IB Maths AI SL Topic 3 — Modelling with Functions Paper 1 & 2 GDC-essential ~7 min read

Cubic Models

A cubic model f(x) = ax³ + bx² + cx + d describes anything with two turning points in its domain — a local max and a local min — or none at all if the quantity is monotonic. Real-world examples: an open box’s volume vs the size of the corner squares cut from a sheet, a drug’s concentration in the bloodstream over time, a bungee jumper’s depth below the platform, a startup’s profit that dips, recovers, and falls again. Most cubic-model questions in AI SL use the GDC for maxes, mins, and zeros — the algebra rarely factorises neatly.

📘 What you need to know

Standard form: f(x) = ax³ + bx² + cx + d with a ≠ 0. a > 0: tails go from −∞ (left) to +∞ (right). a < 0: opposite direction.
Initial value: f(0) = d. This is the quantity before any time/input has elapsed — initial height, starting volume, baseline concentration.
Turning points: a cubic has either 0 or 2 local extrema. Use the GDC’s maximum and minimum tools to locate them — cubic algebra is rarely clean.
Local ≠ global: cubic tails go to ±∞, so the model has no global max/min. Always say “local” when describing turning points.
When to choose a cubic: the data rises then falls then rises again (or vice versa) within the observed range. Two turning points is the signature.
Domain restriction is essential: cubics predict negative values or unbounded growth outside the meaningful range. Restrict to where the quantity makes physical sense (often a closed interval [0, t_end]).

Reading the parameters

For a cubic model f(x) = ax³ + bx² + cx + d:

d = initial value. The output at x = 0. This is the only parameter you can read off without computation — the rest just shape the curve.

Sign of a sets the direction. Positive a: the curve rises overall (left to right), with a local max and then a local min in the middle. Negative a: the curve falls overall, with a local min then a local max.

Size of |a| controls how sharply the curve bends. Large |a| means dramatic peaks and troughs; small |a| means a gentler curve.

Cubic model f(x) = ax³ + bx² + cx + d, a ≠ 0

initial value f(0) = d · turning points: 0 or 2 (use GDC)

Within the meaningful domain 0 ≤ t ≤ 6, the cubic captures one full bungee bounce. Outside that range the model predicts the jumper rising above the platform — non-physical, so the domain is restricted.

When to choose a cubic — and when not to

Use a cubic when the real-world data shows two turning points within the observation window: a quantity that goes up, peaks, comes down past a trough, then goes up again (or the reverse). If you only see one turning point, a quadratic is simpler and better; if there are three or more, a higher-degree polynomial is needed.

Monotonic cubic case: not every cubic has two turning points. If the derivative (a quadratic) has no real roots, the cubic is always increasing or always decreasing — useful when modelling something that grows steadily but with varying speed. The “S-shape” cubic without obvious peaks falls into this category.

Spotting a cubic problem: open-box volume problems (cut squares from a sheet and fold up) always give a cubic. So do “volume of a container” problems where one dimension is fixed in terms of another. The shape of the curve (rise-then-fall-then-rise within the domain) confirms the cubic choice.

Limitations — why domain matters

The mathematical cubic continues to ±∞ on both tails. Real-world quantities have natural cut-offs: a bungee jumper doesn’t rise infinitely above the platform; a drug concentration doesn’t go negative; a box can’t have a negative dimension. State the domain in context — usually a closed interval starting at 0 and ending where the model first hits a boundary (zero output, physical impossibility, etc.).

🧭 Recipe — using a cubic model

Identify the initial value: f(0) = d. State its meaning in context (initial volume, baseline level, starting position).
Plot the model on your GDC over a sensible viewing window. Make sure both turning points (if any) are visible.
Use GDC tools for turning points: maximum and minimum commands give exact coordinates. There will be 0 or 2 — never just 1.
Find any boundary values: when does the model hit zero (run out, return to start)? Use the GDC’s zero/root tool.
State the domain in context: usually [0, t_end] where t_end is where the model first hits a physical boundary. Outside that range, the model is just symbols.

