AQA Level 2 Further Maths · Level 2 Certificate · Chapter 3: Algebra III · Lesson 6

Inverse functions

About 80 minutesNo calculator70 marks

Hints online, answers free with an account:
mathswariz.uk/worksheets/further-maths-inverse-functions/

Date

Do Now

Quick questions to start.

A1
Needed today
Make $x$ the subject of $y = \frac{3 x - 4}{5}$ .
[2 marks]
Answer
A2
Needed today
Make $x$ the subject of $y = 2 x^{2} + 7$ , where $x > 0$ .
[2 marks]
Answer
A3
Last lesson
Work out the coordinates of the turning point of $y = x^{2} + 8 x + 11$ .
[2 marks]
Answer $(\dots\dots, \dots\dots)$
A4
From chapter 3
$f (x) = x^{2} - 1$ and $g (x) = 2 x + 3$ . Work out $f g (- 2)$ .
[2 marks]
Answer

Example, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it.

Example 1

$f (x) = 4 x - 7$ . Work out an expression for $f^{- 1} (x)$ .

Hint Write

y = 4 x - 7

, swap

x

and

y

, then make

y

the subject.

B1
$f (x) = \frac{5 x + 2}{3}$ . Work out an expression for $f^{- 1} (x)$ .
[2 marks]
Answer $f^{- 1} (x) = \dots\dots\dots\dots$

Example 2

$f (x) = x^{2} + 3$ for $x \geq 0$ . Work out an expression for $f^{- 1} (x)$ .

Hint Swap

x

and

y

. Only the positive square root is used, because

x \geq 0

f

B2
$f (x) = 2 x^{2} - 5$ for $x \geq 0$ . Work out an expression for $f^{- 1} (x)$ .
[3 marks]
Answer $f^{- 1} (x) = \dots\dots\dots\dots$

Example 3

$f (x) = (x - 3)^{2} + 1$ for $x \geq 3$ . Work out $f^{- 1} (x)$ and state the domain of $f^{- 1}$ .

Hint The domain of

f^{- 1}

is the range of

f

. The least value of

(x - 3)^{2} + 1

1

B3
$f (x) = (x + 2)^{2} - 6$ for $x \geq - 2$ . Work out $f^{- 1} (x)$ and state the domain of $f^{- 1}$ .
[4 marks]
Answer $f^{- 1} (x) = \dots\dots\dots\dots$ domain $\dots\dots\dots\dots$

Example 4

$f (x) = \frac{6}{x - 1}$ for $x > 1$ . Work out $f^{- 1} (x)$ and state the domain of $f^{- 1}$ .

Hint For

x > 1

the denominator is positive, so every value of

f (x)

is positive.

B4
$f (x) = \frac{12}{x + 3}$ for $x > 0$ . Work out $f^{- 1} (x)$ and state the domain of $f^{- 1}$ .
[4 marks]
Answer $f^{- 1} (x) = \dots\dots\dots\dots$ domain $\dots\dots\dots\dots$

Example 5

$f (x) = 3 x - 8$ . Solve $f (x) = f^{- 1} (x)$ .

Hint Find

f^{- 1} (x)

first. Then put the two expressions equal.

B5
$f (x) = 5 x + 12$ . Solve $f (x) = f^{- 1} (x)$ .
[3 marks]
Answer $x = \dots\dots$

Practice

Level 2

C1
$f (x) = 7 - 2 x$ . Work out $f^{- 1} (3)$ by first finding an expression for $f^{- 1} (x)$ .
[2 marks]
Answer $f^{- 1} (3) = \dots\dots$
C2
$f (x) = x + 4$ for $x \geq - 4$ . Work out $f^{- 1} (x)$ and state the domain of $f^{- 1}$ .
[3 marks]
Answer $f^{- 1} (x) = \dots\dots\dots$ domain $\dots\dots\dots$

