Inverse Functions (AA HL)
The inverse of a function \(f\), written \(f^{-1}\), undoes what \(f\) does: if \(f\) sends \(a\) to \(b\), then \(f^{-1}\) sends \(b\) back to \(a\). This page covers how to find \(f^{-1}\) algebraically, when an inverse actually exists, and the mistakes that cost the most marks. It's part of the broader Composite & Inverse Functions topic.
24 questions on this sub-topic.
Definition and method
Covered under IB syllabus reference SL2.5: finding the inverse function \(f^{-1}(x)\), including the existence of an inverse for one-to-one functions, and the link to \(f^{-1}\) as a reflection in the line \(y=x\).
Self-inverse functions
\[(f\circ f)(x)=x\]
If composing \(f\) with itself gives back \(x\), then \(f=f^{-1}\) and the graph is symmetric about \(y=x\).
Not in the formula booklet - prior knowledgeFinding an inverse
\(y=f^{-1}(x) \iff x=f(y)\)
Write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject. The inverse satisfies \((f^{-1}\circ f)(x) = (f\circ f^{-1})(x) = x\).
Not in the formula booklet - prior knowledgeNeed the full syllabus wording and formula-booklet reference table? See Composite & Inverse Functions.
Worked examples
\(f(x)=\dfrac{2x-1}{x+3}.\) Find \(f^{-1}(x).\)
Worked solution
\(y=\dfrac{2x-1}{x+3}\Rightarrow y(x+3)=2x-1.\) M1
\(yx+3y=2x-1\)
\(yx-2x=-1-3y\)
\( x(y-2)=-1-3y.\) M1
\(x=\dfrac{-1-3y}{y-2}=\dfrac{3y+1}{2-y}.\) A1 Hence \(f^{-1}(x)=\dfrac{3x+1}{2-x}.\) A1
\(f(x)=x^2-4x+1.\)
(a) State the largest domain \(x\ge a\) on which \(f\) is one-to-one.
(b) Find \(f^{-1}(x)\) on that domain.
Worked solution
(a) \(f(x)=x^2-4x+1=(x-2)^2-3\), a parabola with vertex \(x=2.\) It is one-to-one on a half-line from the vertex, so the largest such domain is \(x\ge2\), giving \(a\) R1
\(=2.\) A1
(b) Set \(y=(x-2)^2-3\): \((x-2)^2=y+3.\) M1 Since \(x\ge2\), take the positive root: \(x=2+\sqrt{y+3}.\) A1 So \(f^{-1}(x)=2+\sqrt{x+3},\quad x\ge-3.\) A1
Find the inverse of \(f(x)=\dfrac{x-4}{3}.\)
Worked solution
\(y=\dfrac{x-4}{3}\Rightarrow 3y=x-4.\) M1
\(x=3y+4.\) A1
So \(f^{-1}(x)=3x+4.\) A1
\(f(x)=2x-3.\) Solve \(f(x)=f^{-1}(x).\)
Worked solution
\(y=2x-3\Rightarrow x=\dfrac{y+3}{2}\), so \(f^{-1}(x)\) M1 \(=\dfrac{x+3}{2}.\) A1
\(f\) is increasing and linear, so any intersection of \(f\) and \(f^{-1}\) lies on \(y=x.\) R1 Solve \(2x-3=x\Rightarrow x\) A1 \(=3.\) A1
Check: \(f(3)=3\), confirming the fixed point. A1
Common mistakes
- Writing \(f^{-1}(x)\) as \(\dfrac{1}{f(x)}\). The \(-1\) in \(f^{-1}\) is inverse-function notation, not a power - it never means "reciprocal of \(f\)".
- Forgetting to restrict the domain before inverting. A many-to-one function like \(f(x)=x^2\) has no inverse over its full domain; you must restrict to \(x\ge0\) (or \(x\le0\)) first.
- Stating the inverse but not its domain. The domain of \(f^{-1}\) equals the range of \(f\), not the domain of \(f\) - this is worth its own mark and is easy to skip.
Ready to practise properly?
24 inverse-function questions, marked instantly like the real exam.
Quick answers
How do you find the inverse of a function?
Write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject. The result is \(f^{-1}(x)\).
What is the domain of an inverse function?
The domain of \(f^{-1}\) equals the range of \(f\), and the range of \(f^{-1}\) equals the domain of \(f\).