Composite & inverse functions — notes, question types & practice · Functions, Graphs & Logarithms

01

Two machines in a line

A tiffin service: the first machine packs rice, the second adds a sweet. Feed $x$ through $g$ first, then push the result through $f$ . The combined rule is the composite, written $f(g(x))$ or $f \circ g$ .

Take $f(x) = x + 1$ and $g(x) = 2x$ .

$f(g(3))$ : $g(3) = 6$ , then $f(6) = 7$ .

Rule: In $f(g(x))$ , always work from the inside out. $g$ acts first, $f$ acts on what comes out.

02

Order matters

Swap the machines and watch the answer change.

$g(f(3))$ : $f(3) = 4$ , then $g(4) = 8$ .

So $f(g(3)) = 7$ but $g(f(3)) = 8$ . In general the two composites differ. Never run $f$ first unless the question says $g \circ f$ .

Building the full rule is plain substitution. With $f(x) = x + 3$ and $g(x) = 2x$ :

f(g(x)) = f(2x) = 2x + 3

The reverse: $g(f(x)) = 2(x + 3) = 2x + 6$ . Same parts, different result.

03

The inverse machine

The inverse $f^{-1}$ runs the machine backwards: output in, original input out.

If $f(x) = 3x - 2$ , then $f^{-1}(x) = \dfrac{x+2}{3}$ .

Find it by swap-and-solve:

Write $y = 3x - 2$ .
Swap: $x = 3y - 2$ .
Solve for $y$ : $y = \dfrac{x+2}{3}$ .

Check: $f(2) = 4$ , then $f^{-1}(4) = 2$ . Back where we started.

Watch: $f^{-1}(x)$ is not $\dfrac{1}{f(x)}$ . The $-1$ means "undo", never "divide by".

04

When an inverse exists

Undoing needs each output to point back to exactly one input.

$f(x) = x^2$ fails: $2$ and $-2$ both give $4$ , so $4$ cannot choose a way back. Restrict to $x \ge 0$ and it works: $f^{-1}(x) = \sqrt{x}$ .

Fractional linear functions always pass. For $f(x) = \dfrac{ax+b}{cx+d}$ , swap-and-solve gives

f^{-1}(x) = \frac{b - dx}{cx - a}

Tip: When $d = -a$ , the function is its own inverse. Check: $f(x) = \dfrac{3x+1}{x-3}$ undoes itself, so $f^{-1}(t) = f(t)$ for every $t$ .

05

Functional equations

The question gives a rule connecting values of $f$ and asks for one value. Feed small numbers and climb.

Take $f(x+y) = f(x) + f(y)$ with $f(1) = 5$ . Each step adds $5$ : $f(2) = 10$ , $f(3) = 15$ , so $f(7) = 35$ . In general $f(x) = 5x$ .

The product version $f(x+y) = f(x) \cdot f(y)$ with $f(1) = 3$ multiplies at every step: $f(2) = 9$ , $f(3) = 27$ , $f(4) = 81$ . Such a function is $f(x) = 3^x$ .

A shifted rule needs the same climbing, one step at a time. For $f(x+y) = f(x) + f(y) + 3$ with $f(1) = 2$ :

$f(2) = 2 + 2 + 3 = 7$ , then $f(3) = f(2) + f(1) + 3 = 12$ .

Note: In every functional equation, build outward from the given value using the given rule. Two or three steps almost always reach the asked value.

06

Reading the question

"Find $f(g(2))$ " — evaluate the inside rule first.
"Find $f^{-1}(3)$ " — swap-and-solve, or substitute into the inverse formula.
"Is $f^{-1}$ possible?" — look for two inputs sharing one output.
" $f(x+y) = \ldots$ , find $f(10)$ " — iterate from $f(1)$ ; the steps repeat.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Composite at a point

How to spot it:

Two rules $f$ and $g$ are given and $f(g(a))$ or $g(f(a))$ is asked for a number $a$ .

f(g(a)) = f\big(g(a)\big)

Method

Spot the inner function (the one next to $a$ ).
Evaluate the inner rule at $a$ .
Feed the result into the outer rule.
Check which composition the question names.

