A-Level Maths (Edexcel 9MA0) · Pure Year 2 · Chapter 2: Functions and graphs · Lesson 3

Inverse functions

124 marks

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Quick questions to start.

A1
Needed today
Make $x$ the subject of $y = \frac{3 x + 1}{x - 2}$ .
[3 marks]
Answer
A2
Last lesson
$f (x) = 2 x^{2} - 12 x + 7$ , $x \in R$ . Find the range of $f$ .
[3 marks]
Answer
A3
Last lesson
$f (x) = x^{2} - 1$ and $g (x) = 3 x + 2$ , $x \in R$ . Find $g f (2)$ .
[2 marks]
Answer
A4
From Year 1
Solve $ln (2 x - 1) = 3$ , giving your answer in exact form.
[2 marks]
Answer

Example, then your turn

A-Level

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

Example 1

The function $f$ is defined by $f (x) = 4 x - 7$ , $x \in R$ . Find $f^{- 1} (x)$ and $f^{- 1} (5)$ .

Hint Write

y = f (x)

, make

x

the subject, then swap the letters.

B1
The function $f$ is defined by $f (x) = 3 x + 8$ , $x \in R$ . Find $f^{- 1} (x)$ and $f^{- 1} (2)$ .
[3 marks]
Answer

Example 2

$f (x) = \frac{x + 5}{x - 1}$ , $x > 1$ . Find $f^{- 1} (x)$ and state its domain.

Hint The domain of

f^{- 1}

is the range of

f

. Write

f (x)

1 + \frac{6}{x - 1}

to see the range.

B2
$f (x) = \frac{2 x + 3}{x - 4}$ , $x > 4$ . Find $f^{- 1} (x)$ and state its domain.
[4 marks]
Answer

Example 3

$f (x) = (x - 2)^{2} + 3$ , $x \geq 2$ . Find $f^{- 1} (x)$ , and state the domain and range of $f^{- 1}$ .

Hint On

x \geq 2

the bracket

x - 2

0

or more, so take the positive square root.

B3
$f (x) = x^{2} + 6 x + 1$ , $x \geq - 3$ . Find $f^{- 1} (x)$ , and state the domain and range of $f^{- 1}$ .
[5 marks]
Answer

Example 4

$f (x) = x^{2} + 1$ , $x \geq 0$ . Find $f^{- 1} (x)$ and sketch $y = f (x)$ and $y = f^{- 1} (x)$ on the same axes.

Hint The graph of

y = f^{- 1} (x)

is the reflection of

y = f (x)

in the line

y = x

B4
$f (x) = e^{x} + 2$ , $x \in R$ .
(a) Find $f^{- 1} (x)$ and state its domain.
(b) Give the coordinates of the point where $y = f^{- 1} (x)$ crosses the $x$ -axis.
[4 marks]
Answer

Example 5

$f (x) = x^{2} - 2$ , $x \geq 0$ . Find the point where the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ meet.

Hint

f

is increasing on its domain, so the two graphs meet on the line

y = x

. Solve

f (x) = x

B5
$f (x) = x^{2} - 6$ , $x \geq 0$ . Find the coordinates of the point where the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ meet.
[4 marks]
Answer

Practice

A-Level

C1
$f (x) = 2 x + 5$ , $x \in R$ . Sam writes: " $f^{- 1} (x) = \frac{1}{2 x + 5}$ ."
(a) Find the correct $f^{- 1} (x)$ .
(b) Show that Sam's function gives the wrong value of $f^{- 1} (7)$ .
[4 marks]
Answer

C2
For each function, state whether it has an inverse. Give a reason each time.
(a) $f (x) = x^{2} - 4 x$ , $x \in R$
(b) $g (x) = x^{2} - 4 x$ , $x \geq 2$
(c) $h (x) = \frac{1}{x}$ , $x \neq = 0$
[5 marks]
C3
$f (x) = \frac{3 x - 5}{x - 3}$ , $x \in R$ , $x \neq = 3$ . Show that $f$ is self-inverse, and hence write down $f f (7)$ .
[4 marks]
Answer
C4
$f (x) = x^{2} + 4 x$ , $x \geq - 2$ . Leo says: "The domain of $f$ is $x \geq - 2$ , so the domain of $f^{- 1}$ is also $x \geq - 2$ ."
Explain why Leo is wrong. Find $f^{- 1} (x)$ and its domain.
[5 marks]
Answer
C5
$f (x) = ln (2 x + 1)$ , $x > - \frac{1}{2}$ .
(a) Find $f^{- 1} (x)$ and state its domain.
(b) Solve $f^{- 1} (x) = 4$ , giving your answer in the form $a ln b$ where $a$ and $b$ are integers.
[5 marks]
Answer
C6
$f (x) = 2 x - 1$ and $g (x) = x^{3}$ , $x \in R$ .
(a) Find $(f g)^{- 1} (x)$ .
(b) Show that $(f g)^{- 1} (x) = g^{- 1} f^{- 1} (x)$ .
[5 marks]
Answer

