Composite and Inverse Functions (AI HL)
A composite function chains two functions together, feeding the output of one into the other - and the order you apply them in changes the answer. An inverse function undoes what the original function did, but only exists cleanly where that original function is one-to-one. This page covers both skills with worked examples and the mistakes that cost the most marks. It's part of the broader Function Concepts topic.
24 questions on this sub-topic.
Composition and inverses
Covered under IB syllabus reference AHL2.7. Composite function notation and the inverse relationship both appear in the formula booklet - restricting the domain so an inverse exists is prior knowledge you're expected to apply yourself.
Composite function notation
\((f\circ g)(x) = f(g(x))\)
Work from the inside out: evaluate \(g(x)\) first, then substitute the result into \(f\).
Inverse function relationship
\(f^{-1}(f(x)) = x = f(f^{-1}(x))\)
To find \(f^{-1}\): write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject.
Making an inverse exist
Restrict the domain so \(f\) is one-to-one
Not in the formula booklet - prior knowledge. A many-to-one function like \(f(x)=x^2\) needs a restricted domain before \(f^{-1}\) is a genuine function.
Need the full syllabus wording and formula-booklet reference table? See Function Concepts.
Worked examples
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 \(f\). \(g(3)=3^2=9.\) M1
\(f(9)=2(9)+1=19.\) A1
Let \(f(x) = x^2 - 4\), \(x \geq 0\), and \(g(x) = 2x + 1\).
(a) Find \(f(g(2))\).
Find \(f^{-1}(x)\).
(b)(i) State \(f^{-1}(x)\).
(b)(ii) State its domain.
Worked solution
(a) \(g(2) = 5\) M1
\(f(g(2)) = f(5) = 25 - 4 = 21\) A1
(b)(i) Let \(y = x^2 - 4\), so \(x = \sqrt{y+4}\) (taking positive root since \(x \geq 0\)) M1
\(f^{-1}(x) = \sqrt{x+4}\) A1
(b)(ii) Domain: \(x \geq -4\) A1
Common mistakes
- 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.
- Applying the functions in the wrong order. \((f\circ g)(x)\) means \(g\) acts first - swapping the order gives \((g\circ f)(x)\), which is generally a different function entirely.
- Confusing \(f^{-1}(x)\) with \(\dfrac{1}{f(x)}\). The inverse function reverses \(f\); it is not the reciprocal. Writing \(f^{-1}(x)\) as \(1/f(x)\) is one of the most common notation slips at this level.
Ready to practise properly?
20 composite and inverse function questions, marked instantly like the real exam.
Quick answers
What order do you apply the functions in (f o g)(x)?
The inner function acts first. \((f\circ g)(x) = f(g(x))\) means you evaluate \(g(x)\) first, then substitute that result into \(f\).
How do you find an inverse function?
Write \(y = f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject. State \(f^{-1}(x)\) and, if \(f\) was not one-to-one over its original domain, restrict the domain so it is. For more on domain and GDC-based checks, see Function Concepts.