Functions (AI HL) — IB Math AI HL AI HL Study Guide

For: IB Math AI HL candidates sitting IB Math: Applications & Interpretation HL.

Covers: standard linear/quadratic/exponential/logarithmic/sinusoidal models, piecewise functions, composite and inverse functions in applied contexts, and optimisation for real-world function modelling.

You should already know: IGCSE / pre-DP math; comfort with applied problems and tech.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Math AI HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Functions (AI HL)?

Functions are mathematical relationships that map every input value from a defined domain to exactly one output value in a corresponding range. For IB Math AI HL, functions are the primary tool for modelling real-world phenomena, from population growth and corporate profit to periodic tide patterns and pharmaceutical decay. You will see them tested across all three papers, often in 10+ mark applied questions that require you to select, adjust, and interpret models for given contexts. Standard notation uses $f(x)$ to denote the output of function $f$ for input $x$, and functions are often referred to as mappings or mathematical models in exam wording.

2. Linear, quadratic, exp, log, sinusoidal models

These five core function types make up 90% of the modelling questions you will encounter on the exam, each suited to a specific type of real-world pattern:

• Linear model: $f(x)=mx+c$, where $m$ is the constant rate of change and $c$ is the initial value. Used for scenarios with fixed per-unit change, e.g., hourly wage, constant speed travel.
• Quadratic model: $f(x)=a{x}^2+bx+c$, with a single turning point (vertex) at $x=-\frac{b}{2a}$. Used for projectile motion, profit functions, and area optimisation, with a maximum if $a<0$ and minimum if $a>0$.
• Exponential model: $f(x)=A(1+r{)}^x$, where $A$ is the initial value and $r$ is the percentage growth/decay rate per unit $x$. Used for population growth, compound interest, and radioactive decay.
• Logarithmic model: $f(x)=A{\log }_b(x)+C$, the inverse of the exponential function. Used for scaling phenomena like pH levels, decibel ratings, and earthquake magnitude.
• Sinusoidal model: $f(x)=A\sin (B(x-C))+D$, where $A$ = amplitude, period = $\frac{2\pi }{|B|}$, $C$ = horizontal phase shift, and $D$ = vertical midline. Used for periodic patterns like tides, seasonal temperature, and circular motion.

Worked Example: A small business has 1200 customers at the start of 2023, growing at a fixed 4.5% annual rate. What year will the business reach 2500 customers? Use the exponential model $C(t)=1200(1.045{)}^t$, where $t$ = years after 2023. Set equal to 2500: $$1200(1.045{)}^t=2500⟹(1.045{)}^t=\frac{25}{12}⟹t=\frac{\ln \left(\frac{25}{12}\right)}{\ln (1.045)}\approx 17.2$$ The business will reach 2500 customers in 2023 + 17 = 2040.

3. Piecewise functions

Piecewise functions are defined by multiple separate sub-functions, each applying to a specific interval of the input domain. They are used for scenarios where the relationship between input and output changes at a threshold value, such as tiered pricing, overtime pay, or distance-time graphs with changes in speed. When evaluating piecewise functions, always first identify which interval your input falls into before selecting the correct sub-function to use.

Worked Example: A theme park ticket pricing scheme is as follows: children under 12 pay $15,adultsaged12to64pay$35, and seniors aged 65 or over pay $20.Writethepiecewisefunctionforticketcost$P(a)$where$a$ is age in years, and calculate the total cost for a family with two children aged 8 and 10, two adults aged 38 and 41, and one senior aged 72. The piecewise function is: $$P(a) = \begin{cases} 15 & 0 < a < 12 \ 35 & 12 \leq a < 65 \ 20 & a \geq 65 \end{cases}$$ Total cost = $2\ast 15+2\ast 35+1\ast 20=30+70+20=\$120$. Exam tip: Examiners often ask you to evaluate piecewise functions at interval boundaries (e.g., $a=12$ in this example) to test that you select the correct sub-function.

4. Composite and inverse functions in context

Composite functions

A composite function $(f\circ g)(x)=f(g(x))$ applies the inner function $g$ to the input first, then uses the output of $g$ as the input for the outer function $f$. Composite functions are used for multi-step conversions, e.g., converting currency first, then calculating import tax for an international purchase.

