Math Core

A phone plan charges a flat fee up to a data limit and then a per-gigabyte rate after that. A parking garage charges one price for the first hour and another for each hour after. In situations like these, one formula isn't enough: the rule changes depending on the input. Functions built this way are called piecewise functions, and the absolute value function is the most famous one.

What a piecewise function is

Definition

Piecewise function

A piecewise function uses different formulas on different parts of its domain. Each formula comes with a condition that says which inputs it applies to:

f(x) = \begin{cases} \text{formula 1}, & \text{if } x \text{ is in interval 1} \\ \text{formula 2}, & \text{if } x \text{ is in interval 2} \end{cases}

To evaluate a piecewise function, first decide which condition the input satisfies, then use only that formula. Each input gets exactly one output, so the conditions never overlap.

Worked example: Evaluating a piecewise function

Let

f(x) = \begin{cases} 2x + 1, & x < 1 \\ x^2, & 1 \le x \le 3 \\ 10 - x, & x > 3 \end{cases}

Find $f(-2)$ , $f(1)$ , $f(3)$ and $f(5)$ .

$-2 < 1$ , so use the first piece: $f(-2) = 2(-2) + 1 = -3$ .
$1$ satisfies $1 \le x \le 3$ (not $x < 1$ ), so use the second piece: $f(1) = 1^2 = 1$ .
$3$ satisfies $1 \le x \le 3$ : $f(3) = 3^2 = 9$ .
$5 > 3$ , so use the third piece: $f(5) = 10 - 5 = 5$ .

Common mistake

Pay close attention to the boundary values. At $x = 1$ above, it's tempting to use $2x + 1$ because it's listed first, but the condition $x < 1$ does not include $1$ . Check $<$ versus $\le$ every time you land exactly on a boundary.

Graphing piecewise functions

Graph each piece as if it were a whole function, but draw only the part over its own interval. Then mark each endpoint:

a closed dot (filled) when the endpoint is included ( $\le$ or $\ge$ ),
an open dot (hollow) when it is not included ( $<$ or $>$ ).

Worked example: Graphing three pieces

Graph

f(x) = \begin{cases} x + 4, & x < -1 \\ x^2, & -1 \le x \le 2 \\ 1, & x > 2 \end{cases}

Left piece. The line $y = x + 4$ for $x < -1$ . At $x = -1$ it would reach $y = 3$ , but $-1$ is not included, so draw an open dot at $(-1, 3)$ and draw the line to the left.

Middle piece. The parabola $y = x^2$ from $x = -1$ to $x = 2$ . Both ends are included: closed dots at $(-1, 1)$ and $(2, 4)$ .

Right piece. The horizontal line $y = 1$ for $x > 2$ , with an open dot at $(2, 1)$ .

The three pieces of f. Hollow circles are open dots; solid dots are included endpoints.Open in grapher →

The graph "jumps" at $x = -1$ and at $x = 2$ . That's allowed. What's not allowed is two filled dots stacked at the same $x$ -value, because then one input would have two outputs.

Absolute value as a piecewise function

The absolute value of a number is its distance from $0$ . For a nonnegative number, that's the number itself. For a negative number, it's the opposite of the number (so $\lvert -5 \rvert = -(-5) = 5$ ). That description is already a piecewise function:

\lvert x \rvert = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}

The left piece y = −x (x < 0) and the right piece y = x (x ≥ 0) join at the origin to form y = |x|.Open in grapher →

The two pieces are lines with slopes $-1$ and $1$ that meet at the vertex $(0, 0)$ .

Graphing absolute value functions

From the previous lesson, every absolute value function can be written in the transformation form below.

Vertex form of an absolute value function

The graph of $f(x) = a\lvert x - h \rvert + k$ is a V with

vertex $(h, k)$ ,
slopes $a$ to the right of the vertex and $-a$ to the left,
opening up if $a > 0$ (the vertex is a minimum) and down if $a < 0$ (the vertex is a maximum).

