Functions Questions, Worksheets and Revision

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

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 AQA Edexcel OCR WJEC

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}

Level 8-9 GCSE AQA Edexcel OCR WJEC

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}

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