Inverse functions

An inverse function $f^{-1}(y)=x$ if and only if $f(x)=y$, and reverses the input and output of the original function. In context, inverse functions let you solve for the input required to get a specific target output, e.g., calculating how many units you can produce for a fixed budget given a cost function. Note that only one-to-one functions have valid inverses, so you will often need to restrict the domain of the original function to match the context (e.g., only positive time values for projectile motion).

Worked Example: Let $e(r)=0.92r$ convert British pounds to euros, and $t(e)=1.08e$ calculate the total cost of an item in euros including 8% import tax. (a) Find the composite function $t(e(r))$ and explain what it represents, (b) If you have a budget of £300, what is the maximum pre-tax price in euros you can afford? (a) Substitute $e(r)$ into $t(e)$: $t(e(r))=1.08\ast (0.92r)=0.9936r$. This function gives the total cost of an item in euros for a given price in pounds including import tax. (b) First, find the inverse of the cost function $C(e)=1.08e$: solve for $e$ to get $e=\frac{C}{1.08}$. Convert £300 to euros: $0.92\ast 300=276$ EUR. Max pre-tax price = $\frac{276}{1.08}\approx 255.56$ EUR.

5. Optimisation in modelling

Optimisation is the process of finding the maximum or minimum value of a function for a given real-world context, one of the most frequently tested high-mark skills in AI HL functions. For basic models:

1. For quadratic functions, the vertex is the global maximum or minimum.
2. For differentiable non-quadratic functions, find critical points by setting the first derivative equal to zero, then confirm if it is a max or min using the second derivative test.
3. For piecewise functions, always evaluate the function at all critical points and the boundaries of the domain, as the global max/min may lie at a threshold value. You are permitted to use your GDC to find turning points for complex functions on Paper 2 and 3, but you must show your working to demonstrate you selected the correct domain and interpreted the output correctly.

Worked Example: A manufacturer produces cardboard boxes with a square base, and the total surface area of each box is fixed at 600 cm². Find the maximum possible volume of the box. Let $x$ = side length of the square base, $h$ = height of the box. Surface area: $2x^2+4xh=600$. Rearrange to isolate $h$: $h=\frac{600-2x^2}{4x}=\frac{300-x^2}{2x}$. Volume function: $V(x)=x^{2}h=x^2\ast \frac{300-x^2}{2x}=\frac{x(300-x^2)}{2}=150x-0.5x^3$. Domain: $x>0$ and $300-x^2>0⟹0<x<\sqrt{300}\approx 17.32$ cm. Find critical points: $V^{\prime }(x)=150-1.5x^2=0⟹x^2=100⟹x=10$ cm. Second derivative test: $V^{\prime \prime }(x)=-3x$, so $V^{\prime \prime }(10)=-30<0$, so it is a maximum. Maximum volume: $V(10)=150\ast 10-0.5\ast (1000)=1500-500=1000$ cm³.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Using a linear model for exponential growth/decay problems. Why: Students confuse fixed per-unit change (linear) with fixed percentage change (exponential). Correct move: If the problem says "grows by 20 units per year" use linear, if "grows by 2% per year" use exponential.
• Wrong move: Forgetting to restrict the domain of inverse functions to make them one-to-one. Why: Students assume all functions have inverses without checking context. Correct move: For quadratic or sinusoidal functions, restrict the domain to the relevant interval (e.g., time from launch to landing for projectile motion) before calculating the inverse.
• **Wrong move: Applying the wrong sub-function when evaluating piecewise functions at interval boundaries. Why: Students rush and select the first interval they see. Correct move: Circle the interval that includes your input value first, e.g., for age 12 in the theme park ticket example, use the adult sub-function, not the child sub-function.
• Wrong move: Mixing up the order of composite functions, using $g(f(x))$ instead of $f(g(x))$. Why: Students confuse inner and outer functions. Correct move: Map the context to the order of operations: if you convert currency first then calculate tax, the currency conversion is the inner function.
• Wrong move: Forgetting to check domain boundary values when optimising. Why: Students only look for turning points, but global maxima/minima often lie at the edges of the valid domain. Correct move: List all critical points (turning points + domain boundaries) and evaluate the function at each to find the global max/min.

7. Practice Questions (IB Math AI HL Style)

Question 1

The height of water in a harbour over a 24-hour period follows a sinusoidal model. Low tide of 1.2 m occurs at 3 AM, and high tide of 5.8 m occurs at 9 AM. (a) Write a sinusoidal function $H(t)$ where $t$ is hours after midnight. (b) Find the height of the water at 1 PM. (c) A boat can only enter the harbour when the water height is at least 4 m. Find all time intervals in a 24-hour period when the boat can enter.

