Mathematics Advanced • Year 11 • Module 1 • Lesson 8
Working with Functions — Synthesis
Apply multiple function tools — composition, inverse, domain, and symmetry — to integrated, real-world style problems.
Problem 1 — Temperature, time, and a composite scale
A cooling liquid has temperature T (°C) as a function of time t (minutes) given by T(t) = 80 − 5t. The Celsius → Fahrenheit conversion is F(C) = (9/5) C + 32.
T(t) = 80 − 5t, F(C) = (9/5) C + 32
Set up: What are we solving for?
(i) Find a single formula for temperature in °F as a function of time t, by forming (F ∘ T)(t) and simplifying. 3 marks
(ii) Use your formula from (i) to find the temperature in °F at t = 0, t = 4 and t = 8 minutes. 2 marks
(iii) The liquid stops cooling at room temperature, 68 °F. Use your formula to find the time at which T reaches 68 °F. 2 marks
Stuck on (iii)? Set 176 − 9t = 68 and solve.Problem 2 — Consulting fee: function and its inverse
A consultant's cost for h hours is C(h) = 50 + 20h dollars (flat fee plus hourly rate). A client has a budget of $250.
Set up: What are we solving for?
(i) Find C⁻¹(x) by swap-and-solve, and explain in one sentence what C⁻¹(x) gives. 2 marks
(ii) Use C⁻¹ to find the maximum number of whole hours the client can afford on their $250 budget. 2 marks
(iii) The consultant introduces a new package P(h) = 80 + 15h (higher flat fee, lower rate). Set up and solve the equation that finds the number of hours at which the two pricing options cost the same. 3 marks
Problem 3 — Domain of a composite (root + rational)
An engineer uses the composite (f ∘ g)(x) where f(x) = √(x − 1) and g(x) = 2/x.
Set up: What are we solving for?
(i) Write (f ∘ g)(x) as a single simplified surd expression. 2 marks
(ii) Find the domain of (f ∘ g)(x). Write the two conditions separately, solve the second using sign-of-(2 − x)/x, and give the domain in interval notation. 3 marks
(iii) Show by direct substitution that x = 2 is in the domain and x = 3 is not, listing the domain check that fails for x = 3. 2 marks
Stuck on (ii)? Critical values at x = 0 and x = 2; test the sign of (2 − x)/x in each interval.Problem 4 — Restricted quadratic, then inverse (engineering)
The kinetic energy of a particle (in joules) as a function of its speed v (m/s) is modelled by E(v) = ½ · v² for v ≥ 0. An instrument reports E and the engineer needs the inverse to find the speed.
Set up: What are we solving for?
(i) Explain in one sentence why the restriction v ≥ 0 is essential before finding E⁻¹(x). 1 mark
(ii) Find E⁻¹(x) and state its domain in interval notation. 3 marks
(iii) Verify your inverse by computing E(E⁻¹(x)) and showing it simplifies to x. 2 marks
Problem 5 — Choose the right tool (symmetry vs evaluation)
A signal processor produces three candidate response functions:
A. r(x) = x⁴ − 2x² B. r(x) = x³ − 3x C. r(x) = x² + 5x
Set up: What are we solving for?
(i) Classify each as even, odd, or neither. Show the line r(−x) = ... for each. 3 marks
(ii) The engineer wants a response whose graph is symmetric about the y-axis (so they can sketch only x ≥ 0 and reflect). Which option(s) qualify, and why? 1 mark
(iii) For option B (odd), the value at x = 2 is 2³ − 3(2) = 2. Use the odd-function property to write down the value at x = −2 without substituting, and explain the geometric meaning in one sentence. 2 marks
Stuck? Recall: even ⇒ y-axis symmetry; odd ⇒ 180° rotational symmetry about the origin.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Temperature composite
Set up. Build a single formula for °F as a function of time by composing the temperature model with the unit-conversion function.
(i) (F ∘ T)(t) = F(80 − 5t) = (9/5)(80 − 5t) + 32 = (9/5)(80) − (9/5)(5t) + 32 = 144 − 9t + 32 = 176 − 9t °F.
(ii) t = 0: 176 °F. t = 4: 176 − 36 = 140 °F. t = 8: 176 − 72 = 104 °F.
(iii) 176 − 9t = 68 ⇒ 9t = 108 ⇒ t = 12 minutes.
Problem 2 — Consulting fee
Set up. Inverting a linear cost function to convert "dollars available" into "hours purchasable", and comparing two pricing options.
(i) y = 50 + 20h ⇒ swap: x = 50 + 20y ⇒ y = (x − 50)/20. C⁻¹(x) = (x − 50)/20. It gives the number of hours that can be purchased for $x.
(ii) C⁻¹(250) = (250 − 50)/20 = 200/20 = 10 hours.
(iii) Equal cost: 50 + 20h = 80 + 15h ⇒ 5h = 30 ⇒ h = 6 hours. (At this point both packages cost $170.)
Problem 3 — Domain of composite (root + rational)
Set up. Apply both domain conditions and combine using a sign-of-rational analysis.
(i) (f ∘ g)(x) = f(2/x) = √((2/x) − 1) = √((2 − x)/x).
(ii) Condition 1 (inner): x ≠ 0. Condition 2 (outer): (2 − x)/x ≥ 0. Critical values: x = 0 and x = 2. Sign of (2 − x)/x: on (−∞, 0) numerator + and denominator − ⇒ negative; on (0, 2] numerator ≥ 0 and denominator + ⇒ ≥ 0 ✓; on (2, ∞) numerator − ⇒ negative. So the rational expression is ≥ 0 on (0, 2]. Combined with x ≠ 0: Domain = (0, 2].
(iii) x = 2: g(2) = 1, then f(1) = √(1 − 1) = 0. ✓ defined. x = 3: g(3) = 2/3, then we need (2/3) − 1 = −1/3 ≥ 0, which fails. The outer (square-root) condition fails at x = 3.
Problem 4 — Restricted quadratic, then inverse
Set up. Restrict, invert, verify — the canonical workflow for a non-injective polynomial.
(i) Without restriction, E(v) = ½ v² is not one-to-one because E(v) = E(−v); the inverse would not be a function. The restriction v ≥ 0 makes E one-to-one.
(ii) y = ½ v² ⇒ swap: x = ½ y² ⇒ y² = 2x ⇒ y = √(2x) (positive root, because v ≥ 0). E⁻¹(x) = √(2x). Domain of E⁻¹ = range of E = [0, ∞).
(iii) E(E⁻¹(x)) = E(√(2x)) = ½ · (√(2x))² = ½ · 2x = x. ✓
Problem 5 — Symmetry decision
Set up. Classify three candidate response functions and use symmetry to halve sketching/computation effort.
(i) A. r(−x) = (−x)⁴ − 2(−x)² = x⁴ − 2x² = r(x) → even.
B. r(−x) = (−x)³ − 3(−x) = −x³ + 3x = −(x³ − 3x) = −r(x) → odd.
C. r(−x) = (−x)² + 5(−x) = x² − 5x; not r(x), not −r(x) → neither.
(ii) Option A only, because evenness is the algebraic statement of y-axis (mirror) symmetry. (Option B has 180°-rotational symmetry about the origin; option C has no useful symmetry.)
(iii) For odd r, r(−2) = −r(2) = −2. Geometrically: the point (2, 2) on the graph is paired with the point (−2, −2), reflecting the 180° rotational symmetry about the origin.