# Functions Questions, Worksheets and Revision

GCSE 4 - 5GCSE 6 - 7AQAEdexcelOCRWJECFoundationHigherAQA 2022Edexcel 2022WJEC 2022

## Functions

In maths, a function is something that takes an input and produces an output. Functions may be given in the form of function machines – or they may be given as mathematical expressions.

Make sure you are happy with the following topics before continuing.

Level 4-5 GCSE

## Skill 1: Evaluating Functions

Evaluating functions involves putting numbers into the function to get the result.

Example: A function is given by $f(x) = 3x+1$, Find $f(10)$

All this requires is to replace $x$ with $10$ and calculate the result.

When we input 10 into this function that would look like:

$f(\textcolor{red}{10}) = 3\times \textcolor{red}{10} + 1 = 31$.

Level 6-7 GCSE
Level 6-7 GCSE

## Type 2: Composite Functions

A composite function is the result of one function being applied immediately after the other.

Example: Let $f(x)=\textcolor{red}{2x-3}$ and $g(x)=\textcolor{blue}{x+1}$, find $fg(x)$

To find $fg(x)$ we replace $x$ in $f(x)$ with $g(x)$

$fg(x) = f(g(x)) = \textcolor{red}{2(}\textcolor{blue}{x+1}\textcolor{red}{) - 3}$

Next we can expand the brackets and simplify if required.

$\textcolor{red}{2(}\textcolor{blue}{x+1}\textcolor{red}{) - 3} = 2x+2-3 = 2x-1$

Level 8-9 GCSE

## Type 3: Inverse Functions

An inverse function is a function acting in reverse. The inverse function of $f(x)$ is given by $f^{-1}(x)$, and it tells us how to go from an output of $f(x)$ back to its input.

Example: Given that $f(x) = \dfrac{x+8}{3}$, find $f^{-1}(x)$

Step 1: Write the equation in the form $x = f(y)$

For this we need to replace all the $x$‘s in the equation with $y$‘s and set the equation equal to $x$

$f(x) = \dfrac{x+8}{3}$ becomes $x= \dfrac{y+8}{3}$

Step 2: Rearrange the equation to make $y$ the subject.

\begin{aligned}x&= \dfrac{y+8}{3} \\ 3x& = y+8 \\ 3x-8 &= y \end{aligned}

Step 3: Replace $y$ with $f^{-1}(x)$

\begin{aligned}y & = 3x-8 \\ f^{-1}(x) & = 3x-8\end{aligned}

Level 8-9 GCSE
Level 6-7 GCSE

## Example 1: Composite Functions

Let $f(x)=x-3$ and $g(x)=x^2$

[4 marks]

Find:

a) $fg(10)$ – we must find $g(10)$ then apply $f(x)$ to the answer.

$g(10) = 10^2 = 100$ so $fg(10) = f(100) = 100 - 3 = 97$.

b) $gf(-4)$ – we must find $f(-4)$ then apply $g(x)$ to the answer.

$f(-4) = -4-3 = -7$ so $gf(-4) = g(-7) = (-7)^2 = 49$

c) an expression for $fg(x)$ – we need to input $g(x)$ into $f(x)$. So, we get

$fg(x) = f\left(g(x)\right) = g(x) - 3 = x^2 - 3$

Level 8-9 GCSE

## Example 2: Inverse Functions

Given that $f(x) = 3x - 9$, find $f^{-1}(x)$

[3 marks]

Step 1: Write the equation in the form $x = f(y)$

$f(x) = 3x- 9$ becomes $x = 3y-9$

Step 2: Rearrange to make $y$ the subject

\begin{aligned}x &= 3y-9 \\ x+9 &= 3y \\ \dfrac{x+9}{3} &=y\end{aligned}

Step 3: Replace $y$with $f^{-1}(x)$

\begin{aligned} \dfrac{x+9}{3} & = y \\ f^{-1}(x) & = \dfrac{x+9}{3}\end{aligned}

Level 8-9 GCSE

## Example Questions

a) Substituting $x=10$ into $f(x)$, we find,

$f(10) = \dfrac{10}{3(10)-5} = \dfrac{10}{25}= \dfrac{2}{5}=0.4$

b) Substituting $x=2$ into $f(x)$, we find,

$f(10) = \dfrac{10}{3(2)-5} = \dfrac{10}{1}= 10$

c) Substituting $x=-1$ into $f(x)$, we find,

$f(10) = \dfrac{10}{3(-1)-5} = \dfrac{10}{-8}=-\dfrac{5}{4} =1.25$

a) Substituting $x=4$ into $g(x)$, then substituting the result into $f(x)$,

$g(4) = (2\times 4) - 5 = 8 - 5 = 3$

$fg(4) = f(3) = \dfrac{15}{3} = 5$

b) For $gf(-30)$ we must first find $f(-30)$ and then substitute the result into $g(x)$,

$f(-30) = \dfrac{15}{-30} = -\dfrac{1}{2}$

$gf(-30) = g(-\dfrac{1}{2}) = 2(-\dfrac{1}{2}) - 5 = -1 - 5 = -6$

c) To find an expression for $gf(x)$, substitute $f(x)$ in for every instance of $x$ in $g(x)$,

$gf(x) = 2(f(x)) - 5 = 2\times(\dfrac{15}{x}) - 5 = \dfrac{30}{x} - 5$

So, we need to write the function as $y=\frac{5}{x-4}$ and rearrange this equation to make $x$ the subject. Then, we will swap every $y$ with an $x$ – and vice versa.

We won’t be able to get $x$ on its own whilst it’s in the denominator, so our first step will be multiplying both sides by $(x-4)$:

$y(x-4)=5$

Then, divide both sides by $y$:

$x-4=\dfrac{5}{y}$

Finally, add 4 to both sides to make $x$ the subject:

$x=\dfrac{5}{y}+4$

Now, swap each $x$ with a $y$ and vice versa to get

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

So, we need to write the function as $g=\frac{4}{x}+3$ and rearrange this equation to make $x$ the subject. Then, we will swap every $g$ with an $x$ – and vice versa.

The first step is to subtract $3$ from both sides,

$g-3=\dfrac{4}{x}+\cancel{3}-\cancel{3}$

Then, multiply both sides by $x$:

$x(g-3)=4$

Finally, divide both sides by $(g-3)$ to make $x$ the subject:

$x=\dfrac{4}{g-3}$

Now, simply swap each $x$ with a $g$ and vice versa to get,

$g^{-1}(x)=\dfrac{4}{x-3}$

Level 1-3GCSEKS3

Level 6-7GCSE

## Worksheet and Example Questions

### (NEW) Functions (The basics) Exam Style Questions - MME

Level 4-5 GCSENewOfficial MME

### (NEW) Functions (Composite and inverse) Exam Style Questions - MME

Level 6-7 GCSENewOfficial MME

Level 6-7 GCSE

## You May Also Like...

### GCSE Maths Revision Cards

Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC.

£8.99

### GCSE Maths Revision Guide

The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!

From: £14.99

### Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99