Solution

(a) Midline $D=\frac{1.2+5.8}{2}=3.5$ m, amplitude $A=5.8-3.5=2.3$ m, period = 12 hours, so $B=\frac{2\pi }{12}=\frac{\pi }{6}$. High tide at $t=9$, so use a cosine function with phase shift 9: $$H(t)=2.3\cos \left(\frac{\pi }{6}(t-9)\right)+3.5$$ (b) At $t=13$ (1 PM): $$H(13)=2.3\cos \left(\frac{\pi }{6}(4)\right)+3.5=2.3\cos \left(\frac{2\pi }{3}\right)+3.5=2.3\ast (-0.5)+3.5=2.35\text{ m}$$ (c) Solve $2.3\cos \left(\frac{\pi }{6}(t-9)\right)+3.5\ge 4$: $$\cos \left(\frac{\pi }{6}(t-9)\right)\ge \frac{0.5}{2.3}\approx 0.2174$$ $$\frac{\pi }{6}(t-9)\in [-1.351+2\pi k,1.351+2\pi k]$$ $$t\in [6.42+12k,11.58+12k]$$ The boat can enter between 6:25 AM and 11:35 AM, and 6:25 PM and 11:35 PM.

Question 2

A freelance writer charges the following rates for blog posts: first 500 words cost $40,eachadditionalwordupto2000wordscosts$0.06, any word over 2000 costs $0.04. (a) Write the piecewise function for cost $C(w)$ where $w$ is the number of words. (b) Calculate the cost of a 2800 word blog post. (c) Find the inverse function for the interval $500<w\le 2000$, and explain what it represents.

Solution

(a) $$C(w) = \begin{cases} 40 & 0 < w \leq 500 \ 40 + 0.06(w-500) & 500 < w \leq 2000 \ 40 + 0.061500 + 0.04(w-2000) & w > 2000 \end{cases}$$ Simplified: second line = $0.06w+10$, third line = $0.04w+50$ (b) $w=2800$ falls in the third interval: $0.042800 +50 = 112 +50 = $162$ (c) Solve $C=0.06w+10$ for $w$: $w=\frac{C-10}{0.06}$. This function gives the maximum number of words you can order for a budget between $40and$130 (the cost of a 2000 word post). For example, a budget of $100getsyou$\frac{90}{0.06} = 1500$ words.

Question 3

A café sells 120 muffins per day at $3each.Forevery$0.25 increase in price, they sell 8 fewer muffins per day. (a) Write the daily revenue function $R(p)$ where $p$ is the price of a muffin. (b) Find the price that maximises daily revenue, and the maximum revenue. (c) Find the range of prices that give a daily revenue of at least $370.

Solution

(a) Let $n$ = number of $0.25increases:$p = 3 + 0.25n$,quantitysold=$120 -8n$.Solvefor$n = \frac{p-3}{0.25} = 4p -12$.Substituteintoquantity:$q = 120 -8*(4p -12) = 120 -32p +96 = 216 -32p$.Revenue:$R(p) = pq = p(216 -32p) = -32p^2 +216p$. (b) Vertex at $p=-\frac{216}{2\ast (-32)}=3.375$. The optimal price is $\$3.38$ (rounded to nearest cent). Maximum revenue: $R(3.375)=-32\ast (3.375{)}^2+216\ast (3.375)=\$364.50$. (c) Solve $-32p^2+216p\ge 370$: $32p^2-216p+370\le 0$. Roots at $p=\frac{216\pm \sqrt{216^2-4\ast 32\ast 370}}{64}\approx 2.71$ and $4.04$. The price range is $\$2.71$ to $\$4.04$.

8. Quick Reference Cheatsheet

9. What's Next

Functions are the foundational building block for almost all remaining topics in IB Math AI HL: you will use them to model probability distributions, calculate derivatives and integrals for calculus applications, analyse time series data, and solve optimisation problems for operations research topics like linear programming. Mastery of function modelling is also required for your internal assessment, where you will apply these concepts to a real-world research question of your choice, so solidifying these skills now will save you significant time later in the course.

If you have any questions about function notation, model selection, or optimisation steps, you can ask Ollie for step-by-step explanations or extra practice problems at any time on the homepage. You can also move on to our study guides for differential calculus or statistical modelling next, as both rely heavily on the function concepts covered in this guide.