Functions are how IB Mathematics Applications & Interpretation HL turns a real situation into mathematics you can compute with. AI treats a function less as abstract algebra and more as a model: a rule that takes an input — a time, a price, a temperature — and returns exactly one output you can graph, invert, evaluate or optimise. Because a calculator is always allowed, the marks live in choosing the right model and reading its features correctly — the intercepts, the vertex, the asymptote — not in grinding algebra by hand.
This cheatsheet condenses the whole of SL 2.1–2.6 and AHL 2.7 — function notation and domain/range, composite and inverse functions, the key AI models (linear, quadratic, exponential, cubic and piecewise) and how to pull every graph feature off the GDC — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Function notation & evaluating $f(x)$ SL 2.2
A function is a rule that gives each input exactly one output; evaluating it just means substituting.
Function:$f:x\mapsto f(x)$ assigns to each input exactly one output
Evaluate:to find $f(a)$, replace every $x$ in the rule by $a$, then compute
Cubic model:$f(x)=ax^3+bx^2+cx+d$ — substitute term by term, watching the sign of each
TrickOn the GDC, type the rule into $Y_1$ and read $f(a)$ straight from a table or the value tool — faster and safer than hand arithmetic for a cubic or an exponential.
Trap"One input, one output" is the whole definition. A rule that returns two values for one $x$ — like $y=\pm\sqrt{x}$ — is not a function, so it must pass the vertical-line test.
§2 — Domain & range SL 2.2
The domain is the set of allowed inputs; the range is the set of outputs the rule can actually produce.
Domain:the inputs $x$ the rule permits — exclude anything that divides by $0$ or square-roots a negative, and respect context (e.g. time $t\ge0$)
Range:the outputs $f(x)$ produced as $x$ runs over the domain
On $a\le x\le b$:for a linear (monotonic) model the extreme outputs sit at the ends of the interval
TrickFor a decreasing model the largest input gives the smallest output — so the minimum of the range is at the right-hand end $x=b$, and the maximum at the left end $x=a$.
TrapContext shrinks the domain. If $x$ counts people or measures time, negative and fractional values are excluded — and the range you quote must match that restricted domain, not the whole real line.
§3 — Composite functions $(f\circ g)$ AHL 2.7
Composition chains two functions, feeding the output of one straight into the other.
Composite:$(f\circ g)(x)=f\big(g(x)\big)$ — put $x$ into the inner function $g$ first, then its output into $f$
Order matters:in general $(f\circ g)(x)\ne(g\circ f)(x)$
TrickFor a single value, work inside-out: compute $g(x_0)$ to one number first, then substitute that into $f$. You never need to build the full composite formula.
Trap$f\circ g$ means "$f$ of $g$", so $g$ acts first. Reading it left-to-right and applying $f$ first is the classic slip — and because order matters, it usually gives a different answer.
§4 — Inverse functions $f^{-1}$ AHL 2.7
The inverse undoes a function, and its graph is the reflection of the original in the line $y=x$.
Reverses $f$:$f(a)=b\iff f^{-1}(b)=a$, and $f^{-1}\big(f(x)\big)=x$
Linear inverse:if $f(x)=mx+c$ then $f^{-1}(x)=\dfrac{x-c}{m}$ — undo the operations in reverse: subtract $c$, then divide by $m$
Graphically:$y=f^{-1}(x)$ is the mirror image of $y=f(x)$ in the line $y=x$
To find an inverse, set $y=f(x)$, swap $x$ and $y$, then solve for $y$.
The inverse f−1 is the mirror image of f in the line y = x — shown here for f(x) = x2 (restricted to x ≥ 0) and its inverse f−1(x) = √x. Reflecting swaps every (x, y) to (y, x).
Trap$f^{-1}$ exists only when $f$ is one-to-one. A many-to-one rule such as $f(x)=x^2$ must first have its domain restricted to a single branch (here $x\ge0$) before it can be inverted. And $f^{-1}(x)$ is the inverse, never the reciprocal $\tfrac{1}{f(x)}$.
NoteAHL synthesis with differentiation. Reflecting a curve in $y=x$ turns each slope into its reciprocal: if $f(a)=b$ then $\left(f^{-1}\right)'(b)=\dfrac{1}{f'(a)}$, where $a=f^{-1}(b)$. Find $a$ first, differentiate $f$ (not $f^{-1}$), then take the reciprocal — on a decreasing branch $f'(a)<0$, so the answer is negative.
§5 — Linear models SL 2.1
The straight line is the simplest AI model: a constant rate of change on top of a starting value.
Linear model:$f(x)=mx+c$ — gradient $m$ is the constant rate of change, intercept $c$ is the value when $x=0$
Gradient:$m=\dfrac{y_2-y_1}{x_2-x_1}$ — "rise over run" between any two points on the line
TrapWhen you evaluate, the answer is $mx$ plus $c$, not just $mx$. Dropping the fixed term $c$ — the start-up cost, the initial population — is the single most common careless slip on linear models.
