Mathematics 9709 · Pure Mathematics 1

Composite functions

A composite function joins two machines. The output of the first becomes the input of the second.
The main thing to get right is the order. The notation hides the order, so almost everyone gets it wrong the first time.

$\mathrm{gf}(x)$ means $\mathrm{g}(\mathrm{f}(x))$ . The inner (right-most) function acts first. Work it out, then put that output into the outer one.

Does it exist? $\mathrm{fg}$ only exists if the range of $\mathrm{g}$ is inside the domain of $\mathrm{f}$ . The inner machine's outputs must be allowed inputs for the outer machine. If not, there is nothing to put in.

The right-most function acts first

In $\mathrm{gf}(x)$ , the function nearest the $x$ goes first. So $\mathrm{gf}(x)$ says: put $x$ through $\mathrm{f}$ , then put that result through $\mathrm{g}$ . Read it from the inside out.

So the order matters. In most cases, $\mathrm{gf}(x)\neq \mathrm{fg}(x)$ .
Doing $\mathrm{f}$ then $\mathrm{g}$ usually gives a different rule.
So you can never just swap them.

Why you read it from the inside out (the bracket reason)

It works like brackets inside brackets in algebra. To find $\mathrm{g}(\mathrm{f}(x))$ , you must know the inner value $\mathrm{f}(x)$ first. Only then can you put it into $\mathrm{g}$ . It is like $\sqrt{3+\mathrm{f}(x)}$ : you work out the inside before the square root. So the function next to $x$ has to happen first.

How to build one

To find $\mathrm{gf}(x)$ : take the rule for $\mathrm{g}$ . Then put the whole expression $\mathrm{f}(x)$ in for its input, inside brackets.

Worked example

Textbook 2.4: $\mathrm{f}(x)=2x+3$ and $\mathrm{g}(x)=x^2-1$ , find $\mathrm{fg}(x)$ and $\mathrm{gf}(x)$

$\mathrm{fg}(x)$ : do $\mathrm{g}$ first, so put $x^2-1$ into the “double and add 3” machine: $\mathrm{fg}(x)=2(x^2-1)+3$ .
Expand and tidy: $2x^2-2+3=2x^2+1$ .
$\mathrm{gf}(x)$ : do $\mathrm{f}$ first, so put $2x+3$ into the “square and subtract 1” machine, keeping it in a bracket: $\mathrm{gf}(x)=(2x+3)^2-1$ .
Expand the square first, then subtract: $(4x^2+12x+9)-1=4x^2+12x+8$ .

Answer

\mathrm{fg}(x)=2x^2+1,\quad \mathrm{gf}(x)=4x^2+12x+8

Two completely different rules. This shows the order matters.

Worked example

Real exam: $\mathrm{f}(x)=\dfrac{x+3}{x-1}$ for $x>1$ , find $\mathrm{ff}(5)$ (9709/12 Oct/Nov 2021 Q3a)

$\mathrm{ff}(5)$ means do $\mathrm{f}$ twice, and with a number you do not need a general formula, you just work out the inner one first.

Inner first: $\mathrm{f}(5)=\dfrac{5+3}{5-1}=\dfrac{8}{4}=2$ .
Put that answer back in: $\mathrm{f}(2)=\dfrac{2+3}{2-1}=\dfrac{5}{1}=5$ .

Answer

\mathrm{ff}(5)=5

(9709/12 Oct/Nov 2021 Q3(a)) The mark scheme wants clear proof that you used f twice, so write the middle step $\mathrm{f}(5)=2$ , not just the final $5$ . The answer comes back to $5$ , which is a sign this $\mathrm{f}$ is its own inverse (more in §2.3).

Find the smallest k so the composite can be formed

A common exam question gives you a smaller domain with an unknown $k$ . It asks for the smallest $k$ that lets the composite exist.
This is just the rule above. Fit the range of the inner inside the domain of the outer.

Worked example

$\mathrm{f}(x)=2x-3$ for $x\ge k$ , $\mathrm{g}(x)=x^2-4$ for $x\ge-4$ : find the smallest $k$ for which $\mathrm{gf}$ can be formed

For $\mathrm{gf}$ , $\mathrm{f}$ is the inner machine, so its outputs must be allowed inputs of $\mathrm{g}$ , which means the range of $\mathrm{f}$ must be inside the domain of $\mathrm{g}$ , which is $x\ge-4$ .

$\mathrm{f}(x)=2x-3$ is an increasing line, so on $x\ge k$ its smallest output is at $x=k$ : $\mathrm{f}(k)=2k-3$ .
Force that lowest output into $\mathrm{g}$ 's domain $x\ge-4$ : $2k-3\ge -4$ .
Solve: $2k\ge -1\;\Rightarrow\; k\ge -\tfrac12$ , so the smallest allowed value is the boundary.

Answer

k=-\tfrac12

(9709-style)

Where the marks go

Common mistake

\mathrm{gf}

does not mean “do

\mathrm{g}

first”. The right-most function (the one next to the

x

) acts first. Getting the order wrong loses more marks than anything else in this topic. (9709/12 May/June 2020 Q5)

Common mistake

Keep the bracket when you put it in. It is

(2x+3)^2

, not

2x^2+3

. Put the whole inner expression in a bracket first, then expand.