Why it works:

A composite is two substitutions done one after another, so point-wise evaluation is fastest.

Try this

If $f(x) = 2x + 3$ and $g(x) = x^2$ , find $f(g(2))$ .

Show solution

$g(2) = 4$ .
$f(4) = 11$ .

Answer

11

Type 2common2 practice Q

Build or compare the composite rule

How to spot it:

'Find $f(g(x))$ ' as a rule, or the difference $f(g(x)) - g(f(x))$ is asked.

f(g(x)) = \text{(outer rule with inner rule substituted)}

Method

Substitute the inner rule into the outer one.
Expand and simplify.
Build the reverse composite too if asked.
Subtract or compare as the question says.

Why it works:

Composition is substitution of one whole rule into the other, so algebra finishes it.

Try this

If $f(x) = 3x - 4$ and $g(x) = x + 2$ , find $f(g(x)) - g(f(x))$ .

Show solution

$f(g(x)) = 3(x+2) - 4 = 3x + 2$ .
$g(f(x)) = (3x - 4) + 2 = 3x - 2$ .
Difference $= 4$ .

Answer

4

Type 3very common3 practice Q

Find the inverse function

How to spot it:

$f(x)$ is given and $f^{-1}(x)$ is asked — often for the fractional linear shape $\dfrac{ax+b}{cx+d}$ .

f^{-1}(x) = \frac{b - dx}{cx - a}

Method

Write $y = f(x)$ .
Swap $x$ and $y$ .
Solve for $y$ .
Check one value: $f(f^{-1}(a))$ must return $a$ .

Why it works:

The inverse swaps the roles of input and output, so interchanging the letters and re-solving does the job.

Try this

Find $f^{-1}(x)$ for $f(x) = \dfrac{x+5}{x-2}$ .

Show solution

$y(x-2) = x + 5$ , so $xy - 2y = x + 5$ .
$x(y - 1) = 2y + 5$ .
$x = \dfrac{2y+5}{y-1}$ .

Answer

$f^{-1}(x) = \dfrac{2x+5}{x-1}$

Type 4occasional2 practice Q

Self-inverse and inverse values

How to spot it:

$f(x) = \dfrac{ax+b}{x-a}$ (bottom constant is minus the top coefficient of $x$ ), or $f^{-1}(t)$ is asked for one number $t$ .

d = -a \Rightarrow f^{-1}(t) = f(t)

Method

Compare the rule with $\dfrac{ax+b}{cx+d}$ .
If $d = -a$ , the function is self-inverse.
Then $f^{-1}(t)$ is just $f(t)$ .
Otherwise compute $f^{-1}$ first.

Why it works:

A self-inverse function undoes itself, so the inverse value equals the function value.

Try this

If $f(x) = \dfrac{4x+1}{x-4}$ , find $f^{-1}(5)$ .

Show solution

Here $a = 4$ , $d = -4$ , so $f$ is self-inverse.
$f^{-1}(5) = f(5) = \dfrac{21}{1} = 21$ .

Answer

21

Type 5common2 practice Q

Functional equation of sum type

How to spot it:

A rule like $f(x+y) = f(x) + f(y)$ (possibly with a constant added) and one value such as $f(1)$ are given.

f(x+y) = f(x) + f(y),\ f(1) = k \Rightarrow f(x) = kx

Method

Get $f(2)$ from $x = y = 1$ .
Climb one step at a time.
For the shifted rule, add the constant at every step.
State the general form if asked.

Why it works:

The rule tells how outputs combine, so known small values generate all the others.

Try this

If $f(x+y) = f(x) + f(y) + 3$ for all $x, y$ and $f(1) = 2$ , find $f(3)$ .

Show solution

$f(2) = 2 + 2 + 3 = 7$ .
$f(3) = f(2) + f(1) + 3 = 7 + 2 + 3$ .

