How do I decide which model a question wants?

Read for the shape and the story. 'Levels off', 'carrying capacity', 'maximum population the environment supports' means logistic. 'Different rates', 'thresholds', 'for the first ... then ...' means piecewise. 'Cycle', 'periodic', 'repeats every' means sinusoidal. 'Grows by a fixed percentage' with no ceiling means exponential. In AI the model is nearly always given to you or strongly signalled — your job is to interpret and use it, not invent it.

Do I have to memorise the logistic formula?

No. The general form $f(x)=\frac{L}{1+Ce^{-kx}}$ is not in the formula booklet, but any specific logistic function you need will be printed in the question. What you must know is how to read it: $L$ is the carrying capacity (top of the fraction), $f(0)=\frac{L}{1+C}$ is the start value, and the curve is S-shaped rising to the asymptote $y=L$.

What is the difference between $<$ and $\le$ in a piecewise domain?

The sign decides whether the boundary point belongs to that piece. On $0<x\le 5$ the rule covers every value above $0$ and up to and including $5$. So to evaluate the function at $x=5$ you use this rule, but at $x=0$ you do not. Confusing the two is the most penalised piecewise error, especially at the exact boundary.

How do I solve a logistic equation like $4000=\frac{5000}{1+9e^{-0.4t}}$ on Paper 2?

You do not need to rearrange the exponential by hand. Graph $y_1=\frac{5000}{1+9e^{-0.4t}}$ and the horizontal line $y_2=4000$ on your GDC, then use the intersect tool to read $t$. But write the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ on your page first — that line is the method mark; the GDC gives the accuracy mark.

How do I judge whether a model fits the data?

Plot the data and the model together and look across the whole range, not just the middle. A good fit hugs the points at the start, in the bulk, and at the extremes, and behaves sensibly at the ends — a logistic model should flatten toward its ceiling, an exponential should not turn negative. If the curve matches early points but overshoots a known limit, it is the wrong family of model however close it looks near the start.

Modelling with logistic, piecewise and other…

Revise Modelling with logistic, piecewise and other functions for IB Mathematics: Applications and Interpretation (2.6)

Looking for free IB Mathematics: Applications and Interpretation notes on Modelling with logistic, piecewise and other functions? This Standard Level lesson covers official syllabus point 2.6 on Mathematics: Applications and Interpretation SL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Modelling with logistic, piecewise and other functions. Built for students searching "IB Maths AI 2.6", "Modelling with logistic, piecewise and other functions IB revision", or a free alternative to paid IB note sites.

Modelling with logistic, piecewise and other functions

Applications and Interpretation is, at heart, a modelling course: you meet real data and real situations and describe them with functions. You have already used linear, quadratic and exponential models. This lesson adds two more that appear constantly in the real world. The logistic model captures growth that speeds up, then slows, then flattens against a natural ceiling — populations, epidemics, the uptake of a new product. The piecewise model captures situations governed by different rules on different intervals — taxi fares, tax bands, delivery charges. The lesson finishes with the judgement that defines AI: given a set of data, which model should you choose, and how well does it fit? Because 2.6 is worked on a GDC, the calculator does the arithmetic and the skill being marked is choosing the model and writing the method — so, as in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Interpret the logistic model $f(x)=\frac{L}{1+Ce^{-kx}}$ : identify the carrying capacity $L$ , describe the S-shaped behaviour, and read the initial value and limit.
2Evaluate a logistic model at a given input and use the GDC to solve for the input that produces a target output.
3Evaluate and interpret a piecewise-defined function, selecting the correct sub-rule for the interval and reading the inequality signs precisely.
4Choose an appropriate model (linear, quadratic, exponential, logistic, piecewise, sinusoidal) for a data set and evaluate its fit.
5Use a GDC to graph these models, read key features, and solve equations, presenting working so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Beyond Straight Lines: Growth With a Ceiling, and Rules That Switch

Real situations rarely follow one tidy rule. Some grow quickly and then flatten out against a natural limit — that is the logistic model. Others obey one rule up to a threshold and a different rule beyond it — that is a piecewise function. This lesson is about recognising which model fits, using it on a GDC, and reading the answer back into the real situation.

