🎯⭐ INTERACTIVE LESSON

Piecewise Functions

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Piecewise Functions - Complete Interactive Lesson

Part 1: Reading the Notation

🧩 Piecewise Functions

Part 1 of 5 — Reading the Notation

Topics in This Part

Section
What Is a Piecewise Function?
The Big Curly Brace
Choosing the Right Piece

🔑 Key Concept: A piecewise function is one function built from several rules — each rule applies only on its own stretch of the number line. The whole game is figuring out which rule to use for which input.

What Is a Piecewise Function?

Most functions you've met use a single formula for every input — like $f(x) = 2x + 1$ . A piecewise function uses different formulas on different intervals.

Think of a parking garage:

The first hour costs a flat $5.
After that it's $2 per additional hour.

That's two rules glued together. Mathematically we stack them under a single curly brace:

$f(x) = \begin{cases} 5 & 0 \le x \le 1 \\ 5 + 2(x - 1) & x > 1 \end{cases}$

Each line is a piece: a formula on the left and the interval where it applies on the right.

💡 The function still passes the vertical line test — for any single input $x$ , exactly one piece gives the output. The intervals are designed so they never overlap.

The Anatomy of the Notation

Read a piecewise function right-to-left: first check the condition, then use the matching formula.

$f(x) = \begin{cases} x^2 & x < 0 \\ x + 1 & x \ge 0 \end{cases}$

Concept Check 🎯

Match Input to Piece 🔽

Use $f(x) = \begin{cases} x^2 & x < 0 \\ x + 1 & x \ge 0 \end{cases}$ . For each input, choose the formula you'd use.

From "Which Piece" to "What Value"

Choosing the right piece is half the job. The other half is just plugging in — substitute your input into that one formula and simplify.

$g(x) = \begin{cases} x^2 & x < 0 \\ 2x + 3 & x \ge 0 \end{cases}$

Pick the Piece, Then Evaluate 🧮

Use $g(x) = \begin{cases} x^2 & x < 0 \\ 2x + 3 & x \ge 0 \end{cases}$

Recap

To work with any piecewise function:

Read the conditions on the right of each piece.
Locate your input $x$ on the number line — find the interval it belongs to.
Use only that piece's formula.

You've got the reading skill down. In Part 2 we zoom in on the trickiest spot: the boundary point where two intervals meet, and what those $<$ , $\le$ , $>$ , $\ge$ symbols really decide.

Part 2: Boundary Points & Evaluating Carefully

🧩 Piecewise Functions

Part 2 of 5 — Boundary Points & Evaluating Carefully

🔑 The Idea: Two pieces meet at a boundary. The inequality symbol ( $<$ vs $\le$ ) decides which piece owns that exact boundary value. Getting this right is the #1 skill for evaluating and graphing piecewise functions.

Who Owns the Boundary?

Look at where the intervals "touch":

$f (x) = {\begin{cases} x + 1 & x \leq 3 \\ 2 x - 2 & x > \end{cases}$

Part 3: Graphing Piecewise Functions

🧩 Piecewise Functions

Part 3 of 5 — Graphing Piecewise Functions

🔑 Big Payoff: A piecewise graph is just each piece's graph, trimmed to its own interval. The only new skill is marking endpoints correctly: a closed dot ● for "included," an open dot ○ for "excluded."

Open vs. Closed Endpoints

At every boundary, the inequality tells you the dot:

Symbol at the boundary	Endpoint dot	Meaning
$\le$ or $\ge$	● closed (filled)	the point is on the graph
$<$ or

Part 4: Absolute Value, Step Functions & Continuity

🧩 Piecewise Functions

Part 4 of 5 — Absolute Value, Step Functions & Continuity

🔑 The Reveal: Some "famous" functions are secretly piecewise. Absolute value is the classic example, and step functions model real costs. We'll also learn to spot whether a piecewise graph is continuous (no gaps) or has a jump.

Absolute Value Is Piecewise

The definition of absolute value is a piecewise rule:

$|x| = \begin{cases} -x & x < 0 \\ x & x \ge 0 \end{cases}$

Part 5: Applications & Mastery Check

🧩 Piecewise Functions

Part 5 of 5 — Applications & Mastery Check

You can now read the notation, evaluate carefully across boundaries, graph the pieces with correct dots, recognize absolute-value and step functions, and test for continuity. Let's apply it to the real world — then finish with an Exit Quiz.

