Function Notation, Domain and Inverse (AI SL)
Before you can model anything with functions, you need to be fluent in the notation itself: reading \(f(x)\) correctly, stating where a function is defined, and reversing it to find an inverse. This page isolates that narrow skill - notation, domain, range and inverses - with worked examples and the mistakes that cost the most marks. It sits inside the wider Function Concepts topic.
18 questions on this sub-topic.
Notation you need to know
Covered under IB syllabus reference SL2.2: the concept of a function, its domain, range and graph, and function notation such as \(f(x)\), \(v(t)\) or \(C(n)\). None of this is a "formula" in the booklet sense - it's notation and definitions you're expected to already know.
Function notation
\(f(x)\), \(v(t)\), \(C(n)\)
Names the rule and its input variable. Not in the formula booklet - it's notation, not a formula, so \(f(3)\) means "substitute \(x=3\)", never "\(f\) times 3".
Domain and range
Domain: the set of valid \(x\). Range: the set of resulting \(f(x)\).
The largest set of \(x\)-values for which the function is defined, unless the question restricts it further. Not in the booklet - this is prior knowledge you apply to each new function.
Inverse function \(f^{-1}(x)\)
Swap \(x\) and \(y\) in \(y=f(x)\), then rearrange for \(y\).
The inverse undoes what \(f\) does. Not a booklet formula - it's a method, and you apply the same three steps (write as \(y=\), swap, rearrange) to any function you're asked to invert.
Need the full syllabus wording, GDC screenshots and the rest of the Function Concepts formula table? See Function Concepts, including its GDC guidance for graphing and tracing a function's domain and range.
Worked examples
Let \(f(x) = 3x - 1\).
(a) Find \(f(4)\).
(b) Find \(f^{-1}(x)\).
(c) State the domain and range of \(f\).
Worked solution
(a) f(4) = 3(4) - 1 = 11 A1
(b) Let y = 3x - 1, so x = (y+1)/3. M1
f⁻¹(x) = (x+1)/3 A1
(c) Domain: x ∈ ℝ A1
; Range: f(x) ∈ ℝ A1
Let \(f(x)=2x+1\) and \(g(x)=x^2\).
Find \((f\circ g)(3)\).
Worked solution
\((f\circ g)(x)=f(g(x))\): the inner function \(g\) acts first, then its output is fed into \(f\). Order matters. M1
\(g(3)=3^2=9.\) A1
\(f(9)=2(9)+1=19.\) A1
Common mistakes
- Reading \(f(x)\) as multiplication. \(f(x)\) is notation for "the output of \(f\) at \(x\)" - it isn't \(f\) times \(x\), and \(f(3)\) means substitute \(x=3\), not "\(f\) times 3".
- Stating the range using the domain's numbers. The range is a set of \(y\)-values, not \(x\)-values - substitute the domain endpoints (or check the graph's turning points) into the function to get the actual range.
- Forgetting to swap \(x\) and \(y\) when inverting. Rearranging \(y=f(x)\) for \(x\) without then swapping the letters gives you \(x\) in terms of \(y\), not \(f^{-1}(x)\) - the swap is what turns the rearranged rule into the inverse function.
Ready to practise properly?
41 function-notation questions, marked instantly like the real exam.
Quick answers
What does \(f(x)\) notation actually mean?
\(f(x)\) names the output of the function \(f\) when the input is \(x\). It is notation, not multiplication - \(f(3)\) means substitute \(x=3\) into the rule, it does not mean \(f\) times 3.
How do you find the inverse of a function?
Write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject again. The result, relabelled \(f^{-1}(x)\), reverses what the original function does.