C3
$f (x) = 2 x^{3} + 1$ . Work out $f^{- 1} (x)$ . Hence work out $f^{- 1} (55)$ .
[3 marks]
Answer $f^{- 1} (x) = \dots\dots\dots$ $f^{- 1} (55) = \dots\dots$
C4
$f (x) = (x - 4)^{2}$ for all values of $x$ . Explain why $f$ has no inverse function. Then give a domain of the form $x \geq k$ on which $f$ does have an inverse.
[2 marks]
Answerdomain $x \geq \dots\dots$
C5
$f (x) = 3 x + a$ , where $a$ is a constant. $f^{- 1} (14) = 4$ . Work out the value of $a$ .
[2 marks]
Answer $a = \dots\dots$
C6
$g (x) = \frac{x + 1}{x - 2}$ for $x \neq = 2$ . Work out an expression for $g^{- 1} (x)$ .
[3 marks]
Answer $g^{- 1} (x) = \dots\dots\dots\dots$

Exam-style questions

AQA exam style

D1
$f (x) = \frac{x - 6}{2}$ . Circle the expression for $f^{- 1} (x)$ .
$2 x - 62 x + 6 \frac{x + 6}{2} \frac{2}{x - 6}$
[1 mark]
Answer

D2
$f (x) = x^{2} - 9$ for $x \geq 0$ . Priya says that $f^{- 1} (x) = x + 3$ , because $9 = 3$ .
Priya is wrong.
(a)
Use $f (4)$ to show that Priya is wrong.
[1 mark]
(b)
Work out the correct expression for $f^{- 1} (x)$ and state its domain.
[3 marks]
Answer $f^{- 1} (x) = \dots\dots\dots$ domain $\dots\dots\dots$
Total for D2: 4 marks
D3
$f (x) = x^{2} + 5$ for $x \geq 0$
$g (x) = 3 x - 1$ for all $x$
(a)
Work out $f^{- 1} (x)$ and state its domain.
[3 marks]
Answer $f^{- 1} (x) = \dots\dots\dots$ domain $\dots\dots\dots$
(b)
Show that $g^{- 1} f (x)$ can be written as $\frac{x ^{2} + a}{3}$ , where $a$ is an integer.
[2 marks]
Answer $a = \dots\dots$
(c)
Show that the equation $f^{- 1} (x) = g^{- 1} (x)$ has no solutions. You must show your working.
[3 marks]
Total for D3: 8 marks
D4
$f (x) = 2 x^{2} - a$ for $x \geq 0$ , where $a$ is a positive constant.
$f^{- 1} (3 a) = 4$
Work out the value of $a$ . You must show your working.
[4 marks]
Answer $a = \dots\dots$
D5
$f (x) = \frac{8}{x - 2}$ for $x > 2$
(a)
State the range of $f$ .
[1 mark]
Answer $\dots\dots\dots\dots$
(b)
Work out $f^{- 1} (x)$ .
[2 marks]
Answer $f^{- 1} (x) = \dots\dots\dots\dots$
(c)
Solve $f (x) = f^{- 1} (x)$ . You must show your working.
[4 marks]
Answer $x = \dots\dots$
Total for D5: 7 marks
D6
The function $f$ has inverse $f^{- 1} (x) = 2 + x - 1$ for $x \geq 1$ .
Work out $f (x)$ and state the domain of $f$ .
[3 marks]
Answer $f (x) = \dots\dots\dots$ domain $\dots\dots\dots$

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
$f (x) = \frac{a x + 7}{x - 3}$ for $x \neq = 3$ , where $a$ is a constant. The inverse function $f^{- 1}$ is the same function as $f$ . Work out the value of $a$ .
[4 marks]
Hint 1 · What to try
Find $f^{- 1} (x)$ in terms of $a$ , then compare it with $f (x)$ .
Hint 2 · The first line
$x = \frac{a y + 7}{y - 3}$ gives $x y - 3 x = a y + 7$ , so $y (x - a) = 3 x + 7$ .
Hint 3 · The full method
$f^{- 1} (x) = \frac{3 x + 7}{x - a}$ . This is $\frac{a x + 7}{x - 3}$ for every $x$ only when $a = 3$ .
Answer $a = \dots\dots$

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can find the inverse of a function by swapping $x$ and $y$ and rearranging.
I can choose the right square root when a function is on a restricted domain.
I can state the domain of $f^{- 1}$ as the range of $f$ .
I can solve equations that use $f^{- 1}$ , such as $f (x) = f^{- 1} (x)$ .

Answers and mark scheme

The worked answers and the full mark scheme are free with a Mathswariz account. Teachers and students can both sign up.

Get the answers