Function Concepts (AI HL)
A function is a rule that pairs every allowed input with exactly one output, and it's the language every model in this course is written in. This topic covers function notation, domain and range, building composite functions from simpler ones, and finding an inverse function - including why some functions need their domain restricted before an inverse exists at all.
What the syllabus says
This topic maps onto two points in the official IB Applications & Interpretation syllabus - one shared with SL, one added at HL.
| Code | Syllabus content |
|---|---|
| SL2.2 | Concept of a function, domain, range and graph. Function notation, for example \(f(x)\), \(v(t)\), \(C(n)\). The concept of a function as a mathematical model. Informal concept that an inverse function reverses or undoes the effect of a function. Inverse function as a reflection in the line \(y=x\), and the notation \(f^{-1}(x)\). |
| AHL2.7 | Composite functions in context: \((f\circ g)(x) = f(g(x))\). Inverse function \(f^{-1}\), including domain restriction. Finding an inverse function. |
SL2.2 is common to both AA and AI; AHL2.7 extends it to composite functions and formally finding an inverse.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a function?
A function is a rule that assigns exactly one output to every input in its domain. It's usually written \(f(x)\), and can represent a real-world quantity like cost, volume or population as a mathematical model.
e.g. \(f(x) = 2x+1\): the input \(x=3\) gives the single output \(f(3)=7\).
What is domain and range?
The domain is the complete set of input values a function accepts; the range is the complete set of output values it can produce. In a modelling context, both are usually restricted to values that make physical sense.
e.g. \(P(t)=4+2.5t\) for \(0\le t\le8\): domain \(0\le t\le8\), range \(4\le P\le24\).
What is a composite function?
A composite function applies one function to the output of another. \((f\circ g)(x) = f(g(x))\) means "do \(g\) first, then do \(f\) to the result" - the order matters, and it's not generally the same as \((g\circ f)(x)\).
e.g. \(f(x)=2x+3,\ g(x)=x^2-1\): \((f\circ g)(4) = f(15) = 33\).
What is an inverse function?
An inverse function \(f^{-1}\) undoes what \(f\) does - if \(f(a)=b\) then \(f^{-1}(b)=a\). Graphically, \(f^{-1}\) is the reflection of \(f\) in the line \(y=x\), and its domain is the range of \(f\).
e.g. \(f(x)=2x+3\): solving \(y=2x+3\) for \(x\) gives \(f^{-1}(x)=\dfrac{x-3}{2}\).
Why does an inverse need a restricted domain?
An inverse only exists where the original function is one-to-one - each output must come from exactly one input. A full parabola fails this (a horizontal line crosses it twice), so the domain is restricted to one side of the vertex before \(f^{-1}\) can be defined.
e.g. \(h(x)=(x-2)^2+1,\ x\ge2\): one-to-one, so \(h^{-1}(x)=2+\sqrt{x-1}\) exists.
Key formulas
There's no long list of formulas here - function concepts are mostly definitions and a couple of algebraic techniques. The tables below summarise the key relationships, and the explanations underneath cover the reasoning in more depth.
Formula reference
Composite and inverse function notation are given in the formula booklet's list of notation, but the working - substituting, rearranging, checking one-to-one - is prior knowledge and technique rather than a formula to look up.
| Notation / relationship | Used for | Booklet? |
|---|---|---|
| \((f\circ g)(x) = f(g(x))\) | Composite function notation | ✓ Yes |
| \(f^{-1}(f(x)) = x = f(f^{-1}(x))\) | Inverse function relationship | ✓ Yes |
| Swap \(x\) and \(y\), then solve for \(y\) | Method for finding \(f^{-1}(x)\) | Not in the formula booklet - technique, not a formula |
| Restrict the domain so \(f\) is one-to-one | Making an inverse exist | Not in the formula booklet - prior knowledge |
Function vs inverse function
A function and its inverse are mirror images of each other in every sense - this table lines up the key properties.
| Feature | \(f(x)\) | \(f^{-1}(x)\) |
|---|---|---|
| Relationship | \(f(a)=b\) | \(f^{-1}(b)=a\) |
| Domain | Domain of \(f\) | Range of \(f\) |
| Range | Range of \(f\) | Domain of \(f\) |
| Graph | Original curve | Reflection in \(y=x\) |
Working with functions
These skills are tested constantly in context - a model dressed up as a taxi fare or a water tank is still just a function.
Evaluating \(f(x)\)
Substitute the given input value for \(x\) (or \(t\), or whatever the variable is called) and simplify.
Finding domain and range
The domain comes from the context (what inputs make sense); the range comes from evaluating \(f\) at the domain's endpoints, or from the shape of the graph.
Interpreting \(f(0)\) or an endpoint value
In a model, \(f(0)\) is usually the starting value before anything has changed - always translate the number back into the real-world quantity it represents.
Composite and inverse functions
Composite functions build new relationships; inverse functions reverse a single one.
Building \((f\circ g)(x)\)
Substitute the entire expression for \(g(x)\) everywhere \(x\) appears in \(f(x)\), then simplify.
Finding \(f^{-1}(x)\)
Write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject again. Relabel as \(f^{-1}(x)\).
Checking one-to-one
A horizontal line should cross the graph at most once. If it crosses twice, the domain needs restricting before an inverse exists.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A parking garage charges \(P(t) = 4 + 2.5t\) dollars, where \(t\) is the number of hours parked, for \(0 \le t \le 8.\)
(a) Find \(P(3),\) the cost of parking for 3 hours.