THINK OF IT LIKE…

Think of filling a stadium for a concert. Word spreads slowly at first, then explosively as friends tell friends, then slowly again as almost everyone who is coming has already bought a ticket — and it can never exceed the number of seats. That ceiling is the carrying capacity $L$ , and the S-shaped curve of ticket sales is a logistic model. Now think of the car park outside: free for the first 30 minutes, then a fixed rate per hour after that. The cost obeys one rule below the threshold and another above it — a piecewise function.

1
Identify the model. S-shaped growth toward a ceiling means logistic; different rules on different intervals means piecewise; a repeating cycle means sinusoidal. Let the shape of the data or the wording of the problem decide.
2
Read the parameters. For a logistic model the carrying capacity $L$ is the number on top of the fraction; the initial value is found by putting the input to zero. For a piecewise model, note which interval each rule owns and whether the endpoints are included.
3
Use the GDC. Graph the model, then use 'value' to evaluate it and 'intersect' to solve for the input that gives a target output. On Paper 2 the calculator does the arithmetic; you supply the method.
4
Interpret in context. Every final answer is a quantity in the real situation — a population, a time, a cost — so state it with units and check it is sensible against the model's limit.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$f(x)=\frac{L}{1+Ce^{-kx}},\qquad L,\,C,\,k>0.$

Full topic notes

Formal explanation with the rigour you need for the exam.

Piecewise functions: different rules for different intervals

A piecewise function is built from several pieces, each a separate rule that applies to its own slice of the domain. The slices do not overlap, so at any input exactly one rule is in charge. Everyday pricing is full of them: a delivery service might charge a flat fee up to a certain weight and then a rate per extra kilogram; a car park might be free for thirty minutes and then charge by the hour. To use a piecewise function you first ask which interval the input falls in, then apply that interval's rule — and only that rule.

A piecewise function with rules $g$ and $h$ on non-overlapping intervals $D_1$ and $D_2$ is written with a brace:\n\n $f(x)=\begin{cases} g(x) & \text{if } x\in D_1 \\ h(x) & \text{if } x\in D_2 \end{cases}$

One rule per input. The intervals do not overlap, so every input is governed by exactly one sub-rule.
Watch the boundary. At an endpoint, use the rule whose interval includes it — the one written with $\le$ or $\ge$ , not $<$ or $>$ .
Graphing convention. Draw an open circle where an endpoint is excluded and a filled circle where it is included.
Continuity is not automatic. The pieces meet smoothly only if both rules give the same value at the shared boundary.

Logistic functions: growth with a ceiling

A plain exponential model grows without limit, which is unrealistic for anything living in a finite world. Bacteria run out of nutrients, a rumour runs out of people who have not heard it, a new phone runs out of buyers. The logistic model captures this: growth is nearly exponential at first, then slows as the quantity approaches a ceiling, and finally levels off. The result is the S-shaped (sigmoid) curve. The ceiling — the value the model approaches but never passes — is called the carrying capacity.

The logistic model is\n $f(x)=\frac{L}{1+Ce^{-kx}},\qquad L,\,C,\,k>0.$ \n- $L$ is the carrying capacity — the limiting value, read straight from the numerator.\n- $C$ is fixed by the starting conditions; it sets the initial value $f(0)=\frac{L}{1+C}$ .\n- $k$ controls how steeply the curve rises.

Carrying capacity $L$ : the horizontal asymptote $y=L$ ; the model approaches it but never reaches or exceeds it.
Initial value: at $x=0$ , $f(0)=\frac{L}{1+C}$ , because $e^{0}=1$ — this is the start size, not $L$ .
Shape: an increasing S-curve between the asymptotes $y=0$ and $y=L$ .
Fastest growth: at the point of inflection, which sits at half the carrying capacity, $y=\frac{L}{2}$ .

Choosing an appropriate model and evaluating its fit

The signature skill of AI is not calculation but choice: faced with a set of data, which family of function should describe it? Match the shape and the story. Steady change points to a linear model; a single peak or valley to a quadratic; fixed-percentage growth or decay with no ceiling to an exponential; growth toward a ceiling to a logistic; a repeating cycle to a sinusoidal $f(x)=a\sin\!\big(b(x-c)\big)+d$ ; and different rules on different intervals to a piecewise model. Having chosen, you must judge the fit — and a good judgement looks across the whole range, not just the middle.