Worked examples

WE 1

Open box from cardboard — maximise volume

A square piece of cardboard 30 cm by 30 cm has equal squares of side x cm cut from each corner. The sides are then folded up to make an open box. (a) Show that the volume can be written as V(x) = 4x³ − 120x² + 900x. (b) State the domain of x in context. (c) Use your GDC to find the value of x that maximises the volume, and state the maximum volume.

(a) base side = 30 − 2x, height = x V(x) = x × (30 − 2x)² = x(900 − 120x + 4x²) = 4x³ − 120x² + 900x ✓ (b) need positive base AND positive height x > 0 AND 30 − 2x > 0 so 0 < x < 15 (c) plot on GDC, use maximum tool max at x = 5 cm V(5) = 4(125) − 120(25) + 900(5) = 500 − 3000 + 4500 = 2000 domain 0 < x < 15 · max volume 2000 cm³ at x = 5 cm classic AI SL setup: any “cut squares from a sheet, fold a box” problem gives a cubic V(x) and an obvious domain. The maximum is always strictly inside the domain (not at a boundary).

WE 2

Drug concentration in the bloodstream

The concentration of a drug in the bloodstream (mg/L) at time t hours after taking it is modelled by C(t) = −0.4t³ + 3t² + 0.5 for t ≥ 0. (a) Find C(0) and interpret. (b) Use your GDC to find the maximum concentration and when it occurs. (c) Find when the concentration returns to baseline.

(a) initial value C(0) = 0.5 mg/L means: baseline level in the body BEFORE the dose (b) plot on GDC; use maximum tool max at t = 5 hours C(5) = −0.4(125) + 3(25) + 0.5 = −50 + 75 + 0.5 = 25.5 mg/L (c) back to baseline: C(t) = 0.5 −0.4t³ + 3t² + 0.5 = 0.5 −0.4t³ + 3t² = 0 ⇒ t²(3 − 0.4t) = 0 t = 0 or t = 3/0.4 = 7.5 (a) 0.5 mg/L baseline · (b) 25.5 mg/L at t = 5 h · (c) returns at t = 7.5 h “baseline” recognises that drugs don’t start (or end) at zero in the body. C(0) is the level before the dose; the concentration peaks at 5 h, then drops back to 0.5 mg/L at 7.5 h. Useful domain: 0 ≤ t ≤ 7.5.

WE 3

Bungee jumper — depth below platform

A bungee jumper’s depth below the platform (m) at time t seconds is D(t) = 0.5t³ − 6t² + 18t. (a) Find D(0). (b) Use your GDC to find the deepest point reached and when. (c) Find the next time the jumper is at the platform again, and state the meaningful domain.

(a) initial depth D(0) = 0 (jumper is AT the platform, no depth yet) (b) GDC maximum tool on D(t) max at t = 2 s D(2) = 0.5(8) − 6(4) + 18(2) = 4 − 24 + 36 = 16 deepest = 16 m at t = 2 s (c) back at platform: D(t) = 0 0.5t³ − 6t² + 18t = 0 0.5t(t² − 12t + 36) = 0 0.5t(t − 6)² = 0 t = 0 or t = 6 (double root) (a) D(0) = 0 · (b) deepest 16 m at t = 2 s · (c) back at platform at t = 6 s, domain 0 ≤ t ≤ 6 the model only makes physical sense for one bounce: 0 ≤ t ≤ 6. Beyond t = 6 it would predict NEGATIVE depth (the jumper rising above the platform), which the real bungee won’t do.

WE 4

Startup profit — local min & local max

A startup’s monthly profit (in thousand dollars) at t months after launch is P(t) = −t³ + 9t² − 15t + 25, for 0 ≤ t ≤ 8. Find the initial profit, the worst month (local min), and the best month (local max), stating the profit in each case.

