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.

## 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

## 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

## 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

## Maths Exam Worksheets

£3.99

### 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

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