Piecewise-defined functions and absolute value

Warm-up

Three quick questions from earlier lessons.

Discover

A parking lot charges 5 dollars for up to 2 hours, and 9 dollars for a longer stay. What does a stay of exactly 2 hours cost?

5 dollars
9 dollars

Key idea

A piecewise function uses a different formula on different parts of its domain.

To find

f(a)

: first find which condition

a

satisfies, then use that piece's formula. Check the boundary carefully (

<

or

\le

).

The absolute value is piecewise: $|x| = x$ if $x \ge 0$ , and $|x| = -x$ if $x < 0$ .

Worked example

$f(x) = \begin{cases} x^2 & \text{if } x < 1 \\ 2x + 3 & \text{if } x \ge 1 \end{cases}$ . Find $f(1)$ .

Use the second piece,

2x + 3

Check the conditions first: 1 is not less than 1, but 1 ≥ 1.

f(1) = 2 \cdot 1 + 3 = 5

Replace x by 1.

Your turn, guided

Your turn, with the same $f$ : find $f(-2)$ .

Use the first piece,

x^2

−2 < 1.

f(-2) = 4

(−2)² = 4.

Practice ★

$f(x) = \begin{cases} x + 4 & \text{if } x < 8 \\ 4x & \text{if } x \ge 8 \end{cases}$ . Find $f(0)$ .
$f(x) = \begin{cases} x + 8 & \text{if } x < 7 \\ 4x & \text{if } x \ge 7 \end{cases}$ . Find $f(2)$ .
$f(x) = \begin{cases} x + 7 & \text{if } x < 3 \\ 3x & \text{if } x \ge 3 \end{cases}$ . Find $f(1)$ .
$f(x) = \begin{cases} x + 6 & \text{if } x < 6 \\ 4x & \text{if } x \ge 6 \end{cases}$ . Find $f(1)$ .

Watch out!

Practice ★★

$f(x) = \begin{cases} x + 3 & \text{if } x < 5 \\ 3x & \text{if } x \ge 5 \end{cases}$ . Find $f(5)$ .
$f(x) = \begin{cases} x + 4 & \text{if } x < 3 \\ 2x & \text{if } x \ge 3 \end{cases}$ . Find $f(3)$ .
$f(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ 5x & \text{if } x \ge 1 \end{cases}$ . Find $f(1)$ .
$f(x) = \begin{cases} x + 2 & \text{if } x < 5 \\ 2x & \text{if } x \ge 5 \end{cases}$ . Find $f(5)$ .

For $x < 7$ , which expression is equal to $|x - 7|$ ?
1. $7 - x$
2. $x - 7$
3. $-x - 7$
For $x < 5$ , which expression is equal to $|x - 5|$ ?
1. $5 - x$
2. $x - 5$
3. $-x - 5$

Exit ticket

Three questions to check you have got it.

$f(x) = \begin{cases} x + 3 & \text{if } x < 5 \\ 3x & \text{if } x \ge 5 \end{cases}$ . Find $f(5)$ .
$f(x) = \begin{cases} x + 4 & \text{if } x < 3 \\ 2x & \text{if } x \ge 3 \end{cases}$ . Find $f(3)$ .
$f(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ 5x & \text{if } x \ge 1 \end{cases}$ . Find $f(1)$ .

Summary card

Piecewise functions

Find the piece whose condition the input satisfies.
Use that piece's formula.
$|x| = x$ for $x \ge 0$ , and $-x$ for $x < 0$ .

Watch out: at a boundary, read $<$ and $\ge$ carefully.

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