Read the shape. Plot the data first; the outline usually rules whole families in or out at a glance.
Respect the ends. A logistic model must flatten toward a ceiling; an exponential must not turn negative; a population must not go below zero.
Check the extremes, not just the bulk. A curve can hug early points and still overshoot a known limit — that reveals the wrong family.
Interpret parameters in context. The carrying capacity, the start value, the period must all be sensible for the real situation.

Common mistakes examiners penalise

Reading $L$ as the start value — $L$ is the carrying capacity (the ceiling), read from the numerator. The initial value is $f(0)=\frac{L}{1+C}$ , found by putting the input to zero; the two are almost never equal.
Treating logistic growth as unlimited exponential — the whole point of a logistic model is the ceiling. Do not extend it past $y=L$ , and do not choose an exponential when the data flattens.
Evaluating a piecewise function with the wrong rule at a boundary — the endpoint belongs to the piece carrying $\le$ or $\ge$ . Check the inequality before substituting.
Solving a logistic equation without writing it down — on Paper 2 you may use the GDC intersect tool, but you must show the equation (e.g. $4000=\frac{5000}{1+9e^{-0.4t}}$ ) to earn the method mark; a bare answer risks losing it.
Choosing a model from the middle of the data only — a curve can fit early points and still overshoot a known ceiling. Judge the fit at the extremes too.
Dropping units or context — a logistic or piecewise answer is a population, a time or a cost. State it with units and check it against the model's limit.
Over-rounding mid-calculation — carry the GDC's full figures and round only the final answer, to 3 significant figures unless told otherwise.

Model answer — marked the way our engine marks it

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and an accuracy mark depends on the method mark it follows. Follow-through (FT) means a value that is wrong earlier need not cost you the marks that depend on it, provided the later step is done correctly on your own figure. The engine also ignores subsequent working (ISW) once a correct answer appears, and accepts any equivalent form and any correctly-rounded value. But that protection only exists if the method is on the page. Study how each mark below is earned by a specific line — and how using the GDC is expected, not penalised, as long as you show the equation you solved.

Where this leads

Logistic and piecewise models are the modelling core of AI, and the habits you build here carry through the course. Reading a carrying capacity, setting a model equal to a target and solving on the GDC is exactly the routine you will reuse for exponential and sinusoidal models, and later when you fit trends to bivariate data and judge them by their residuals. The discipline is always the same: choose the family that matches the shape and the story, read the parameters in context, show the equation you solve, and quote the answer with units against the model's limits. Master that and every 'modelling with functions' question — whatever family it names — becomes a variation on the method you have just practised.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A courier charges a cost $C$ , in dollars, to deliver a package of weight $w$ kg, given by\n\n $C(w)=\begin{cases} 10 & 0<w\le 2 \\ 10+4(w-2) & w>2 \end{cases}$ \n\n(a) Find the cost of delivering a 1.5 kg package.\n(b) Find the cost of delivering a 5 kg package.\n(c) A customer was charged $24. Find the weight of their package. [5]

1
(a) Here $w=1.5$ . Since $0<1.5\le 2$ , the first rule applies. [M1: correct interval]\n $C(1.5)=10$ , so the cost is $10. [A1]\n\n**(b)** Here $w=5$ . Since $5>2$ , the second rule applies. [M1: correct interval]\n $C(5)=10+4(5-2)=10+12=22$ , so the cost is $22. [A1]\n\n**(c)** A charge of $24 exceeds $10, so the package is heavier than 2 kg and the second rule applies. Solve\n $24=10+4(w-2)\;\Rightarrow\;14=4(w-2)\;\Rightarrow\;w-2=3.5\;\Rightarrow\;w=5.5.$ \nThe package weighed 5.5 kg. [A1]\n\nThe whole skill is choosing the interval first; once the right rule is selected the arithmetic is routine.

✓ FULLY WORKED

EXAMPLE 2

The number of students who have heard a rumour, $N$ , after $t$ days is modelled by $N(t)=\frac{800}{1+79e^{-0.9t}}$ .\n\n(a) How many students had heard it at the start ( $t=0$ )?\n(b) State the maximum number who will ever hear it according to the model.\n(c) Find the number who have heard it after 2 days.\n(d) Use your GDC to find how long it takes for 400 students to have heard it. [6]

