Piecewise Functions

5 min read

Reference: Stewart §1.1

One Function, Many Rules

Real-world situations often behave differently in different conditions. Tax brackets charge different rates at different income levels. Shipping costs jump at weight thresholds. Cell phone plans charge nothing for data up to a limit, then charge per gigabyte.

Piecewise functions let us model these situations with one mathematical object that uses different formulas on different parts of its domain.

Prerequisite Map

Quick Reference

Property	Value
Chapter	1 - Functions and Limits
Section	1.1
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

Piecewise Function Notation

A piecewise function is written using a brace that groups different formulas with their conditions:

$$f(x) = \begin{cases} \text{formula}_1 & \text{if condition}_1 \\ \text{formula}_2 & \text{if condition}_2 \\ \vdots & \vdots \end{cases}$$

Example:

$$f(x) = \begin{cases} x + 3 & \text{if } x < 1 \\ 5 & \text{if } x = 1 \\ x^2 & \text{if } x > 1 \end{cases}$$

Evaluating Piecewise Functions

Process:

Look at the input value $x$
Find which condition it satisfies
Use that formula to compute the output

Example: For the function above, find $f(-2)$, $f(1)$, and $f(3)$.

$f(-2)$: Since $-2 < 1$, use the first formula: $f(-2) = (-2) + 3 = 1$
$f(1)$: Since $x = 1$ exactly, use the second piece: $f(1) = 5$
$f(3)$: Since $3 > 1$, use the third formula: $f(3) = (3)^2 = 9$

Graphing Piecewise Functions

Process:

Graph each piece on its restricted domain
Use solid dots (●) for included endpoints
Use open circles (○) for excluded endpoints

        y
        |
    5   |     ●   (the point (1,5))
        |
    4   |
        |   ○     (excluded: x=1 gives y=4 from x^2, but we use 5 instead)
    3   |
        | ●
    2   |/        (line y = x + 3 for x < 1)
        |
    1 ○-+-○-------→ x
        |   parabola y = x² for x > 1

The Absolute Value Function

The absolute value is a classic piecewise function:

$$\vert x\vert = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$

Why $-x$ for negatives? If $x$ is negative (like $x = -3$), then $-x$ is positive ($-(-3) = 3$), which gives the correct absolute value.

Graph shape: V-shaped, with vertex at the origin.

        y
        |
    3   |●     ●
        | \   /
    2   |  \ /
        |   \
    1   |  / \
        | /   \
        ●-------→ x
       -3 -2 -1  1  2  3

Practice Problems

Level 1 Basic Evaluation

For the function $$g(x) = \begin{cases} 2x + 1 & \text{if } x \leq 2 \\ 7 - x & \text{if } x > 2 \end{cases}$$

Evaluate: $g(0)$, $g(2)$, $g(5)$.

Thought Process

For each input, check which condition it satisfies:

Is the input $\leq 2$ or $> 2$?

Note that $x = 2$ satisfies the first condition (it includes equality).

Show Answer

$g(0)$: Since $0 \leq 2$, use $2x + 1$: $g(0) = 2(0) + 1 = 1$
$g(2)$: Since $2 \leq 2$, use $2x + 1$: $g(2) = 2(2) + 1 = 5$
$g(5)$: Since $5 > 2$, use $7 - x$: $g(5) = 7 - 5 = 2$

Level 2 Graphing a Two-Piece Function

Sketch the graph of: $$f(x) = \begin{cases} x + 2 & \text{if } x < 1 \\ 3 & \text{if } x \geq 1 \end{cases}$$

Be sure to indicate open and closed circles at the boundary.

Thought Process

For $x < 1$: Graph the line $y = x + 2$, stopping just before $x = 1$
At $x = 1$ from the line: $y = 1 + 2 = 3$ (but this is excluded: open circle)
For $x \geq 1$: Graph the horizontal line $y = 3$
At $x = 1$ for the constant: $y = 3$ (included: closed circle)

Notice both pieces give $y = 3$ at $x = 1$, but from different formulas!

