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.