# Functions

## Functions Revision

**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.

**Type 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

**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

**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}

**Note:**

Sometimes functions are displayed in the following form:

f:x \rightarrow 3x-5

This is the same as f(x)=3x-5

**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

**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 ywith f^{-1}(x)

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

## Functions Example Questions

**Question 1:** Let f(x) = \dfrac{10}{3x-5}

a) Find f(10)

b) Find f(2)

c) Find f(-1)

**[4 marks]**

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

**Question 2:** Let f(x) = \dfrac{15}{x} and g(x) = 2x - 5

a) Find fg(4)

b) Find gf(-30)

c) Find gf(x)

**[4 marks]**

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

**Question 3:** Find the inverse function of f(x) = \dfrac{5}{x-4}

**[3 marks]**

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

**Question 4:** Find the inverse function of g(x) = \dfrac{4}{x}+3

**[3 marks]**

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}

## Functions Worksheet and Example Questions

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

Level 6-7NewOfficial MME### (NEW) Functions (Composite and inverse) Exam Style Questions - MME

Level 8-9GCSENewOfficial MME## Functions Drill Questions

### Functions - Drill Questions

Level 6-7GCSE## MME Premium Membership

£19.99

/monthLearn an entire GCSE course for maths, English and science on the most comprehensive online learning platform. With revision explainer videos & notes, practice questions, topic tests and full mock exams for each topic on every course, it’s easy to Learn and Revise with the MME Learning Portal.

Sign Up Now