AQA Level 2 Further Maths · Level 2 Certificate · Chapter 3: Algebra III · Lesson 1

Function notation, domain and range

About 75 minutesNo calculator66 marks

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Date

Do Now

The first questions are what today's lesson needs. The others bring back earlier topics.

A1
Needed today
Work out the value of $2 x^{2} - 3 x$ when $x = - 4$ .
[2 marks]
Answer
A2
Needed today
Solve $- 3 < 2 x + 1 ⩽ 9$ .
[2 marks]
Answer
A3
Last lesson
Write $x^{2} + 16 x + 70$ in the form $(x + a)^{2} + b$ .
[2 marks]
Answer
A4
From chapter 1
Expand and simplify $(3 + 2) (5 - 22)$ .
[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

$f (x) = x^{2} - 4 x + 1$ . Work out $f (- 3)$ , and write $f (x + 2)$ in its simplest form.

Hint Put

- 3

in place of every

x

, in brackets. For

f (x + 2)

, put

(x + 2)

in place of every

x

, then expand.

B1
$h (x) = 2 x^{2} + 3 x - 5$ . Work out $h (- 2)$ , and write $h (x - 1)$ in its simplest form.
[3 marks]
Answer

Example 2

$f (x) = \frac{2 x + 6}{x - 1}$ , $x \neq = 1$ . Solve $f (x) = 4$ , and then solve $f (2 x) = 3$ .

Hint Multiply both sides by the denominator. For

f (2 x)

, write

2 x

in place of every

x

first.

B2
$g (x) = \frac{5 x - 1}{x + 3}$ , $x \neq = - 3$ . Solve $g (x) = 2$ , and then solve $g (2 x) = 1$ .
[4 marks]
Answer

Example 3

$f (x) = a x^{2} + b$ . $f (1) = 5$ and $f (3) = 21$ . Work out the values of $a$ and $b$ .

Hint Each value of

f

gives an equation in

a

and

b

. Solve the two equations together.

B3
$f (x) = p x^{2} + q x$ . $f (2) = 10$ and $f (- 1) = 7$ . Work out the values of $p$ and $q$ .
[3 marks]
Answer

Example 4

$f (x) = 9 - 2 x$ with domain $- 3 ⩽ x < 5$ . Work out the range of $f$ . Give your answer as an inequality.

Hint Work out

f

at each end of the domain. The line goes down, so the left end gives the largest value. An end that is not in the domain gives a value that is not in the range.

B4
$g (x) = 10 - 4 x$ with domain $- 1 ⩽ x < 3$ . Work out the range of $g$ . Give your answer as an inequality.
[2 marks]
Answer

Example 5

$f (x) = x^{2} - 6 x + 5$ with domain $0 ⩽ x < 7$ . Work out the range of $f$ .

Not drawn accurately

Hint Complete the square to find the minimum point. Check whether it is inside the domain, then work out

f

at each end.

B5
$g (x) = x^{2} + 4 x - 1$ with domain $- 5 < x ⩽ 1$ . Work out the range of $g$ .
[3 marks]
Answer

Practice

Level 2

C1
$f (x) = x^{2} - 1$ . Write $[f (x)]^{2} - f (x^{2})$ in its simplest form.
[3 marks]
Answer
C2
$f (x) = 4 x + 7$ . The range of $f$ is $- 5 < f (x) ⩽ 19$ . Work out the domain of $f$ .
[2 marks]
Answer

C3
$f (x) = 3 x^{2} - 18 x + 31$ for all values of $x$ . Work out the range of $f$ .
[3 marks]
Answer
C4
$g (x) = 2 x - 10$ . Write down the largest possible domain of $g$ , and the range of $g$ for that domain.
[2 marks]
Answer
C5
$f (x) = 1 - x^{3}$ with domain $- 1 < x ⩽ 2$ . Work out the range of $f$ .
[2 marks]
Answer
C6
$f (x) = \frac{12}{x}$ with domain $x ⩾ 4$ . Work out the range of $f$ .
[2 marks]
Answer

Exam-style questions

AQA exam style

D1
Sam works out the range of $f (x) = x^{2} - 2 x$ with domain $- 1 < x ⩽ 4$ . Sam writes " $f (- 1) = 3$ and $f (4) = 8$ , so the range is $3 < f (x) ⩽ 8$ ".
Explain why Sam is wrong. Work out the correct range.
[3 marks]
Answer

D2
This question is about three functions.
(a)
$f (x) = \frac{4 x + 1}{5 x - 2}$ . Which value of $x$ cannot be in the domain of $f (x)$ ? Circle your answer. $- \frac{1}{4} 0 \frac{2}{5} \frac{5}{2}$
[1 mark]
Answer
(b)
$g (x) = 18 - 5 x$ with domain $- 3 < x ⩽ 4$ . Work out the range of $g$ . Give your answer as an inequality.
[2 marks]
Answer
(c)
$h (x) = x^{2} + 6 x + 4$ with domain $- 5 ⩽ x ⩽ 1$ . Work out the range of $h$ .
[3 marks]
Answer
Total for D2: 6 marks
D3
$f (x) = 3 x + k$ with domain $2 ⩽ x ⩽ 6$ , where $k$ is a constant. The range of $f$ is $1 ⩽ f (x) ⩽ c$ . Work out the values of $k$ and $c$ .
[3 marks]
Answer
D4
$f (x) = x^{2} - x - 6$
(a)
Show that $f (x + 3) = x^{2} + 5 x$ .
[2 marks]
(b)
Hence solve $f (x + 3) = f (x)$ .
[2 marks]
Answer
(c)
Solve $f (2 x) = 14$ .
[2 marks]
Answer
Total for D4: 6 marks
D5
The sketch shows the graph of $y = f (x)$ . The curve has a minimum point at $(1, - 4)$ . Its end points are $(- 2, 5)$ and $(6, 21)$ .
Not drawn accurately
(a)
Write down the domain and the range of $f$ .
[2 marks]
Answer
(b)
The equation $f (x) = k$ has exactly two solutions. Work out the range of values of $k$ .
[2 marks]
Answer
(c)
$f (x) = x^{2} + p x + q$ . Work out the values of $p$ and $q$ .
[2 marks]
Answer
Total for D5: 6 marks

Extension

Stretch

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

E1
$f (x) = x^{2} - 4 x + 7$ with domain $a ⩽ x ⩽ 5$ , where $a$ is a constant. The range of $f$ is $3 ⩽ f (x) ⩽ 12$ . Work out all the possible values of $a$ .
[5 marks]
Hint 1 · What to try
Complete the square and sketch the curve. Which value of $x$ gives the smallest value $3$ , and which gives $12$ ?
Hint 2 · The first line
$f (x) = (x - 2)^{2} + 3$ . The smallest value $3$ is at $x = 2$ , so $x = 2$ must be in the domain: $a ⩽ 2$ .
Hint 3 · The full method
$f (5) = 12$ , so the largest value is $12$ as long as $f (a) ⩽ 12$ . $(a - 2)^{2} + 3 ⩽ 12$ gives $(a - 2)^{2} ⩽ 9$ , so $- 1 ⩽ a ⩽ 5$ . With $a ⩽ 2$ , the answer is $- 1 ⩽ a ⩽ 2$ .
Answer

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can work out $f (a)$ and write $f$ of an expression in its simplest form.
I can solve equations written in function notation, including with fractions.
I can find the range of a function from its domain, checking for a turning point inside it.
I can say which values cannot be in a domain, and find a domain from a range.

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