(b) State the domain and range of \(P.\)
Worked solution
(a) \(P(3) = 4 + 2.5(3) = 4 + 7.5\) M1
\(= $11.50.\) A1
(b) Domain: \(0 \le t \le 8.\) A1
\(P(0)=4,\ P(8)\) M1
\(=24,\) so \(4 \le P \le 24.\) A1
Let \(f(x)=2x+3\) and \(g(x)=x^2-1.\)
(a) Find \((f\circ g)(x),\) and hence evaluate \((f\circ g)(4).\)
Let \(h(x)=(x-2)^2+1,\) for \(x\ge2.\)
(b) Explain why the domain restriction \(x\ge2\) is needed for \(h^{-1}\) to exist.
(c) Find \(h^{-1}(x).\)
Worked solution
(a) \((f\circ g)(x)=f(g(x))=2(x^2-1)+3\) M1
\(=2x^2+1.\) A1
\((f\circ g)(4)=33.\) A1
(b) Without the restriction, \(h\) is not one-to-one (a horizontal line would cross the parabola twice), so \(h^{-1}\) would not be a function; restricting to \(x\ge2\) makes \(h\) one-to-one. R1 R1
(c) \(y=(x-2)^2+1 \Rightarrow x-2=\sqrt{y-1}\) (taking the positive root since \(x\ge2\)). \(h^{-1}(x)\) M1
\(=2+\sqrt{x-1}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Composing in the wrong order. \((f\circ g)(x)\) means "\(g\) first, then \(f\)" - substituting \(f(x)\) into \(g\) instead gives \((g\circ f)(x)\), which is usually a different function entirely.
- Forgetting to restrict the domain when finding an inverse. If the original function isn't one-to-one over its stated domain, \(f^{-1}\) doesn't exist there - the restriction has to be stated as part of the answer, not left implicit.
- Giving the range as a copy of the domain. The range comes from the function's outputs, not its inputs - always evaluate \(f\) at the domain's endpoints (or find the vertex, for a quadratic) rather than assuming they match.
- Losing context when interpreting a value. \(P(0)=4\) means nothing on its own - the mark is for stating what it represents in the real-world situation, such as "the minimum parking charge, or drive-in fee."
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
The starting point for almost every graphing task - if you can't get the graph on screen, nothing else works.
- Use x as the variable - type it with the dedicated x key, not ALPHA + X.
- Press Y= and type the function next to Y1=. Use X,T,θ,n for x. Press GRAPH to draw it. Clear old functions by moving to them and pressing CLEAR.TI-84
- Open a Graphs page (press ctrl + I → Add Graphs, or press the Graphs app). Type the function in the entry bar at the bottom and press ENTER.Nspire
- Press MENU → Graph (or press the Graph icon). Press SHIFT → F3 (TYPE) to choose the graph type (Y= is the default). Type the function next to Y1 and press F6 (DRAW).Casio
- If the graph looks blank or wrong: check the window (see 'Set a good viewing window') and check the angle mode (degrees vs radians).
- To graph multiple functions, enter them as Y1, Y2, Y3 etc. - useful for finding intersections.
Tip: The most common reason a graph doesn't appear is a bad viewing window, not a mistake in the function. Try ZoomFit or ZStandard first.
Read off the exact y-value for a given x - useful for checking answers and for questions that ask for a specific coordinate.
- After graphing the function, you can evaluate it at any x-value.
- Method 1 - Table: 2nd → GRAPH (TABLE), scroll to the x you want. Method 2 - Trace: press TRACE, then type the x-value and press ENTER. Method 3 - Home screen: type Y1(value) e.g. Y1(3) using VARS → Y-VARS → 1:Function.TI-84
- On the graph, press TRACE and type the x-value, then ENTER. Or on a Calculator page type f1(3) to evaluate the stored function at x = 3.Nspire
- Press TRACE (F1) on the graph, then type the x-value and EXE. Or use the Table view (MENU → Table) to see multiple values.Casio
- For models: substitute the x-value into the regression equation stored in the calculator (see 'Use a fitted model to make predictions').
Tip: Trace gives an approximate value by cursor position - typing the x-value after pressing TRACE gives the exact value.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Function concepts questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between domain and range?
The domain is every input value a function accepts; the range is every output value it can produce. For a real-world model, both are usually restricted to whatever makes sense in context - you can't have negative time, for example.
How do I find a composite function?
To find \((f \circ g)(x)\), substitute the whole expression for \(g(x)\) in place of every \(x\) in \(f(x)\), then simplify. Read it from the inside out: \(g\) acts on \(x\) first, then \(f\) acts on the result.
Why does an inverse function sometimes need a restricted domain?
An inverse only exists if the original function is one-to-one - each output comes from exactly one input. Many functions, like a parabola, fail this test over their full domain, so the domain is restricted (e.g. to one side of the vertex) to make the inverse well-defined.
Can I use my GDC for function questions?
Yes - your GDC can graph a function, evaluate it at any point, and read off values from a table, which is useful for checking domain, range and composite function values quickly on Paper 1 or Paper 2. See the GDC guide for model-specific instructions.
Sub-topics
Function Concepts broken down into its individual skills, each with its own focused page.
Related topics
More Functions & Modelling topics from the same AI HL syllabus unit, in case you want to keep going.