1
(a) At $t=0$ : $N(0)=\frac{800}{1+79e^{0}}=\frac{800}{1+79}=\frac{800}{80}=10$ . [M1: substitute $t=0$ ] [A1]\nSo 10 students started the rumour.\n\n**(b)** The maximum is the carrying capacity, the number on top of the fraction: $L=800$ students. [A1]\n\n**(c)** At $t=2$ : $N(2)=\frac{800}{1+79e^{-0.9(2)}}=\frac{800}{1+79e^{-1.8}}$ . [M1: substitute $t=2$ ]\nBy GDC, $N(2)=61.1\ldots$ , so about 61 students. [A1]\n\n**(d)** Solve $400=\frac{800}{1+79e^{-0.9t}}$ . Graph $y_1=\frac{800}{1+79e^{-0.9t}}$ and $y_2=400$ and use the intersect tool. [M1: correct equation set up]\nThe intersection is at $t=4.85$ days (3 s.f.). This is when the rumour reaches half its carrying capacity — the moment of fastest spread. [A1]

✓ FULLY WORKED

EXAMPLE 3

A biologist records the number of fish, $P$ , in a newly stocked lake over several months:\n\n| Month, $t$ | 0 | 2 | 4 | 6 | 8 | 10 |\n|---|---|---|---|---|---|---|\n| Fish, $P$ | 40 | 120 | 340 | 700 | 900 | 960 |\n\nThe lake can support at most about 1000 fish.\n\n(a) Explain why a logistic model is more appropriate than an exponential model.\n(b) The data is modelled by $P(t)=\frac{1000}{1+24e^{-0.6t}}$ . Verify the fit at $t=4$ .\n(c) Comment on the fit at $t=10$ . [5]

1
(a) The counts rise quickly at first but the growth slows as $P$ nears about 1000 — the lake's ceiling. An exponential model would keep growing without limit and would soon exceed what the lake can support, so it is unsuitable. A logistic model levels off at a carrying capacity, matching the biology. [A1: growth slows toward a ceiling] [A1: exponential has no limit / would overshoot]\n\n**(b)** At $t=4$ : $P(4)=\frac{1000}{1+24e^{-0.6(4)}}=\frac{1000}{1+24e^{-2.4}}$ . [M1: substitute $t=4$ ]\nBy GDC, $P(4)=316$ (3 s.f.), close to the recorded 340, so the model fits well here. [A1]\n\n**(c)** At $t=10$ : $P(10)=\frac{1000}{1+24e^{-6}}=985$ (3 s.f.), against a recorded 960. The model sits just above the data and is approaching its carrying capacity $L=1000$ , so the fit remains good and behaves sensibly at the ceiling. [A1]\n\nThe model is judged by comparing it with the data across the range and confirming it respects the known limit of 1000.

✓ FULLY WORKED

EXAMPLE 4

A population is modelled by $P(t)=\frac{5000}{1+9e^{-0.4t}}$ , where $t$ is measured in years. State the carrying capacity, find the population at $t=0$ , and find when the population reaches 4000. [5]

1
Model answer — full working.\n\nCarrying capacity. The limiting value is the number on top of the fraction:\n $L=5000.$ The population approaches 5000 but never exceeds it.\n\nPopulation at $t=0$ . Substitute $t=0$ , using $e^{0}=1$ :\n $P(0)=\frac{5000}{1+9e^{0}}=\frac{5000}{1+9}=\frac{5000}{10}=500.$ \n\nTime to reach 4000. Set $P(t)=4000$ :\n $4000=\frac{5000}{1+9e^{-0.4t}}.$ \nGraph $y_1=\frac{5000}{1+9e^{-0.4t}}$ and $y_2=4000$ on the GDC and use the intersect tool (or solve the equation numerically):\n $t=8.96$ years (3 s.f.).\n\nInterpretation. The population starts at 500, climbs along an S-curve, and passes 4000 after about 8.96 years, on its way to the ceiling of 5000.\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — carrying capacity. Awarded for stating $L=5000$ , read correctly from the numerator. This is a stand-alone accuracy mark.\n- M1 — substitute $t=0$ . A method mark for putting $t=0$ into the model, i.e. writing $P(0)=\frac{5000}{1+9e^{0}}=\frac{5000}{1+9}$ . It is the substitution that is rewarded, so it survives an arithmetic slip in the division.\n- A1 — value 500. Awarded for $P(0)=500$ . This accuracy mark depends on the M1 above; FT applies, so a candidate who substitutes correctly but slips arithmetically still earns credit on their own figure if consistent.\n- M1 — set $P=4000$ and solve. A method mark for forming the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ and solving it (GDC intersect is expected on Paper 2). The engine checks that the target 4000 was equated to the model, not confused with the carrying capacity.\n- A1 — value 8.96. Awarded for $t=8.96$ years (3 s.f.). This is FT on the candidate's own equation: a student whose earlier value differed but who set up and solved their equation correctly keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the carrying capacity written as $5000$ or 'the model levels off at 5000', accepts $P(0)$ as $\frac{5000}{10}$ or $500$ , and accepts the time as $8.96$ , $8.959$ or a correctly-rounded equivalent. Once a correct final value appears, ISW means later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the substitution and the equation are written on the page. A student who writes just '8.96' with no equation risks the final method mark; a student who shows $L$ , the $t=0$ substitution, and the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ keeps the method regardless of a slip in the final number.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAModelling with logistic, piecewise and other functions