Step 1 — initial profit P(0) = 25 ⇒ $25 000 at launch Step 2 — GDC max/min tools local min: t = 1 P(1) = −1 + 9 − 15 + 25 = 18 local max: t = 5 P(5) = −125 + 225 − 75 + 25 = 50 Step 3 — interpret worst month: t = 1, profit $18 000 best month: t = 5, profit $50 000 initial $25K · worst $18K at month 1 · best $50K at month 5 a < 0 cubic: the LEFT turning point is the local MIN and the RIGHT is the local MAX. After month 5 the profit declines (the curve falls towards −∞ on the right tail), so the model is only useful up to whatever month the business stays profitable.

WE 5

Water tank — fill, peak, drain

Water is added to and later drained from a tank. The volume (litres) at time t minutes is V(t) = −0.5t³ + 6t² for t ≥ 0. Find: (a) the maximum volume the tank reaches and when, (b) the time when the tank is empty again, (c) the meaningful domain of t.

Step 1 — GDC maximum tool max at t = 8 minutes V(8) = −0.5(512) + 6(64) = −256 + 384 = 128 L Step 2 — tank empty: V(t) = 0 −0.5t³ + 6t² = 0 0.5t²(12 − t) = 0 t = 0 (start) or t = 12 (empty again) Step 3 — meaningful domain model valid until the tank empties 0 ≤ t ≤ 12 max 128 L at t = 8 min · empty at t = 12 min · domain 0 ≤ t ≤ 12 factoring helps here: V = 0.5t²(12 − t) shows the roots immediately. After t = 12 the cubic predicts negative volume — nonsense for a tank, so we stop at the second root.

WE 6

Find unknown parameters from data

A container’s volume (litres) at time t minutes is modelled by V(t) = at³ + bt² + 50. After 1 minute the volume is 47 L; after 3 minutes it is 5 L. Find a and b, and use the model to predict V(2).

Step 1 — check V(0) = 50 ✓ (matches the constant 50) Step 2 — set up two equations V(1) = a + b + 50 = 47 ⇒ a + b = −3 V(3) = 27a + 9b + 50 = 5 ⇒ 27a + 9b = −45 divide by 9: 3a + b = −5 Step 3 — solve simultaneously (3a + b) − (a + b) = −5 − (−3) 2a = −2 ⇒ a = −1 b = −3 − (−1) = −2 Step 4 — predict V(2) V(t) = −t³ − 2t² + 50 V(2) = −8 − 8 + 50 = 34 a = −1, b = −2 · V(2) = 34 L three unknowns in V(t) = at³ + bt² + d need three data points: V(0) fixes d; then two more equations fix a and b. GDC simultaneous-equation solver handles 3×3 systems instantly.

💡 Top tips

Use GDC for turning points: cubic critical points rarely come out cleanly by hand. Plot the function, then use maximum and minimum tools.
0 or 2 turning points — never 1: if you only see one, you’ve missed the other. Zoom out further to spot it.
State the initial value’s meaning in context: d is not just a number — it’s the baseline level, starting volume, initial profit, etc.
Restrict the domain to where the model is realistic: usually 0 ≤ t ≤ t_end where t_end is when the quantity hits zero or some physical limit.
For “find a, b, c, d” questions: each data point gives one equation. Four parameters need four pieces of information; the GDC solves the system.

⚠ Common mistakes

Claiming the local max is the global maximum: cubic tails go to ±∞, so the local max is NOT the highest point overall (unless the domain is restricted to capture only that part).
Reporting only one turning point: if you find a max, there’s almost always a min nearby. Cubics with two extrema have BOTH.
Forgetting to restrict the domain in a real-world model: the maths continues forever; the physics doesn’t. State 0 ≤ t ≤ t_end explicitly.
Confusing the cubic’s d with an exponential’s k + c: in a cubic, d is the initial value directly. In y = k·a^x + c, the initial value is k + c — different rule.
Predicting outside the meaningful domain: a cubic profit model might say “month 50 has profit −$1M”. That’s the mathematics, not the business — the model probably isn’t valid that far out.

Up next: Exponential Models. Exponentials describe quantities that grow or decay at a percentage rate — compound interest, radioactive decay, cooling, population growth. The shape is monotonic with a horizontal asymptote, so the model never quite reaches its long-term limit. The parameters k, a (or r), and c each carry direct real-world meaning.

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