Domain and Range (AA SL)
The domain of a function is every input value it's allowed to take; the range is every output value it can actually produce. Most domain problems come down to spotting what would break the function - a zero denominator, a negative under a root, a non-positive log argument - while range problems usually mean picturing the shape of the graph. It's part of the broader Composite & Inverse Functions topic.
14 questions on this sub-topic.
Domain, range, and inverses
Covered under IB syllabus reference SL2.2. These two facts aren't formula-booklet entries - they're definitions you're expected to know and apply, especially once inverse functions are involved.
Domain of an inverse
\(\text{Domain of } f^{-1} = \text{Range of } f\)
Not in the booklet - prior knowledge. Swapping input and output when you invert a function swaps their domain and range too.
Swap and rearrange
Write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject again - that's \(f^{-1}(x)\).
Useful whenever a question asks for the domain of an inverse before you've even found the inverse itself.
Need the full syllabus wording? See Composite & Inverse Functions, including the GDC guidance there for checking domains graphically.
Worked examples
State the domain of \(f(x)=\ln(x-3)\).
Worked solution
The argument of a logarithm must be positive: \(x-3>0.\) M1 R1
So the domain is \(x>3.\) A1
Let \(f(x)=\ln x\) (\(x>0\)) and \(g(x)=x^2-4\).
(a) Find \((f\circ g)(x)\).
(b) State the domain of \(f\circ g\).
Worked solution
(a) \((f\circ g)(x)=\ln(x^2-4).\) A1
(b) Need \(x^2-4>0.\) M1
Factorise: \((x-2)(x+2)>0.\) A1
So \(x<-2\) or \(x>2.\) A1
Consider \(f(x) = \sqrt{x - 4}\).
(a) State the domain of \(f\).
(b) State the range of \(f\).
Worked solution
(a) The expression under a square root must be non-negative: \(x-4\ge0\Rightarrow x\ge4.\) Domain: \(x\ge4.\) M1A1
(b) A square-root output is never negative and starts at 0 (at \(x\) R1
\(=4\)). Range: \(f(x)\ge0.\) A1
Common mistakes
- Forgetting to exclude a zero denominator. Any \(x\) that makes the bottom of a fraction zero must be removed from the domain, even if it looks harmless in the original expression.
- Missing the restriction on logs and roots. The argument of \(\ln(\cdot)\) must be strictly positive, and the expression under a square root must be non-negative - both are easy to skip when the function is buried inside a composite.
- Mixing up domain and range. The domain is the set of \(x\)-values you're allowed to put in; the range is the set of \(y\)-values that come out. Reading a graph's vertex or asymptote correctly only helps if you know which one you're being asked for.
Ready to practise properly?
14 domain-and-range questions, marked instantly like the real exam.
Quick answers
How do you find the domain of a function?
Look for values of \(x\) that would break the function - a zero denominator, a negative number under a square root, or a non-positive argument inside a log - and exclude them from the domain.
What is the domain of an inverse function?
The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\) - the two swap places when you invert a function.