Piecewise and Absolute Value Functions

What You’ll Learn

In this lesson you’ll learn how to define, evaluate, graph, and analyze piecewise functions and absolute value functions. If you covered piecewise functions in Algebra 2, this is a recap with some new depth, including writing absolute value expressions as piecewise functions and analyzing continuity at the boundary points.

The Concept

Piecewise Functions

A piecewise function uses different rules for different parts of its domain. It’s written using cases:

f(x) = \begin{cases} \text{rule 1} & \text{if condition 1} \\ \text{rule 2} & \text{if condition 2} \end{cases}

To evaluate: check which condition the input satisfies, then use that rule. The key is paying attention to the boundary points and whether they use < or ≤.

Absolute Value Functions

The absolute value function is the most common piecewise function:

f(x) = |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

Its graph is a V-shape with vertex at the origin. Transformations work the same as with other functions: y = a|x - h| + k has vertex at (h, k), opens up if a > 0 and down if a < 0, and is stretched or compressed by |a|.

Worked Example

Example 1: Evaluate a piecewise function

f(x) = \begin{cases} 2x + 5 & \text{if } x < 0 \\ x^2 - 3 & \text{if } x \geq 0 \end{cases}

Find f(-2) and f(3):

For f(-2): since -2 < 0, use the first rule:

f(-2) = 2(-2) + 5 = -4 + 5 = 1

For f(3): since 3 ≥ 0, use the second rule:

f(3) = 3^2 - 3 = 9 - 3 = 6

Example 2: Graph y = |x - 2| + 1

This is the basic V-shape shifted right 2 and up 1. The vertex is at (2, 1). The left arm has slope -1 and the right arm has slope +1.

Example 3: Write |2x - 6| as a piecewise function

First find the critical point where the expression inside equals zero: 2x - 6 = 0, so x = 3.

|2x - 6| = \begin{cases} 2x - 6 & \text{if } x \geq 3 \\ -(2x - 6) & \text{if } x < 3 \end{cases} = \begin{cases} 2x - 6 & \text{if } x \geq 3 \\ -2x + 6 & \text{if } x < 3 \end{cases}

Example 4: Check continuity at a boundary

g(x) = \begin{cases} x + 4 & \text{if } x < 2 \\ 3x & \text{if } x \geq 2 \end{cases}

At x = 2, check both sides:

From the left: x + 4 = 2 + 4 = 6
From the right: 3x = 3(2) = 6

Both sides give 6, so g is continuous at x = 2. If they didn’t match, there would be a jump discontinuity.

Real-World Application

Piecewise and absolute value functions model many real situations:

Tax brackets: different rates apply to different income ranges
Shipping costs: base fee plus per-pound charge, with different rates above certain weights
Utility pricing: one rate for the first 500 kWh, a higher rate after that
Distance and displacement: absolute value gives total distance regardless of direction
Temperature control: heating kicks in below a setpoint, cooling above it

Example: a phone plan charges $30/month for the first 5 GB, then$ 10 per additional GB. That’s a piecewise function:

\text{Cost}(x) = \begin{cases} 30 & \text{if } x \leq 5 \\ 30 + 10(x - 5) & \text{if } x > 5 \end{cases}

Quiz

A piecewise function is defined by:

A.Only one rule for all $x$
B.Different rules for different parts of the domain
C.Only linear terms
D.Only quadratic terms

The graph of $y = |x - 3| + 2$ has its vertex at:

A.$(2, 3)$
B.$(-3, 2)$
C.$(3, -2)$
D.$(3, 2)$

When evaluating a piecewise function, the first step is to:

A.Add all pieces together
B.Graph the function
C.Check which condition the input satisfies
D.Find the inverse

$|x|$ written as a piecewise function is:

A.$x$ if $x \geq 0$, $\,-x$ if $x \lt 0$
B.$x$ for all $x$
C.$-x$ for all $x$
D.$x^2$ for all $x$

A piecewise function has a jump discontinuity at $x = a$ when:

A.The function is undefined at $x = a$
B.The function equals zero at $x = a$
C.The slope changes at $x = a$
D.The left and right limits at $x = a$ are different