 Functions Questions | Worksheets and Revision | MME

# Functions Questions, Worksheets and Revision

Level 6-7

## 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

## Test your skills with online exams on the MME Revision Platform

##### 5 Question Types

Our platform contains 5 question types: simple, multiple choice, multiple answers, fraction and image based questions. More question types are coming soon.

##### Written Solutions

Get written solutions for every single exam question, detailing exactly how to approach and answer each one, no matter the difficulty or topic.

Every exam attempt is stored against your unique student profile, meaning you can view all previous exam and question attempts to track your progress over time.

## 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
Level 6-7

## 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

## 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
Level 6-7

## 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

## 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 Maths Revision Cards

(252 Reviews) £8.99

### 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,

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

### Worksheets and Exam Questions

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

Level 4-5 New Official MME

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

Level 6-8 New Official MME