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

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QUESTION · 1 / 12

Logistic model

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ANSWER

$f(x)=\frac{L}{1+Ce^{-kx}}$ , with $L, C, k$ positive constants. It models growth that is fast at first and then slows to level off — an S-shaped (sigmoid) curve — rather than the unlimited growth of a plain exponential.

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Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
One rule per input. The intervals do not overlap, so every input is governed by exactly one sub-rule.
✓
Watch the boundary. At an endpoint, use the rule whose interval includes it — the one written with $\le$ or $\ge$ , not $<$ or $>$ .
✓
Graphing convention. Draw an open circle where an endpoint is excluded and a filled circle where it is included.
✓
Continuity is not automatic. The pieces meet smoothly only if both rules give the same value at the shared boundary.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 2 modelling question marked: read the carrying capacity, evaluate the model and solve for the time with full working

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

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Frequently asked

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Checkpoint

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Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 2 modelling question marked: read the carrying capacity, evaluate the model and solve for the time with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

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Modelling with logistic, piecewise and other functions

BY THE END, YOU CAN…

1Interpret the logistic model $f(x)=\frac{L}{1+Ce^{-kx}}$ : identify the carrying capacity $L$ , describe the S-shaped behaviour, and read the initial value and limit.
2Evaluate a logistic model at a given input and use the GDC to solve for the input that produces a target output.
3Evaluate and interpret a piecewise-defined function, selecting the correct sub-rule for the interval and reading the inequality signs precisely.
4Choose an appropriate model (linear, quadratic, exponential, logistic, piecewise, sinusoidal) for a data set and evaluate its fit.
5Use a GDC to graph these models, read key features, and solve equations, presenting working so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Beyond Straight Lines: Growth With a Ceiling, and Rules That Switch

THINK OF IT LIKE…

1
Identify the model. S-shaped growth toward a ceiling means logistic; different rules on different intervals means piecewise; a repeating cycle means sinusoidal. Let the shape of the data or the wording of the problem decide.
2
Read the parameters. For a logistic model the carrying capacity $L$ is the number on top of the fraction; the initial value is found by putting the input to zero. For a piecewise model, note which interval each rule owns and whether the endpoints are included.
3
Use the GDC. Graph the model, then use 'value' to evaluate it and 'intersect' to solve for the input that gives a target output. On Paper 2 the calculator does the arithmetic; you supply the method.
4
Interpret in context. Every final answer is a quantity in the real situation — a population, a time, a cost — so state it with units and check it is sensible against the model's limit.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$f(x)=\frac{L}{1+Ce^{-kx}},\qquad L,\,C,\,k>0.$

Full topic notes

Formal explanation with the rigour you need for the exam.

Piecewise functions: different rules for different intervals

One rule per input. The intervals do not overlap, so every input is governed by exactly one sub-rule.
Watch the boundary. At an endpoint, use the rule whose interval includes it — the one written with $\le$ or $\ge$ , not $<$ or $>$ .
Graphing convention. Draw an open circle where an endpoint is excluded and a filled circle where it is included.
Continuity is not automatic. The pieces meet smoothly only if both rules give the same value at the shared boundary.

Logistic functions: growth with a ceiling

Carrying capacity $L$ : the horizontal asymptote $y=L$ ; the model approaches it but never reaches or exceeds it.
Initial value: at $x=0$ , $f(0)=\frac{L}{1+C}$ , because $e^{0}=1$ — this is the start size, not $L$ .
Shape: an increasing S-curve between the asymptotes $y=0$ and $y=L$ .
Fastest growth: at the point of inflection, which sits at half the carrying capacity, $y=\frac{L}{2}$ .

