Composite & Inverse Functions (AA SL)
Composite functions chain two functions together so the output of one feeds into the other, while an inverse function undoes what a function does. This topic covers evaluating and forming composite functions, finding an inverse algebraically by swapping variables, checking whether a function is one-to-one, and restricting a domain so an inverse can exist.
What the syllabus says
This topic maps onto two points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| SL2.2 | Concept of a function, domain, range and graph. The informal concept that an inverse function reverses or undoes the effect of a function. Inverse function as a reflection in the line \(y=x\), and the notation \(f^{-1}(x)\). |
| SL2.5 | Composite functions, \((f\circ g)(x)=f(g(x))\). The identity function. Finding the inverse function \(f^{-1}(x)\), including \((f\circ f^{-1})(x)=(f^{-1}\circ f)(x)=x\). The existence of an inverse for one-to-one functions. |
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a composite function?
A composite function applies one function to the output of another. \((f\circ g)(x)=f(g(x))\) means "work out \(g(x)\) first, then substitute that result into \(f\)" - the order matters.
e.g. If \(f(x)=2x+1\) and \(g(x)=x^2\), then \((f\circ g)(3)=f(9)=19\).
What is an inverse function?
An inverse function, \(f^{-1}(x)\), reverses whatever \(f\) does - if \(f\) takes \(a\) to \(b\), then \(f^{-1}\) takes \(b\) back to \(a\). It exists only when \(f\) is one-to-one.
e.g. For \(f(x)=2x+1\), \(f^{-1}(x)=\dfrac{x-1}{2}\), and \(f(f^{-1}(5))=f(2)=5\).
What is a one-to-one function?
A function is one-to-one if every output value corresponds to exactly one input value - no two different inputs give the same output. Only one-to-one functions have an inverse that is itself a function.
e.g. \(f(x)=x^2\) is not one-to-one over all reals, since \(f(2)=f(-2)=4\).
What is the identity function?
The identity function, \(I(x)=x\), leaves every input unchanged. Composing a function with its own inverse, in either order, always produces the identity function.
e.g. \((f\circ f^{-1})(x)=x\) for any invertible \(f\), since the two functions exactly cancel.
What is domain restriction?
Domain restriction means limiting a function's domain to a smaller set of inputs so that it becomes one-to-one, which then allows an inverse function to exist.
e.g. Restricting \(f(x)=x^2\) to \(x\ge0\) gives it the inverse \(f^{-1}(x)=\sqrt{x}\), since \(f(3)=9\) and \(f^{-1}(9)=3\).
Key formulas
This topic is more about relationships than plug-in formulas. The tables below summarise the key ones - the explanations underneath go into more depth on each one.
Formula reference
None of these appear as standalone formulas on the official booklet - they're notation and relationships you're expected to know from the syllabus itself.
| Relationship | Used for | Booklet? |
|---|---|---|
| \((f\circ g)(x) = f(g(x))\) | Composite function notation | Not in booklet - definition |
| \((f\circ f^{-1})(x)=(f^{-1}\circ f)(x)=x\) | Inverse cancels the original function | Not in booklet - prior knowledge |
| Graph of \(f^{-1}\) = reflection of \(f\) in \(y=x\) | Visualising an inverse | Not in booklet - prior knowledge |
| Domain of \(f^{-1}\) = Range of \(f\) | Working out the domain of an inverse | Not in booklet - prior knowledge |
Composite vs inverse
Both combine two functions, but they answer different questions and are built in different ways.
| Feature | Composite function | Inverse function |
|---|---|---|
| Involves | Two different functions \(f\) and \(g\) | One function \(f\) and its reverse |
| Notation | \((f\circ g)(x)\) | \(f^{-1}(x)\) |
| How to build it | Substitute \(g(x)\) into \(f\) | Swap \(x\) and \(y\), then rearrange |
| Order matters? | Yes - \((f\circ g)\neq(g\circ f)\) in general | Not applicable - only one function |
Working with composite functions
The core skill is substituting correctly, in the right order.
Evaluating at a number
Work out the inner function's value first, then substitute that number into the outer function.
Building the algebraic form
Replace every \(x\) in the outer function with the entire expression for the inner function, then simplify.
Order matters
\((f\circ g)(x)\) and \((g\circ f)(x)\) are generally different functions - always check which letter comes first.
Finding an inverse function
The same three-step method works for every invertible function on this topic.
Swap and rearrange
Write \(y=f(x)\), swap \(x\) and \(y\), then rearrange to make \(y\) the subject again - that's \(f^{-1}(x)\).
Verify by composing
Substitute your answer into \(f\) and confirm \((f\circ f^{-1})(x)=x\) - if it doesn't simplify to \(x\), check your algebra.
Check it exists first
An inverse function only exists if the original function is one-to-one - restrict the domain if it isn't.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
\(f(x)=2x+1\) and \(g(x)=x^2.\)
(a) Find \((f\circ g)(3)\).
(b) Find \((f\circ g)(4).\)
Worked solution
(a) \(g(3)=9,\ f(9)\) M1
\(=19.\) A1
(b) \(g(4)=16,\ f(16)=33.\) A1
Let \(f(x)=\dfrac{x-1}{2}\).
Show that \(f^{-1}(x)=2x+1\) and verify \((f\circ f^{-1})(x)=x\).
Worked solution
\(y=\dfrac{x-1}{2}\Rightarrow2y=x-1\Rightarrow x=2y+1\), so \(f^{-1}(x)\) M1 \(=2x+1.\) A1 AG
\(f(2x+1)=\dfrac{(2x+1)-1}{2}=\dfrac{2x}{2}\) M1 \(=x.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Evaluating a composite function in the wrong order. \((f\circ g)(x)\neq(g\circ f)(x)\) in general - work out the inner function first, and check carefully which letter is written first.
- 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.
Using your GDC
Composite and inverse functions are mostly an algebra skill, so your GDC plays a smaller role here than elsewhere. On Paper 2 you can define \(f(x)\) and \(g(x)\) as separate functions and evaluate \(f(g(a))\) directly at a numerical value, which is a fast way to check an answer to a composite-function question. You can also graph a function together with its reflection in the line \(y=x\) to see what its inverse looks like, though the algebraic form of \(f^{-1}(x)\) still has to be found by hand using the swap-and-rearrange method above. Paper 1 tests this topic without a calculator, so the algebra is what matters most.
See the full GDC guide for model-specific button sequences across every topic.
Ready to practise properly?
Composite & inverse functions questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between (f∘g)(x) and (g∘f)(x)?
They apply the same two functions in opposite orders, and generally give different results. \((f\circ g)(x)\) means "do \(g\) first, then \(f\)"; \((g\circ f)(x)\) means "do \(f\) first, then \(g\)". Always check which function is applied first before evaluating.
How do I 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 \(f^{-1}(x)\). You can check your answer by confirming \((f\circ f^{-1})(x)=x\).
Does every function have an inverse?
Only if the function is one-to-one, meaning every output comes from exactly one input. If it isn't - like \(f(x)=x^2\) over all real numbers - you need to restrict the domain first before an inverse function exists.
Can my GDC find a composite or inverse function for me?
Your GDC can evaluate a composite function numerically at a given value, and can graph a function alongside its reflection in \(y=x\) to show the inverse visually. Finding the algebraic form of an inverse is still done by hand. See the GDC guide for more.
Sub-topics
Composite & Inverse Functions broken down into its individual skills, each with its own focused page.
Related topics
More Functions topics from the same AA SL syllabus unit, in case you want to keep going.