Mathematics Advanced • Year 11 • Module 3 • Lesson 11
Increasing and Decreasing Functions
Apply sign-analysis of f′(x) to drug response, reservoir flow, profit functions, particle motion, and parameter searches.
Problem 1 — Drug response curve
The concentration of a drug in a patient's bloodstream (mg/L) after t hours is modelled by
C(t) = 6t − t², for 0 ≤ t ≤ 6.
Set up: What are we solving for?
(i) Find C′(t) and solve C′(t) = 0 to locate the stationary point. 1 mark
(ii) Show using sign analysis that C is increasing on (0, 3) and decreasing on (3, 6). 2 marks
(iii) State the time at which the concentration is greatest, and the peak concentration. Interpret in one sentence in the medical context. 2 marks
Stuck? Revisit lesson § Concept — stationary points separate intervals of increase from intervals of decrease.Problem 2 — Reservoir volume
The volume of water in a reservoir (megalitres) t months after the start of the year is modelled by
V(t) = t³ − 9t² + 24t + 100, for 0 ≤ t ≤ 8.
Set up: What are we solving for?
(i) Show that V′(t) = 3(t − 2)(t − 4). 2 marks
(ii) Identify all intervals on which V is increasing and decreasing in [0, 8]. Use a sign table. 3 marks
(iii) Translate the result for the reservoir manager: state the months in which water is being added, the months in which it is being drained, and the month containing the lowest level inside the year. 2 marks
Problem 3 — Café profit curve
A café's daily profit (in dollars) when it sells x cakes is modelled by
P(x) = −x³ + 30x² − 200x, for 0 ≤ x ≤ 25.
Set up: What are we solving for?
(i) Find P′(x) and factorise. 2 marks
(ii) Find all values of x in [0, 25] for which P is increasing. 3 marks
(iii) The owner reads "P is increasing" as "selling more cakes always makes more money". Explain in one sentence why that interpretation is wrong by referring to the part-(ii) interval. 2 marks
Stuck? Revisit lesson § Trap 03 — f′ = 0 marks where the trend changes, not where the function crosses zero.Problem 4 — Particle on a number line
A particle moves along the x-axis so that its position (in metres) at time t seconds is
x(t) = 2t³ − 15t² + 24t, for t ≥ 0.
Set up: What are we solving for?
(i) Find the velocity v(t) = x′(t) and factorise. 1 mark
(ii) Find the time intervals on which the particle is moving in the positive direction (i.e. x is increasing). 3 marks
(iii) At what times is the particle momentarily stationary? State whether each is a switch from "moving right" to "moving left" or the reverse, and use that to identify the local maximum and local minimum positions. 2 marks
Problem 5 — When is f everywhere increasing?
Consider the family of cubics f(x) = x³ + ax² + bx + 1 with a, b real constants.
Set up: What are we solving for?
(i) Write down f′(x) and state the condition on its discriminant for f to be (weakly) increasing for every real x. 2 marks
(ii) Use part (i) to find all values of b that make f everywhere increasing when a = 0. 2 marks
(iii) Take a = 6. Find all values of b such that f is everywhere increasing. Then sketch the boundary in the (a, b) = (6, b) line: for which b is the cubic strictly increasing, and at which b does it have a horizontal point of inflection rather than a turning point? 3 marks
Stuck on (iii)? f′(x) is a quadratic in x. "Everywhere increasing" needs no two distinct real roots, i.e. discriminant ≤ 0.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Drug response C(t) = 6t − t²
Set up. We need to find where C is increasing/decreasing to locate the peak concentration.
(i) C′(t) = 6 − 2t. C′(t) = 0 ⇒ t = 3 hours.
(ii) For 0 < t < 3: C′(2) = 6 − 4 = 2 > 0 (increasing). For 3 < t < 6: C′(5) = 6 − 10 = −4 < 0 (decreasing). Hence increasing on (0, 3), decreasing on (3, 6).
(iii) Greatest concentration occurs at t = 3 hr, with C(3) = 18 − 9 = 9 mg/L. Medically: the drug peaks 3 hours after dosing, then begins to clear from the bloodstream.
