Function notation and evaluating rules | Year 11

01 / LEARN

The idea, step by step

A rule with named inputs

A function is a rule that assigns exactly one output to every input in its domain. The notation $f (3)$ means the output of the rule f when the input is 3. It does not mean f multiplied by 3. Different letters name different rules; the same input can therefore have different outputs under f and g.

Replace every occurrence of the input variable by the supplied value. Put negative numbers or algebraic expressions in brackets before applying powers and other operations. This makes the order of operations visible and prevents a sign error.

f (x) = 2 x^{2} - 3 x + 1

Values of f(x)=2x²-3x+1
x	f(x)
-2	15
-1	6
0	1
1	0
2	3

Inputs can be expressions

To find $f (a + 1)$ , substitute the whole expression $a + 1$ wherever x appeared. Only expand after the substitution is written correctly. When a question asks you to solve $f (x) = 7$ , the output is known and the input is unknown; this is an equation to solve, rather than an evaluation.

An input-output table is a useful check, but a few entries do not define a rule uniquely. Use the given rule throughout and retain any stated domain restrictions.

A common trap

Writing $f (- 2) = 2 (- 2^{2}) - 3 (- 2) + 1$ and obtaining a negative square.

Instead: The substituted input is the entire number -2, so write ${(- 2)}^{2} = 4$ . Brackets make the intended input unambiguous.

02 / WORKED EXAMPLES

See the thinking

Example 1

A negative input

For $f (x) = 2 x^{2} - 3 x + 1$ , find $f (- 2)$ .

Substitute the whole input: $f (- 2) = 2 {(- 2)}^{2} - 3 (- 2) + 1$ .
Square first: $2 (4) + 6 + 1 = 15$ .

15

Example 2

An expression as input

If $g (x) = 3 x - 4$ , simplify $g (t + 2)$ and solve $g (x) = 11$ .

Replace x by t+2: $g (t + 2) = 3 (t + 2) - 4 = 3 t + 2$ .
For the known output, solve $3 x - 4 = 11$ : $3 x = 15$ , so $x = 5$ .

g (t + 2) = 3 t + 2

;

x = 5

Check your understanding

If $h (x) = x^{2} + 2 x$ , what is $h (- 3)$ ?

Show answer

$3$

${(- 3)}^{2} + 2 (- 3) = 9 - 6 = 3$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $f (x) = 4 x - 7$ , find $f (3)$ .
2 marks
For $f (x) = x^{2} - 5 x$ , find $f (- 2)$ .
2 marks
If $p (x) = 2 x + 5$ , simplify $p (a - 3)$ .
2 marks
Find the input for which $g (x) = 7 x + 2$ has output 30.
2 marks
Let $r (x) = 12 / (x + 1)$ . Find $r (2)$ and state an input that cannot be used.
2 marks
A delivery fee is $C (d) = 5 + 1.8 d$ dollars for d kilometres. Find and interpret $C (10)$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $f (x) = 4 x - 7$ , find $f (3)$ .
2 marks
$f (3) = 4 (3) - 7 = 12 - 7 = 5$ .
$5$
For $f (x) = x^{2} - 5 x$ , find $f (- 2)$ .
2 marks
$f (- 2) = {(- 2)}^{2} - 5 (- 2) = 4 + 10 = 14$ .
$14$
If $p (x) = 2 x + 5$ , simplify $p (a - 3)$ .
2 marks
$p (a - 3) = 2 (a - 3) + 5 = 2 a - 1$ .
$2 a - 1$
Find the input for which $g (x) = 7 x + 2$ has output 30.
2 marks
Solve $7 x + 2 = 30$ .
Then $7 x = 28$ , so $x = 4$ .
$x = 4$
Let $r (x) = 12 / (x + 1)$ . Find $r (2)$ and state an input that cannot be used.
2 marks
$r (2) = 12 / 3 = 4$ .
At x=-1 the denominator is zero.
$r (2) = 4$ ; $x = - 1$ is excluded.
A delivery fee is $C (d) = 5 + 1.8 d$ dollars for d kilometres. Find and interpret $C (10)$ .
2 marks
$C (10) = 5 + 1.8 (10) = 23$ .
The output is the total fee in dollars, not the distance.
$23$ for a 10 km delivery.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Topic 1: Functions and graphs. Prerequisite algebra and coordinate geometry are embedded as supporting lessons.

Official curriculum source ↗

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