Exam-style questions

Edexcel exam style

D1
The function $f$ is defined by $f (x) = \frac{2 x + 1}{x - 3}$ , $x > 3$ .
(a)
Find the range of $f$ .
[2 marks]
Answer
(b)
Find $f^{- 1} (x)$ and state its domain.
[4 marks]
Answer
(c)
Solve $f^{- 1} (x) = 7$ .
[2 marks]
Answer
(d)
The function $g$ is defined by $g (x) = x^{2} + 2$ , $x \geq 0$ . Using your answer to part (b), or otherwise, find the exact value of $a$ for which $f^{- 1} g (a) = 4$ .
[3 marks]
Answer
Total for D1: 11 marks

D2
$f (x) = x^{2} - 8 x + 10$ , $x \geq 2$ . Mia says: "The domain of $f$ has been restricted, so $f$ has an inverse."
Mia is wrong.
(a)
Show that $f$ does not have an inverse.
[2 marks]
(b)
The domain is changed to $x \geq k$ . Find the least value of $k$ for which $f$ has an inverse.
[2 marks]
Answer
(c)
With this value of $k$ , find $f^{- 1} (x)$ and state its domain.
[4 marks]
Answer
Total for D2: 8 marks
D3
The functions $f$ and $g$ are defined by $f (x) = x^{2} - 8 x + 21$ , $x \geq 4$ , and $g (x) = x^{2} - 8 x + 21$ , $x \in R$ . Find $f^{- 1} (x)$ and state its domain. Show that $f^{- 1} g (x) = 4 + ∣ x - 4∣$ . Hence state the set of values of $x$ for which $f^{- 1} g (x) = x$ , and solve $f^{- 1} g (x) = 2 x$ .
[9 marks]
Answer
D4
The functions $f$ and $g$ are defined by $f (x) = e^{2 x} + 3$ , $x \in R$ , and $g (x) = ln (x - 3)$ , $x > 3$ .
(a)
Find $f^{- 1} (x)$ and state its domain.
[3 marks]
Answer
(b)
Show that $g f (x) = 2 x$ .
[2 marks]
(c)
Find $f g (x)$ in its simplest form and solve $f g (x) = 19$ .
[4 marks]
Answer
Total for D4: 9 marks
D5
The function $f$ is defined by $f (x) = \frac{4}{x + 2}$ , $x \geq 0$ .
(a)
Find the range of $f$ .
[2 marks]
Answer
(b)
Find $f^{- 1} (x)$ and state its domain.
[3 marks]
Answer
(c)
Solve $f (x) = f^{- 1} (x)$ , giving your answer in surd form.
[3 marks]
Answer
(d)
Show that $f f (x) = \frac{2 ( x + 2 )}{x + 4}$ .
[3 marks]
Total for D5: 11 marks
D6
The function $f$ is defined by $f (x) = 2 x + 6$ , $x \geq - 3$ .
(a)
Find $f^{- 1} (x)$ and state its domain.
[3 marks]
Answer
(b)
State the range of $f^{- 1}$ .
[2 marks]
Answer
(c)
On the same axes, sketch $y = f (x)$ and $y = f^{- 1} (x)$ . Give the coordinates of the points where each graph meets the coordinate axes.
[4 marks]
Answer
Total for D6: 9 marks

Extension

Stretch

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

E1
The function $f$ is defined by $f (x) = x^{2} - 2 p x + 3$ , $x \geq p$ , where $p$ is a positive constant. Find the set of values of $p$ for which the graphs of $y = f (x)$ and $y = f^{- 1} (x)$ meet at exactly two points.
[9 marks]
Hint 1 · What to try
$f$ is increasing on $x \geq p$ . So the two graphs can only meet on the line $y = x$ . Solve $f (x) = x$ .
Hint 2 · The first line
$x^{2} - (2 p + 1) x + 3 = 0$ needs two different roots, and both roots must be $p$ or more, since the domain of $f$ is $x \geq p$ .
Hint 3 · The full method
The larger root is always more than $p$ . The smaller root is $\frac{2 p + 1 - ( 2 p + 1 ) ^{2} - 12}{2}$ . Find when this is $p$ or more.
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can say whether a function has an inverse, and restrict its domain so that it does.
I can find f⁻¹(x) by writing y = f(x) and making x the subject.
I can state the domain and range of f⁻¹ from the range and domain of f.
I can sketch y = f⁻¹(x) as the reflection of y = f(x) in y = x and solve f(x) = f⁻¹(x).

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