NoteRead the parameters in context: $m$ is a "per unit" rate (dollars per item, °C per minute) and $c$ is the starting amount at $x=0$. Stating those interpretations with correct units earns the modelling marks.
§6 — Quadratic models: vertex, axis & max/min SL 2.5
A quadratic model is a parabola, and its single turning point is where the maximum or minimum lives.
Quadratic model:$f(x)=ax^2+bx+c$ — opens up ($a>0$, has a minimum) or down ($a<0$, has a maximum)
Axis / vertex:the turning point is at $x=-\dfrac{b}{2a}$; the max/min value is $f\!\left(-\dfrac{b}{2a}\right)$
$y$-intercept:$f(0)=c$ — the starting value of the model
A projectile height $h(t)=h_0+vt-ut^2$ is this model with $a=-u,\ b=v$: it peaks at $t=\dfrac{v}{2u}$, and $h_0$ is the launch height.
TrickFind where the peak is first — $t^\ast=\dfrac{v}{2u}$ (or $-\tfrac{b}{2a}$) — then substitute back to get the maximum value. On the GDC, graph the model and use the maximum/minimum tool to confirm both coordinates.
TrapThe vertex has two coordinates. $x=-\tfrac{b}{2a}$ is where the max/min occurs (the time, the price); the maximum itself is the $y$-value $f(-\tfrac{b}{2a})$. Quoting the $x$ when the question wants the $y$ (or vice versa) throws away the answer mark.
§7 — Exponential models: growth, decay & the horizontal asymptote SL 2.5
Exponential models change by a constant ratio each step and, when shifted, level off at a horizontal asymptote.
Exponential model:$f(x)=k\,a^{x}+c$ — growth if $a>1$, decay if $0 < a < 1$; horizontal asymptote $y=c$
Doubling / half-life:$P(t)=P_0\,2^{t/d}$ doubles every $d$ time units (use base $\tfrac12$ for a half-life)
Newton cooling:$T(t)=T_a+(T_0-T_a)\,b^{t}$ — cools towards room temperature $T_a$, the horizontal asymptote
TrapNever drop the added constant. As $x\to\infty$ the $k\,a^{x}$ (or $b^{t}$) term dies away and the model levels off at $y=c$ (or $T_a$) — not at $0$. Reading the asymptote as zero is the signature exponential-model error.
TrickTo find when a target is reached, divide the target by the initial value. If that ratio is a whole power of the base — $2^{t/d}=2^{n}\Rightarrow t=nd$ — match exponents exactly; otherwise take $\log$ on the GDC. For a "first exceeds" question, round the answer up.
§8 — Piecewise functions SL 2.5
A piecewise model uses different rules on different intervals — common for tiered tariffs and step charges.
Piecewise model:$f(x)=\begin{cases}\text{rule 1}, & x < k\\ \text{rule 2}, & x\ge k\end{cases}$ — first decide which interval contains $x$, then apply that piece
Tiered tariff:charge the first block at the base rate; only the excess above the threshold is charged at the next rate
TrapOn a tiered bill do not apply one rate to the whole amount. Everything up to the threshold is charged at the first rate; only the part above it moves to the higher rate.
NoteWatch the endpoints: the inequalities ($<$ versus $\le$) decide which piece owns the boundary value $x=k$. Substitute $k$ into the piece whose interval actually includes it.
§9 — Reading key features off the GDC SL 2.3 · 2.4
AI rewards pulling features straight off the graph — set a sensible window first, then read.
| Graph feature | How to read it off the GDC |
|---|
| $y$-intercept | Evaluate $f(0)$ — where the curve meets the $y$-axis (the model's start value) |
| $x$-intercepts / roots | Graph $y=f(x)$ and use the zero tool; each root solves $f(x)=0$ |
| Maximum / minimum | Use the maximum/minimum tool — read the turning point's $x$ and $y$ |
| Horizontal asymptote | The value the curve levels off to as $x\to\pm\infty$ (the $+c$ in the model) |
| Intersection of two models | Plot both and use the intersect tool, or solve $f(x)=g(x)$ |
TrickAlways set a sensible window and sketch before trusting a value — an autoscaled screen can hide a second root, a turning point, or the asymptote entirely.
NoteQuote GDC answers to 3 significant figures (unless told otherwise) and with the correct units; keep full accuracy inside the calculator until the very last line.
