Inverse Functions (AA SL)
The inverse of \(f\), written \(f^{-1}\), is the function that undoes \(f\) - feed it \(f\)'s output and it hands back the original input. Finding one is mostly algebra: swap \(x\) and \(y\), then rearrange. This page covers that method, how the inverse relates to \(f\) itself, and the traps that cost the most marks. It's part of the broader Composite & Inverse Functions topic.
24 questions on this sub-topic.
How inverses behave
Covered under IB syllabus reference SL2.5. None of these are formula-booklet entries - they're properties you're expected to know and use directly.
Inverse cancels the original
\((f\circ f^{-1})(x)=(f^{-1}\circ f)(x)=x\)
Not in the booklet - prior knowledge. A quick way to check an inverse you've found is correct: composing it with \(f\) either way should just give back \(x\).
Visualising an inverse
The graph of \(f^{-1}\) is the reflection of the graph of \(f\) in the line \(y=x\).
Not in the booklet - prior knowledge. Handy for sketching \(f^{-1}\) quickly once you already have a sketch of \(f\).
Domain of an inverse
\(\text{Domain of } f^{-1} = \text{Range of } f\)
Not in the booklet - prior knowledge. Input and output swap roles when you invert, so their domains and ranges swap too.
Need the full syllabus wording? See Composite & Inverse Functions, including the GDC guidance there for checking an inverse graphically.
Worked examples
Find \(f^{-1}(x)\) for \(f(x)=5x-2\).
Worked solution
Set \(y=5x-2\) and solve for \(x\): \(x=\dfrac{y+2}{5}.\) M1 A1
So \(f^{-1}(x)=\dfrac{x+2}{5}.\) A1
Let \(f(x) = e^{x} + 1\).
(a) Find \(f^{-1}(x)\).
(b) State the domain of \(f^{-1}\).
Worked solution
(a) Set \(y=e^x+1\), isolate \(e^x\): \(e^x=y-1.\) M1
Take the natural log: \(x=\ln(y-1).\) A1
So \(f^{-1}(x)=\ln(x-1).\) A1
(b) The domain of \(f^{-1}\) equals the range of \(f\). Since \(e^x>0\), \(f(x)=e^x+1>1.\) R1
So the domain of \(f^{-1}\) is \(x>1.\) A1
For \(f(x)=(x-1)^{2}+2\), \(x\ge1\), find \(f^{-1}(x)\) and state its domain.
(a)(i) Find \(f^{-1}(x)\).
(a)(ii) State its domain.
Worked solution
(a)(i) Set \(y=(x-1)^2+2\) and isolate the square: \((x-1)^2=y-2.\) M1
Since \(x\ge1\), take the positive root: \(x-1=\sqrt{y-2}\) A1
\(\Rightarrow x=1+\sqrt{y-2}.\) R1
(a)(ii) So \(f^{-1}(x)=1+\sqrt{x-2}.\) A1
The domain is \(x\ge2\) (the range of \(f\)). A1
Common mistakes
- Confusing \(f^{-1}(x)\) with \(\dfrac{1}{f(x)}\). The \(-1\) here means "inverse function", not "reciprocal" - these are almost always different expressions.
- Assuming every function has an inverse. An inverse function only exists for a one-to-one function - if two inputs share an output, the domain needs restricting first.
- Swapping \(x\) and \(y\) inconsistently. When finding an inverse, every \(x\) must become \(y\) and every \(y\) must become \(x\) at the same step - missing one leads to an equation that won't rearrange cleanly.
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 again - the result is \(f^{-1}(x)\).
Does every function have an inverse?
No - only a one-to-one function has an inverse. If two different inputs give the same output, the domain needs restricting before an inverse can exist.