Choosing an appropriate model and evaluating its fit

Read the shape. Plot the data first; the outline usually rules whole families in or out at a glance.
Respect the ends. A logistic model must flatten toward a ceiling; an exponential must not turn negative; a population must not go below zero.
Check the extremes, not just the bulk. A curve can hug early points and still overshoot a known limit — that reveals the wrong family.
Interpret parameters in context. The carrying capacity, the start value, the period must all be sensible for the real situation.

Common mistakes examiners penalise

Reading $L$ as the start value — $L$ is the carrying capacity (the ceiling), read from the numerator. The initial value is $f(0)=\frac{L}{1+C}$ , found by putting the input to zero; the two are almost never equal.
Treating logistic growth as unlimited exponential — the whole point of a logistic model is the ceiling. Do not extend it past $y=L$ , and do not choose an exponential when the data flattens.
Evaluating a piecewise function with the wrong rule at a boundary — the endpoint belongs to the piece carrying $\le$ or $\ge$ . Check the inequality before substituting.
Solving a logistic equation without writing it down — on Paper 2 you may use the GDC intersect tool, but you must show the equation (e.g. $4000=\frac{5000}{1+9e^{-0.4t}}$ ) to earn the method mark; a bare answer risks losing it.
Choosing a model from the middle of the data only — a curve can fit early points and still overshoot a known ceiling. Judge the fit at the extremes too.
Dropping units or context — a logistic or piecewise answer is a population, a time or a cost. State it with units and check it against the model's limit.
Over-rounding mid-calculation — carry the GDC's full figures and round only the final answer, to 3 significant figures unless told otherwise.

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

1
(a) Here $w=1.5$ . Since $0<1.5\le 2$ , the first rule applies. [M1: correct interval]\n $C(1.5)=10$ , so the cost is $10. [A1]\n\n**(b)** Here $w=5$ . Since $5>2$ , the second rule applies. [M1: correct interval]\n $C(5)=10+4(5-2)=10+12=22$ , so the cost is $22. [A1]\n\n**(c)** A charge of $24 exceeds $10, so the package is heavier than 2 kg and the second rule applies. Solve\n $24=10+4(w-2)\;\Rightarrow\;14=4(w-2)\;\Rightarrow\;w-2=3.5\;\Rightarrow\;w=5.5.$ \nThe package weighed 5.5 kg. [A1]\n\nThe whole skill is choosing the interval first; once the right rule is selected the arithmetic is routine.

✓ FULLY WORKED

EXAMPLE 2

1
(a) At $t=0$ : $N(0)=\frac{800}{1+79e^{0}}=\frac{800}{1+79}=\frac{800}{80}=10$ . [M1: substitute $t=0$ ] [A1]\nSo 10 students started the rumour.\n\n**(b)** The maximum is the carrying capacity, the number on top of the fraction: $L=800$ students. [A1]\n\n**(c)** At $t=2$ : $N(2)=\frac{800}{1+79e^{-0.9(2)}}=\frac{800}{1+79e^{-1.8}}$ . [M1: substitute $t=2$ ]\nBy GDC, $N(2)=61.1\ldots$ , so about 61 students. [A1]\n\n**(d)** Solve $400=\frac{800}{1+79e^{-0.9t}}$ . Graph $y_1=\frac{800}{1+79e^{-0.9t}}$ and $y_2=400$ and use the intersect tool. [M1: correct equation set up]\nThe intersection is at $t=4.85$ days (3 s.f.). This is when the rumour reaches half its carrying capacity — the moment of fastest spread. [A1]

✓ FULLY WORKED

EXAMPLE 3

1
(a) The counts rise quickly at first but the growth slows as $P$ nears about 1000 — the lake's ceiling. An exponential model would keep growing without limit and would soon exceed what the lake can support, so it is unsuitable. A logistic model levels off at a carrying capacity, matching the biology. [A1: growth slows toward a ceiling] [A1: exponential has no limit / would overshoot]\n\n**(b)** At $t=4$ : $P(4)=\frac{1000}{1+24e^{-0.6(4)}}=\frac{1000}{1+24e^{-2.4}}$ . [M1: substitute $t=4$ ]\nBy GDC, $P(4)=316$ (3 s.f.), close to the recorded 340, so the model fits well here. [A1]\n\n**(c)** At $t=10$ : $P(10)=\frac{1000}{1+24e^{-6}}=985$ (3 s.f.), against a recorded 960. The model sits just above the data and is approaching its carrying capacity $L=1000$ , so the fit remains good and behaves sensibly at the ceiling. [A1]\n\nThe model is judged by comparing it with the data across the range and confirming it respects the known limit of 1000.

