t | C(t) | Average per hour C(t)/t |
|---|---|---|
| 1 | 2.50 | $2.50 |
| 2 | 4.50 | $2.25 |
| 3 | 6.00 | $2.00 |
| 4 | 7.00 | $1.75 |
Not linear: a linear rule would have constant difference C(t+1) − C(t). Here the differences are $2.00, $1.50, $1.00 — they decrease, so the function is not linear (it is concave — bulk discount).
(iii) The driver is wrong. The definition is: "for every input there is exactly one output". It says nothing about outputs being distinct across inputs. Two different t-values producing the same average rate would only break one-to-one, which is a stricter property. C(t) here is a function regardless.