Answer

12

Type 6occasional2 practice Q

Functional equation of product type

How to spot it:

A rule like $f(x+y) = f(x) \cdot f(y)$ and a value such as $f(1) = 3$ or $f(3) = 8$ are given.

f(x+y) = f(x)f(y),\ f(1) = k \Rightarrow f(x) = k^x

Method

Note the given value.
Each extra unit multiplies once more.
$f(n)$ equals the given value to the power $n$ (or a matching power).
Use powers of powers for big jumps.

Why it works:

The rule multiplies outputs as inputs add, which is exactly how powers behave.

Try this

If $f(x+y) = f(x) f(y)$ and $f(3) = 8$ , find $f(9)$ .

Show solution

$f(9) = f(3 + 3 + 3) = f(3)^3$ .
$= 8^3 = 512$ .

Answer

512

08

Formula sheet

Composite

(f \circ g)(x) = f(g(x))

$g$ acts first; work inside out.

Inverse check

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

Composition both ways must return $x$.

Inverse of (ax+b)/(cx+d)

f^{-1}(x) = \frac{b - dx}{cx - a}

From swap-and-solve; memorise for speed.

Sum-type equation

f(x+y) = f(x) + f(y),\ f(1) = k \Rightarrow f(x) = kx

Each unit step adds $k$.

Product-type equation

f(x+y) = f(x)f(y),\ f(1) = k \Rightarrow f(x) = k^x

Each unit step multiplies by $k$.

09

Shortcuts that save time

⚡ Work inside out

In $f(g(a))$ , compute $g(a)$ first, box it, then feed that number into $f$ . Two tiny calculations, no algebra.

Example

If $f(x) = x + 4$ and $g(x) = 3x$ , find $f(g(2))$ .

Show solution

$g(2) = 6$ .
$f(6) = 10$ .

Answer

10

⚡ Swap-and-solve for inverses

Write $y = f(x)$ , swap $x$ and $y$ , solve for $y$ . Works for every invertible rule, including fractions.

Example

Find the inverse of $f(x) = 5x - 2$ .

Show solution

$y = 5x - 2$ , so $x = 5y - 2$ .
$y = \dfrac{x+2}{5}$ .

Answer

$f^{-1}(x) = \dfrac{x+2}{5}$

⚡ Climb from the given value

Functional equations reward stepping: use $x = y = 1$ to get $f(2)$ , then $f(2)$ and $f(1)$ for $f(3)$ , and so on.

Example

$f(x+y) = f(x) + f(y)$ and $f(1) = 4$ . Find $f(5)$ .

Show solution

Each step adds $4$ : $f(2) = 8$ , $f(3) = 12$ .
$f(5) = 4 \times 5 = 20$ .

Answer

20

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Computing $f(g(x))$ as $g(f(x))$ — the order matters; in $f \circ g$ the inner rule $g$ acts first.

Mistake 02

Treating $f^{-1}(x)$ as $1/f(x)$ — the $-1$ means undo, not divide.

Mistake 03

Swapping $x$ and $y$ but not solving for $y$ — the inverse must end as ' $y =$ rule in $x$ '.

Mistake 04

Assuming every function has an inverse — $x^2$ sends $2$ and $-2$ to the same output, so it cannot be undone on all reals.

Mistake 05

In $f(x+y) = f(x) + f(y)$ , guessing $f(x) = x$ without using the given $f(1)$ — the given value fixes the multiple.

Mistake 06

For a shifted rule like $f(x+y) = f(x) + f(y) + 3$ , forgetting the extra $+3$ at every step.

11

Quick revision

Read this the night before the exam.

$f(g(x))$ : inside first, then outside.
Generally $f(g(x)) \ne g(f(x))$ .
Inverse: swap $x$ and $y$ , solve for $y$ .
$\dfrac{ax+b}{cx+d}$ has inverse $\dfrac{b-dx}{cx-a}$ ; self-inverse when $d = -a$ .
Sum type: $f(x) = kx$ . Product type: $f(x) = k^x$ .
No inverse when two inputs share one output.

12

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 10 min · wrong answers go to your mistake notebook automatically.