§10 — Exam attack plan All sections
| Question cue | What to do | Watch for |
|---|
| "Find $f(a)$" | Substitute $a$ for every $x$ (or use the GDC table/value) | Don't drop the constant term |
| "State the domain / range" | Domain = allowed inputs; range = outputs (read off the graph) | Context limits ($t\ge0$); match the range to the domain |
| "$(f\circ g)(x)$" or "$f(g(2))$" | Inner function first: apply $g$, then $f$ | $f\circ g\ne g\circ f$; $g$ acts first |
| "Find $f^{-1}$" | Set $y=f(x)$, swap $x\leftrightarrow y$, solve; reflect in $y=x$ | Needs one-to-one; restrict the domain to a branch |
| "Maximum / minimum value" (quadratic) | Vertex $x=-\tfrac{b}{2a}$, then substitute (or GDC max/min) | Report the $y$-value, not just where it occurs |
| "Long-run / levels off / asymptote" | Read the constant $c$ (or $T_a$) in the model | It tends to $c$, not $0$ |
| "How long until…" (exponential) | Match exponents, or take $\log$ on the GDC | "First exceeds" $\Rightarrow$ round up |
| "Tiered charge / different rules" | Find the interval, then apply that piece | Charge only the excess in the higher tier |
Worked Example — AI HL-Style Exponential Cooling
Question (AI HL Paper 2 style — 6 marks)
A mug of coffee cools in a room. Its temperature after $t$ minutes is modelled by $$T(t)=21+74(0.93)^{t},$$ where $T$ is measured in °C.
(a) Write down the initial temperature of the coffee.
(b) Find the temperature after 10 minutes.
(c) State the temperature the coffee approaches in the long run, and explain why.
(d) Find the time, to the nearest tenth of a minute, for the coffee to cool to 40 °C.
Solution
- (a) $T(0)=21+74(0.93)^{0}=21+74=95$ °C. (A1) (the initial value is the constant plus the coefficient, not just $74$.)
- (b) $T(10)=21+74(0.93)^{10}=21+74(0.4840)=56.8$ °C (3 s.f.). (M1)(A1)
- (c) As $t\to\infty$, $(0.93)^{t}\to0$, so $T\to21$ °C. This is the horizontal asymptote — room temperature — and the coffee never cools below it. (A1)
- (d) Solve $21+74(0.93)^{t}=40\Rightarrow 74(0.93)^{t}=19\Rightarrow(0.93)^{t}=\dfrac{19}{74}=0.2568$. (M1) Taking logs (or the GDC solver), $t=\dfrac{\ln 0.2568}{\ln 0.93}=18.7$ minutes (3 s.f.). (A1)
Examiner's note: keep the $+21$ throughout — the long-run temperature is the room's $21$ °C, not $0$, and that constant is exactly the horizontal asymptote. Hold full calculator accuracy for $(0.93)^{10}$ and only round the final answers to 3 s.f., and always attach the correct units (°C, minutes). Adding $R$-style terms or dropping the constant is the classic way to lose the reading marks here.
Common Student Questions
What is the difference between $f\circ g$ and $g\circ f$?
They are the two orders of composing the same pair of functions. $f\circ g$ means "$f$ of $g$", so $g$ acts first: $(f\circ g)(x)=f(g(x))$. Swap them and $f$ acts first instead. Because the inner function runs first, the two usually give different results — so always read which function is on the inside and work outward from there.
How do I find an inverse function, and when does one exist?
Write $y=f(x)$, swap $x$ and $y$, and solve for $y$; the result is $f^{-1}(x)$, and its graph is the reflection of $f$ in the line $y=x$. An inverse only exists when $f$ is one-to-one (each output comes from just one input). A many-to-one rule like $f(x)=x^2$ has to be restricted to a single branch — say $x\ge0$ — before it can be inverted.
My exponential model "levels off" — how do I find that value?
It is the added constant in the model. For $f(x)=k\,a^{x}+c$ (with $0 < a < 1$) the $k\,a^{x}$ term shrinks to $0$ as $x$ grows, so the curve approaches the horizontal asymptote $y=c$. In a cooling model $T=T_a+(T_0-T_a)b^{t}$ that constant is the room temperature $T_a$. The one thing it is never is $0$ — writing "levels off at $0$" is the trap.
For a quadratic, does the question want the vertex's $x$ or its $y$?
Read it carefully — they mean different things. The vertex $x$-coordinate, $x=-\tfrac{b}{2a}$, tells you where the maximum or minimum happens (the time, the price, the quantity). The $y$-coordinate is the maximum or minimum value itself. "Find the maximum height" wants the $y$; "at what time" wants the $x$. The GDC's maximum/minimum tool hands you both at once.
Do I have to solve functions by algebra, or can I just use the GDC?
In AI the GDC is expected. Graph the model and use the built-in zero, maximum/minimum and intersect tools to read intercepts, turning points and where two models meet, or use the equation solver for "when does it reach…" questions. Show enough method that the marker can follow you, give answers to 3 significant figures with units, and keep full accuracy in the calculator until the final line.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
- Notes PDF — every definition and model in full, from function notation and domain/range to composite and inverse functions, with worked asymptote and vertex examples.
- Tutorial booklet — 25+ AI HL-style questions from evaluating f(x) to full modelling with quadratic, exponential and piecewise functions.
- Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations and the exact GDC steps for every graph feature.
- Predicted-paper questions — the exact function and modelling question types most likely in the next AI HL session.
Get the printable PDF cheatsheet — free
The same cheatsheet on one A4 page — formulas, methods and traps. No sign-in needed.