Problem 2 — Reservoir V(t) = t³ − 9t² + 24t + 100
Set up. Differentiate, factorise, build a sign table for V′, then interpret the months.
(i) V′(t) = 3t² − 18t + 24 = 3(t² − 6t + 8) = 3(t − 2)(t − 4).
(ii) Stationary at t = 2 and t = 4. V′(0) = 24 (+); V′(3) = 3(1)(−1) = −3 (−); V′(5) = 3(3)(1) = 9 (+). Increasing on (0, 2) ∪ (4, 8); decreasing on (2, 4).
(iii) Water is being added during months 0–2 and months 4–8, drained during months 2–4. The lowest level inside the year occurs at t = 4 months (the local minimum). (For the absolute minimum on [0, 8], also check V(0) = 100 and V(4) = 64 − 144 + 96 + 100 = 116 — so the lowest level is at t = 4 with V = 116 ML.)
Problem 3 — Café P(x) = −x³ + 30x² − 200x
Set up. Differentiate and sign-test to find the range of x where profit grows with sales.
(i) P′(x) = −3x² + 60x − 200. (Optional: factor of 1/3 etc. — keep as a quadratic.)
(ii) Solve P′(x) = 0: x = (60 ± √(3600 − 2400))/6 = (60 ± √1200)/6 = (60 ± 20√3)/6 = 10 ± (10√3)/3. Numerically x ≈ 4.23 and x ≈ 15.77. Since the leading coefficient of P′ is negative (opens down), P′ > 0 between the two roots. Hence P is increasing on approximately (4.23, 15.77) (or exactly: (10 − 10√3/3, 10 + 10√3/3)) intersected with [0, 25].
(iii) "Selling more cakes always makes more money" is wrong because profit is only increasing on the middle interval ≈ (4.23, 15.77). Below 4.23 cakes you are losing money (fixed costs dominate) and above 15.77 each extra cake reduces profit (the cubic curve turns over). The "more is better" intuition fails because the marginal profit P′(x) eventually goes negative.
Problem 4 — Particle x(t) = 2t³ − 15t² + 24t
Set up. Velocity = derivative. The particle moves "in the positive direction" exactly when v(t) > 0.
(i) v(t) = x′(t) = 6t² − 30t + 24 = 6(t − 1)(t − 4).
(ii) v = 0 at t = 1 and t = 4. v(0) = 24 (+); v(2) = 6(1)(−2) = −12 (−); v(5) = 6(4)(1) = 24 (+). Positive direction (x increasing) on [0, 1) ∪ (4, ∞).
(iii) The particle is momentarily stationary at t = 1 s (switch from + to −, so a local maximum of position) and t = 4 s (switch from − to +, so a local minimum). x(1) = 2 − 15 + 24 = 11 m (local max); x(4) = 128 − 240 + 96 = −16 m (local min).
Problem 5 — When is x³ + ax² + bx + 1 everywhere increasing?
Set up. Look at the derivative quadratic and use the discriminant.
(i) f′(x) = 3x² + 2ax + b. f is (weakly) increasing for every real x exactly when 3x² + 2ax + b ≥ 0 for all x, which (since the leading coefficient 3 > 0) requires discriminant ≤ 0: (2a)² − 4(3)(b) = 4a² − 12b ≤ 0, i.e. a² ≤ 3b.
(ii) With a = 0: need 0 ≤ 3b, i.e. b ≥ 0. (Strictly increasing requires b > 0; at b = 0, f′(x) = 3x² ≥ 0 with equality only at x = 0, giving a stationary point of inflection there.)
(iii) With a = 6: need 36 ≤ 3b, i.e. b ≥ 12. For b > 12 the quadratic 3x² + 12x + b has no real roots and is strictly positive, so f is strictly increasing. At b = 12 the discriminant is zero — f′ has a double root at x = (−2a)/(2·3) = −2, and the cubic has a horizontal point of inflection at x = −2 (not a turning point). For b < 12, f has both a local maximum and a local minimum.