Show Answer

        y
        |
    4   |
        |
    3   |  ○●─────────  (horizontal line y = 3 for x ≥ 1)
        | /
    2   |/
        |
    1   |
        |
    ----+----+----→ x
       -1    1   3

    (Line y = x + 2 for x < 1)

Key observations:

At $x = 1$: The line would give $y = 3$ (open circle) and the constant gives $y = 3$ (closed circle)
Since both values coincide at $y = 3$, the function is actually continuous at $x = 1$!
Open and closed circles overlap at the same point

Level 3 Finding Domain and Range

For the function: $$h(x) = \begin{cases} \sqrt{x} & \text{if } 0 \leq x \leq 4 \\ 6 - x & \text{if } x > 4 \end{cases}$$

Find the domain.
Find the range.
Is the function continuous at $x = 4$?

Thought Process

(a) The domain comes from combining the conditions: the first piece covers $[0, 4]$ and the second covers $(4, \infty)$.

(b) For range, find what outputs each piece produces:

$\sqrt{x}$ on $[0, 4]$ gives $[0, 2]$
$6 - x$ on $(4, \infty)$ gives... what?

Show Answer

(a) Domain: The first piece covers $[0, 4]$, the second covers $(4, \infty)$. Together: Domain = $[0, \infty)$

(b) Range:

First piece: $\sqrt{x}$ on $[0, 4]$ produces outputs from $\sqrt{0} = 0$ to $\sqrt{4} = 2$, so $[0, 2]$
Second piece: $6 - x$ on $(4, \infty)$. As $x \to 4^+$, $y \to 2$. As $x \to \infty$, $y \to -\infty$. So this piece gives $(-\infty, 2)$

Combined range: $(-\infty, 2]$ (the pieces overlap at $y = 2$)

(c) Continuity at $x = 4$:

From the first piece: $h(4) = \sqrt{4} = 2$ (this is the actual value since $x = 4$ satisfies $x \leq 4$)
Approaching from the right: $\lim_{x \to 4^+} (6 - x) = 6 - 4 = 2$

Both equal $2$, so the function is continuous at $x = 4$.

Level 4 Absolute Value Transformations

Write $f(x) = \vert 2x - 6\vert $ as a piecewise function (without absolute value signs) and sketch its graph.

Thought Process

The absolute value removes negatives. So:

When is $2x - 6 \geq 0$? When $x \geq 3$.
When is $2x - 6 < 0$? When $x < 3$.

For the negative case, $\vert 2x - 6\vert = -(2x - 6) = -2x + 6$.

Show Answer

Finding the breakpoint: $2x - 6 = 0$ when $x = 3$

Piecewise form: $$f(x) = \vert 2x - 6\vert = \begin{cases} 2x - 6 & \text{if } x \geq 3 \\ -(2x - 6) & \text{if } x < 3 \end{cases} = \begin{cases} 2x - 6 & \text{if } x \geq 3 \\ -2x + 6 & \text{if } x < 3 \end{cases}$$

Graph:

For $x < 3$: Line with slope $-2$, passes through $(0, 6)$ and $(3, 0)$
For $x \geq 3$: Line with slope $2$, passes through $(3, 0)$ and $(6, 6)$

        y
        |
    6   ●           ●
        |\         /
    4   | \       /
        |  \     /
    2   |   \   /
        |    \ /
        +-----●-----→ x
             3    6

V-shape with vertex at $(3, 0)$.

Level 5 Designing a Piecewise Function

A parking garage charges:

\$3 for the first hour or any part of it
\$2 for each additional hour or part of an hour
Maximum daily charge of \$15

Write a piecewise function $C(t)$ for the cost of parking for $t$ hours (where $t > 0$).
What is $C(0.5)$? $C(1)$? $C(4.5)$? $C(10)$?
Sketch the graph of $C(t)$ for $0 < t \leq 12$.

Thought Process

“Any part of an hour” means we round up. Use the ceiling function notation or think in terms of intervals.