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EXAMPLE 4

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Model answer — full working.\n\nCarrying capacity. The limiting value is the number on top of the fraction:\n $L=5000.$ The population approaches 5000 but never exceeds it.\n\nPopulation at $t=0$ . Substitute $t=0$ , using $e^{0}=1$ :\n $P(0)=\frac{5000}{1+9e^{0}}=\frac{5000}{1+9}=\frac{5000}{10}=500.$ \n\nTime to reach 4000. Set $P(t)=4000$ :\n $4000=\frac{5000}{1+9e^{-0.4t}}.$ \nGraph $y_1=\frac{5000}{1+9e^{-0.4t}}$ and $y_2=4000$ on the GDC and use the intersect tool (or solve the equation numerically):\n $t=8.96$ years (3 s.f.).\n\nInterpretation. The population starts at 500, climbs along an S-curve, and passes 4000 after about 8.96 years, on its way to the ceiling of 5000.\n\n---\nHow our marking engine awards the 5 marks:\n\n- A1 — carrying capacity. Awarded for stating $L=5000$ , read correctly from the numerator. This is a stand-alone accuracy mark.\n- M1 — substitute $t=0$ . A method mark for putting $t=0$ into the model, i.e. writing $P(0)=\frac{5000}{1+9e^{0}}=\frac{5000}{1+9}$ . It is the substitution that is rewarded, so it survives an arithmetic slip in the division.\n- A1 — value 500. Awarded for $P(0)=500$ . This accuracy mark depends on the M1 above; FT applies, so a candidate who substitutes correctly but slips arithmetically still earns credit on their own figure if consistent.\n- M1 — set $P=4000$ and solve. A method mark for forming the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ and solving it (GDC intersect is expected on Paper 2). The engine checks that the target 4000 was equated to the model, not confused with the carrying capacity.\n- A1 — value 8.96. Awarded for $t=8.96$ years (3 s.f.). This is FT on the candidate's own equation: a student whose earlier value differed but who set up and solved their equation correctly keeps this mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts the carrying capacity written as $5000$ or 'the model levels off at 5000', accepts $P(0)$ as $\frac{5000}{10}$ or $500$ , and accepts the time as $8.96$ , $8.959$ or a correctly-rounded equivalent. Once a correct final value appears, ISW means later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through — but only if the substitution and the equation are written on the page. A student who writes just '8.96' with no equation risks the final method mark; a student who shows $L$ , the $t=0$ substitution, and the equation $4000=\frac{5000}{1+9e^{-0.4t}}$ keeps the method regardless of a slip in the final number.

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MAIN IDEAModelling with logistic, piecewise and other functions

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QUESTION · 1 / 12

Logistic model

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One rule per input. The intervals do not overlap, so every input is governed by exactly one sub-rule.
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Watch the boundary. At an endpoint, use the rule whose interval includes it — the one written with $\le$ or $\ge$ , not $<$ or $>$ .
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Graphing convention. Draw an open circle where an endpoint is excluded and a filled circle where it is included.
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Continuity is not automatic. The pieces meet smoothly only if both rules give the same value at the shared boundary.

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Revise Modelling with logistic, piecewise and other functions for IB Mathematics: Applications and Interpretation (2.6)

In simple terms

Explore the concept

Key formulas

Full topic notes

Piecewise functions: different rules for different intervals

Logistic functions: growth with a ceiling

Choosing an appropriate model and evaluating its fit

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

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Get a Paper 2 modelling question marked: read the carrying capacity, evaluate the model and solve for the time with full working

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Frequently asked

Checkpoint

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In simple terms

Explore the concept

Key formulas

Full topic notes

Piecewise functions: different rules for different intervals

Logistic functions: growth with a ceiling

Choosing an appropriate model and evaluating its fit

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 modelling question marked: read the carrying capacity, evaluate the model and solve for the time with full working

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Frequently asked

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