B10PureStretch

Composite and Inverse Functions

Q: Does fg(x) mean apply f first or g first?

g first. fg(x) = f(g(x)), so the function written next to the x (the inner one) acts first, and its output becomes the input of the outer function.

Q: Is f^-1(x) the same as 1f(x)?

No. The -1 denotes the inverse function, which undoes f, not a reciprocal power. You find it by rearranging y = f(x) to make x the subject and swapping the letters.

Q: How are the domain and range of an inverse related to the original?

They swap: the domain of f^-1 is the range of f, and the range of f^-1 is the domain of f. This matches the reflection of the graph in the line y = x.

A function is a machine: put a number in, get one out. Composite functions chain two machines together, and inverse functions run a machine backwards. The two ideas you must nail are the order of composition and the domain–range swap of an inverse.

32 min Video by TLMaths Algebra and Functions

The big picture

Thinking of a function as a $process$ — input, transform, output — is one of the most powerful mental shifts in A-Level. It reframes everything: a composite $f g$ is two processes back-to-back, and an inverse $f^{- 1}$ is the process run in reverse. This language underpins the chain rule in calculus, transformations of graphs, and later work on exponentials and logarithms (which are inverse functions of each other). The two habitual pitfalls — reading $f g$ in the wrong order, and forgetting that an inverse swaps domain and range — are worth crushing now, because they echo through the rest of the course.

A-Level Maths B8-08: Introducing Composite FunctionsWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Use function notation, domain and range confidently
Form and evaluate composite functions in the correct order
Find an inverse function by rearranging, and swap domain and range
Know when an inverse exists (one-to-one) and read the $y = x$ reflection

Functions, domain and range

A $function$ takes each input to exactly one output. The set of allowed inputs is the $domain$ ; the set of resulting outputs is the $range$ . For $f (x) = x^{2}$ with domain all real numbers, the range is $y \geq 0$ , because squaring never gives a negative.

Keeping domain and range straight matters enormously for inverses, because — as you will see — an inverse simply swaps them.

Composite functions — mind the order

The composite $f g (x)$ means “do $g$ first, then feed the result into $f$ ”: $f g (x) = f (g (x))$ . The function nearest the $x$ acts first. This order is not a convention you can bend — $f g$ and $g f$ are usually completely different.

f g (x) = f (g (x)) (apply g first)

Read right-to-left: the inside function runs first.

f g (x) = f (g (x)) = f (x^{2}) = x^{2} + 1

g f (x) = g (f (x)) = g (x + 1) = (x + 1)^{2} = x^{2} + 2 x + 1

3They are different — order matters.

Answer

f g (x) = x^{2} + 1

;

g f (x) = (x + 1)^{2}

Why this works

The function written closest to the $x$ is applied first. A useful check: $f g (x)$ substitutes $g$ into $f$ , so $g$ ’s output becomes $f$ ’s input.

Inverse functions — running the machine backwards

The inverse $f^{- 1}$ undoes $f$ : if $f$ turns 3 into 7, then $f^{- 1}$ turns 7 back into 3. To find it, write $y = f (x)$ , rearrange to make $x$ the subject, then swap the letters. Crucially, $f^{- 1}$ is $not$ the reciprocal $\frac{1}{f}$ — the $- 1$ here means “inverse”, not “to the power $- 1$ ”.

Because an inverse reverses inputs and outputs, the $domain of f^{- 1}$ is the $range of f$ , and vice versa. Graphically, $y = f^{- 1} (x)$ is the reflection of $y = f (x)$ in the line $y = x$ .

f (f^{- 1} (x)) = x = f^{- 1} (f (x))

Each undoes the other.

domain of f^{- 1} = range of f

Inverses swap domain and range.

1Write

y = 2 x - 3

2Rearrange for

x

x = \frac{y + 3}{2}

3Swap letters:

f^{- 1} (x) = \frac{x + 3}{2}

Answer

f^{- 1} (x) = \frac{x + 3}{2}

Tip — Check an inverse by composing: $f f^{- 1} (x)$ should simplify to $x$ . If it does not, you have made a slip.