To graph one, plot the vertex, then use the slope to step off a point on each side.

Worked example: A downward V

Graph $f(x) = -2\lvert x + 1 \rvert + 4$ . Find its $x$ -intercepts and range.

The vertex is $(-1, 4)$ and $a = -2$ , so the V opens down. To the right of the vertex, the slope is $-2$ : over $1$ , down $2$ , giving $(0, 2)$ . To the left, the slope is $2$ as you move right, so moving left $1$ also goes down $2$ , giving $(-2, 2)$ .

$x$ -intercepts. Set $f(x) = 0$ :

\begin{aligned} -2\lvert x + 1 \rvert + 4 &= 0 \\ \lvert x + 1 \rvert &= 2 \\ x + 1 = 2 \quad &\text{or} \quad x + 1 = -2 \\ x = 1 \quad &\text{or} \quad x = -3 \end{aligned}

y = -2|x + 1| + 4(-1, 4)(-3, 0)(1, 0)Open in grapher →

Range. The highest point is the vertex, so the range is $y \le 4$ .

Rewriting absolute value without the bars

Sometimes you need an absolute value function as ordinary pieces, for instance to find where it equals a line. Split at the vertex, where the expression inside the bars equals $0$ .

Worked example: Writing an absolute value function in pieces

Write $g(x) = 2\lvert x - 3 \rvert - 1$ as a piecewise function with no absolute value bars.

The inside $x - 3$ is negative when $x < 3$ and nonnegative when $x \ge 3$ .

For $x < 3$ : $\lvert x - 3 \rvert = -(x - 3)$ , so $g(x) = -2(x - 3) - 1 = -2x + 5$ .
For $x \ge 3$ : $\lvert x - 3 \rvert = x - 3$ , so $g(x) = 2(x - 3) - 1 = 2x - 7$ .

g(x) = \begin{cases} -2x + 5, & x < 3 \\ 2x - 7, & x \ge 3 \end{cases}

Both pieces give $g(3) = -1$ at the boundary, which matches the vertex $(3, -1)$ .

Tip

A quick check for any piecewise form of an absolute value function: both pieces must give the same value at the vertex, since the V is one connected graph. If they don't match, a sign is wrong.

Absolute value inequalities in two variables

An inequality like $y \ge \lvert x - 1 \rvert - 2$ describes a region. Graph the boundary V (solid for $\ge$ or $\le$ , dashed for $>$ or $<$ ), then shade the side that works. Test the origin: $0 \ge \lvert 0 - 1 \rvert - 2 = -1$ is true, so shade the side containing $(0, 0)$ , which is the inside of the V.

y >= |x - 1| - 2Open in grapher →

Practice

Practice 1

Let $f(x) = \begin{cases} 3x - 2, & x \le 2 \\ 8 - x, & x > 2 \end{cases}$ . Find $f(2)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Using the same function $f$ from the previous problem, find $f(-1) + f(6)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

What is the vertex of $f(x) = -3\lvert x - 4 \rvert + 2$ ?

Enter a point like (2, -3)

Practice 4

Find both $x$ -intercepts of $f(x) = \lvert x + 2 \rvert - 5$ .

Separate answers with commas, e.g. 2, -5

Practice 5

Which piecewise function is equal to $f(x) = \lvert x + 2 \rvert$ ?

Practice 6

Which function is graphed below?

A piecewise graph with an open dot at (1, 3) and a closed dot at (1, 0).Open in grapher →

Practice 7

A shipping company charges $6.00 for a package weighing up to $2$ pounds. For heavier packages it charges $6.00 plus $1.50 for each pound over $2$ . The cost for a package weighing $w$ pounds is

C(w) = \begin{cases} 6, & 0 < w \le 2 \\ 6 + 1.5(w - 2), & w > 2 \end{cases}

What is the cost, in dollars, to ship a $7$ -pound package?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

The graph of $f(x) = a\lvert x - 1 \rvert + 3$ passes through the point $(4, -3)$ . Find $a$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.