Common mistake

\mathrm{fg}

does not always exist. It only exists if the range of the inner function fits inside the domain of the outer one. “Find the smallest

k

” questions test this, not the algebra. (9709-style)

Now you try

The function $\mathrm{f}(x)=a-2x$ is defined for $x\in\mathbb{R}$ , where $a$ is a constant. Express $\mathrm{ff}(x)$ in terms of $a$ and $x$ . (9709/12 May/June 2020 Q5)

Your turn— tap to reveal the worked answer (9709/12 May/June 2020 Q5(a))

Do $\mathrm{f}$ twice. Put the inner $\mathrm{f}(x)=a-2x$ in for the input, inside a bracket:

$\mathrm{ff}(x)=a-2(a-2x)$

Now expand the bracket and simplify:

$=a-2a+4x$

Answer

\mathrm{ff}(x)=4x-a

PreviousDefinition of a function (domain and range)NextInverse functions

Composite functions

A composite function joins two machines. The output of the first becomes the input of the second.
The main thing to get right is the order. The notation hides the order, so almost everyone gets it wrong the first time.

$\mathrm{gf}(x)$ means $\mathrm{g}(\mathrm{f}(x))$ . The inner (right-most) function acts first. Work it out, then put that output into the outer one.

The right-most function acts first

In $\mathrm{gf}(x)$ , the function nearest the $x$ goes first. So $\mathrm{gf}(x)$ says: put $x$ through $\mathrm{f}$ , then put that result through $\mathrm{g}$ . Read it from the inside out.

So the order matters. In most cases, $\mathrm{gf}(x)\neq \mathrm{fg}(x)$ .
Doing $\mathrm{f}$ then $\mathrm{g}$ usually gives a different rule.
So you can never just swap them.

Why you read it from the inside out (the bracket reason)

How to build one

To find $\mathrm{gf}(x)$ : take the rule for $\mathrm{g}$ . Then put the whole expression $\mathrm{f}(x)$ in for its input, inside brackets.

Worked example

Textbook 2.4: $\mathrm{f}(x)=2x+3$ and $\mathrm{g}(x)=x^2-1$ , find $\mathrm{fg}(x)$ and $\mathrm{gf}(x)$

$\mathrm{fg}(x)$ : do $\mathrm{g}$ first, so put $x^2-1$ into the “double and add 3” machine: $\mathrm{fg}(x)=2(x^2-1)+3$ .
Expand and tidy: $2x^2-2+3=2x^2+1$ .
$\mathrm{gf}(x)$ : do $\mathrm{f}$ first, so put $2x+3$ into the “square and subtract 1” machine, keeping it in a bracket: $\mathrm{gf}(x)=(2x+3)^2-1$ .
Expand the square first, then subtract: $(4x^2+12x+9)-1=4x^2+12x+8$ .

Answer

\mathrm{fg}(x)=2x^2+1,\quad \mathrm{gf}(x)=4x^2+12x+8

Two completely different rules. This shows the order matters.

Worked example

Real exam: $\mathrm{f}(x)=\dfrac{x+3}{x-1}$ for $x>1$ , find $\mathrm{ff}(5)$ (9709/12 Oct/Nov 2021 Q3a)

$\mathrm{ff}(5)$ means do $\mathrm{f}$ twice, and with a number you do not need a general formula, you just work out the inner one first.

Inner first: $\mathrm{f}(5)=\dfrac{5+3}{5-1}=\dfrac{8}{4}=2$ .
Put that answer back in: $\mathrm{f}(2)=\dfrac{2+3}{2-1}=\dfrac{5}{1}=5$ .

Answer

\mathrm{ff}(5)=5

Find the smallest k so the composite can be formed

A common exam question gives you a smaller domain with an unknown $k$ . It asks for the smallest $k$ that lets the composite exist.
This is just the rule above. Fit the range of the inner inside the domain of the outer.

Worked example

$\mathrm{f}(x)=2x-3$ for $x\ge k$ , $\mathrm{g}(x)=x^2-4$ for $x\ge-4$ : find the smallest $k$ for which $\mathrm{gf}$ can be formed

$\mathrm{f}(x)=2x-3$ is an increasing line, so on $x\ge k$ its smallest output is at $x=k$ : $\mathrm{f}(k)=2k-3$ .
Force that lowest output into $\mathrm{g}$ 's domain $x\ge-4$ : $2k-3\ge -4$ .
Solve: $2k\ge -1\;\Rightarrow\; k\ge -\tfrac12$ , so the smallest allowed value is the boundary.

Answer

k=-\tfrac12

(9709-style)

Where the marks go

Common mistake

\mathrm{gf}

does not mean “do

\mathrm{g}

first”. The right-most function (the one next to the

x

) acts first. Getting the order wrong loses more marks than anything else in this topic. (9709/12 May/June 2020 Q5)

Common mistake

Keep the bracket when you put it in. It is

(2x+3)^2

, not

2x^2+3

. Put the whole inner expression in a bracket first, then expand.

Common mistake

\mathrm{fg}

does not always exist. It only exists if the range of the inner function fits inside the domain of the outer one. “Find the smallest

k

” questions test this, not the algebra. (9709-style)

Now you try

The function $\mathrm{f}(x)=a-2x$ is defined for $x\in\mathbb{R}$ , where $a$ is a constant. Express $\mathrm{ff}(x)$ in terms of $a$ and $x$ . (9709/12 May/June 2020 Q5)

Your turn— tap to reveal the worked answer (9709/12 May/June 2020 Q5(a))

Do $\mathrm{f}$ twice. Put the inner $\mathrm{f}(x)=a-2x$ in for the input, inside a bracket:

$\mathrm{ff}(x)=a-2(a-2x)$

Now expand the bracket and simplify:

$=a-2a+4x$

Answer

\mathrm{ff}(x)=4x-a