When does an inverse exist?

An inverse exists only if the function is $one-to-one$ — each output comes from exactly one input. $f (x) = x^{2}$ fails this on all reals, because $2$ and $- 2$ both map to $4$ , so the “inverse” would not know which to return. The fix is to $restrict the domain$ (e.g. $x \geq 0$ ), making it one-to-one so $f^{- 1} (x) = x$ is well defined.

Why this works

The reflection-in- $y = x$ picture explains the one-to-one rule visually: reflecting a curve that fails the “horizontal line test” produces something that fails the vertical line test — i.e. not a function. Restricting the domain removes the ambiguity.

Think like an examiner

For $f g (x)$ , apply $g$ first — the inside function. Getting the order backwards is the most common composite-function error.
When finding an inverse, state the domain of $f^{- 1}$ (it equals the range of $f$ ) if the question asks — it is frequently worth a mark.
Never treat $f^{- 1} (x)$ as $\frac{1}{f ( x )}$ ; the notation means the inverse function.
If a function is not one-to-one, expect to restrict the domain before inverting, and say so explicitly.

Common misconceptions

Reading

f g (x)

as “

f

first, then

g

”.

f g (x) = f (g (x))

: the inner function

g

acts first.

Writing

f^{- 1} (x) = \frac{1}{f ( x )}

f^{- 1}

is the inverse function, not the reciprocal — find it by rearranging.

Giving the inverse of

f (x) = x^{2}

on all reals as

x

without comment.

x^{2}

is not one-to-one on all reals; restrict to

x \geq 0

first, then

f^{- 1} (x) = x

Ignoring the domain and range when inverting.

The domain of

f^{- 1}

is the range of

f

— state it when required.

Functions toolkit

f g (x) = f (g (x))

Composite: inner function first.

f f^{- 1} (x) = x

Inverse undoes the function.

dom (f^{- 1}) = ran (f)

Domain–range swap.

y = f^{- 1} (x) reflects y = f (x) in y = x

Graphical inverse.

Stretch yourself

Given $f (x) = \frac{x + 1}{x - 2}$ for $x \neq = 2$ , find $f^{- 1} (x)$ and show that $f$ is its own inverse.

Hint — Set $y = \frac{x + 1}{x - 2}$ , multiply out, collect the $x$ terms and make $x$ the subject, then swap letters.

Questions students ask

Does

f g (x)

mean apply

f

first or

g

first?

g

first.

f g (x) = f (g (x))

, so the function written next to the

x

(the inner one) acts first, and its output becomes the input of the outer function.

f^{- 1} (x)

the same as

\frac{1}{f ( x )}

No. The

- 1

denotes the inverse function, which undoes

f

, not a reciprocal power. You find it by rearranging

y = f (x)

to make

x

the subject and swapping the letters.

Why do some functions not have an inverse?

Because they are not one-to-one — two inputs share an output, so reversing the process is ambiguous.

x^{2}

is the classic case. Restricting the domain (say

x \geq 0

) makes it one-to-one and invertible.

How are the domain and range of an inverse related to the original?

They swap: the domain of

f^{- 1}

is the range of

f

, and the range of

f^{- 1}

is the domain of

f

. This matches the reflection of the graph in the line

y = x

Key takeaways

A function maps each input (domain) to one output (range).
Composite $f g (x) = f (g (x))$ — apply the inner function first; order matters.
Find an inverse by rearranging $y = f (x)$ for $x$ and swapping letters; $f^{- 1}$ is not a reciprocal.
An inverse swaps domain and range and reflects the graph in $y = x$ .
Inverses exist only for one-to-one functions — restrict the domain if needed.

How this fits the course

Build on

Graphs and Transformations

The Modulus Function
Logarithms as inverses (later)soon

Leads to

The Modulus Function

Test yourself

Ready to lock in Composite and Inverse Functions? Pick a mode and earn XP & Dobloons.