The cost structure:

$0 < t \leq 1$: Pay \$3
$1 < t \leq 2$: Pay \$3 + \$2 = \$5
$2 < t \leq 3$: Pay \$3 + 2(\$2) = \$7
Continue until the total reaches \$15

When does it hit the max? Solve $3 + 2(n-1) = 15$ to find $n = 7$ hours.

Show Answer

(a) Piecewise function:

First, find when the cap kicks in: $3 + 2(n-1) = 15$ gives $n = 7$.

$$C(t) = \begin{cases} 3 & \text{if } 0 < t \leq 1 \\ 5 & \text{if } 1 < t \leq 2 \\ 7 & \text{if } 2 < t \leq 3 \\ 9 & \text{if } 3 < t \leq 4 \\ 11 & \text{if } 4 < t \leq 5 \\ 13 & \text{if } 5 < t \leq 6 \\ 15 & \text{if } t > 6 \end{cases}$$

Or more compactly using ceiling function: For $t \leq 6$, $C(t) = 3 + 2(\lceil t \rceil - 1) = 1 + 2\lceil t \rceil$.

(b) Evaluations:

$C(0.5) = 3$ (first hour)
$C(1) = 3$ (exactly 1 hour, still first tier)
$C(4.5) = 11$ (rounds to 5 hours: $3 + 2(4) = 11$)
$C(10) = 15$ (cap applies)

(c) Graph:

    C($)
    15  |                    ●───────
        |               ●────○
    13  |          ●────○
        |     ●────○
    11  |
        |
     9  |
        |●────○
     7  |
        |
     5  |●────○
        |
     3  ●────○
        +--+--+--+--+--+--+--+--→ t (hours)
           1  2  3  4  5  6  7

Step function with jumps at each integer, capped at \$15.

Common misconception

use whichever formula gives the simplest answer.

This is the input-output-confusion error. The condition in a piecewise function tells you which formula to apply based on the INPUT value, not the output. For the function $f(x) = x^2$ when $x < 0$ and $f(x) = x + 1$ when $x \geq 0$: to find $f(-3)$, check the input: is $-3 < 0$? Yes, so use $f(x) = x^2$ and get $f(-3) = 9$. There is no choice involved and no preference for a simpler answer. The condition governs which formula applies. Using the wrong formula because it is easier produces a value from outside the definition and gives the wrong answer.

Common misconception

if both pieces give the same value at the boundary, the function is automatically continuous everywhere.

This is the continuity-vs-limit-existence error applied to piecewise functions. Matching values at a single boundary point means the function is defined and consistent there -- which is good -- but continuity requires that the pieces approach the same value as the input approaches that point FROM EACH SIDE. A piecewise function can have equal boundary values at $x = 1$ while still having a jump somewhere else, or the limit from one side might not exist at all. Conversely, a piecewise function where the pieces give different outputs at the boundary ($3$ from the left piece, $5$ from the right piece) has a jump discontinuity there regardless of which value is listed as the defined output.

Common Errors to Avoid

Error	Example	Correction
Wrong boundary	Using $x = 1$ with “$x < 1$” formula	Check inequality: is the boundary included?
Forgetting endpoint markers	Drawing continuous line through breakpoint	Use ● for included, ○ for excluded
Mixing up $-x$ in $\\|x\\|$	Thinking $-x$ is always negative	When $x < 0$, $-x$ is positive

Mastery Checklist

I can evaluate a piecewise function by checking which condition the input satisfies
I can graph a piecewise function with correct open/closed circles at boundaries
I can write the absolute value function as a piecewise function
I can transform absolute value expressions like $\vert ax + b\vert $ into piecewise form
I can determine if a piecewise function is continuous at a boundary point

Mental Model

The Road with Multiple Speed Limits:

Think of a piecewise function like a road with different speed limits in different zones. At mile marker 10, the limit might change from 55 to 35. You follow whichever rule applies to your current location. The “breakpoints” are like zone boundaries; you need to know which zone you’re in to follow the right rule.

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Last updated: 2026-01-22