Real-World Models: Tiered Pricing

Piecewise functions are everywhere money is involved: tax brackets, shipping, phone plans, and utility bills all change rate at thresholds.

Example: Cell-Phone Data

A plan charges a flat $30 for up to $2$ GB, then $10 per extra GB beyond $2$ :

$C (g) = {\begin{cases} 30 & 0 \leq g \leq 2 \\ 30 + 10 (g - 2) \end{cases}$

To find...	Ask: which interval is $x$ in?	Then use...
$f(-3)$	$-3 < 0$ ✓	$x^2 = (-3)^2 = 9$
$f(0)$	$0 \ge 0$ ✓	$x + 1 = 0 + 1 = 1$
$f(4)$	$4 \ge 0$ ✓	$x + 1 = 4 + 1 = 5$

⚠️ Don't plug a number into every formula. A piecewise function gives you a menu of formulas; you pick the one whose condition your input satisfies, and ignore the rest.

To find $g(-2)$ : since $-2 < 0$ , use $x^2 \Rightarrow g(-2) = (-2)^2 = 4$ .

💡 Slow down on the sign of the input. A negative input squared becomes positive, but a negative input in $2x+3$ stays negative until you add. Substitution mistakes — not piece-selection mistakes — are where most lost points actually come from.

1) $g(-4) = \,?$ 2) $g(0) = \,?$ 3) $g(6) = \,?$

f (x) = {x + 1 2 x - 2 x \leq 3 x > 3

The boundary is $x = 3$ . Which piece claims it?

The first piece says $x \le 3$ . The $\le$ includes $3$ . ✓
The second piece says $x > 3$ . The strict $>$ excludes $3$ .

So $f(3)$ uses the first piece: $f(3) = 3 + 1 = 4$ .

🔑 Rule of thumb: The piece with the "or-equal-to" symbol ( $\le$ or $\ge$ ) is the one that owns the boundary. A well-defined piecewise function gives exactly one piece the equality at each boundary.

Read the Inequalities 🔽

Use $f(x) = \begin{cases} x + 1 & x \le 3 \\ 2x - 2 & x > 3 \end{cases}$

Functions With Three (or More) Pieces

The same logic scales. Just walk the number line from left to right and find the one interval your input lives in.

$h(x) = \begin{cases} -x & x < -1 \\ 1 & -1 \le x \le 1 \\ x^2 & x > 1 \end{cases}$

Input	Interval test	Output
$h(-4)$	$-4 < -1$ ✓	$-(-4) = 4$

💡 The middle piece here is a constant, $1$ . Constant pieces are perfectly legal — the function just flat-lines across that interval.

Concept Check 🎯

Mind the Boundary 🧮

Use $f(x) = \begin{cases} 4 - x & x < 1 \\ 3x & 1 \le x < 4 \\ x^2 - 10 & x \ge 4 \end{cases}$

1) $f(0) = \,?$ 2) $f(1) = \,?$ 3) $f(4) = \,?$

$f(x) = \begin{cases} x + 1 & x \le 2 \\ 5 & x > 2 \end{cases}$

The first piece, $y = x+1$ , runs up to $x = 2$ . Since $x \le 2$ , put a closed dot ● at its right end: $(2,\, 3)$ because $2 + 1 = 3$ .
The second piece, $y = 5$ , starts just after $x = 2$ . Since $x > 2$ , put an open dot ○ at $(2,\, 5)$ .

💡 At a boundary you often draw two dots stacked vertically — one closed, one open — because the two pieces disagree about the height there. That vertical gap is called a jump.

Concept Check 🎯

Reading Domain & Range From the Graph

Once a piecewise function is graphed, the domain is the set of $x$ -values it covers, and the range is the set of $y$ -values it reaches.

Example

$f(x) = \begin{cases} -2 & -4 \le x < 0 \\ x & 0 \le x \le 3 \end{cases}$

Domain: the pieces cover $-4 \le x < 0$ joined with $0 \le x \le 3$ , which together fill $-4 \le x \le 3$ . In interval notation: .

⚠️ Don't assume the domain is "all real numbers." Piecewise functions are often defined only on a limited set of intervals — read the conditions to find where the function actually lives.

Endpoints & Domain 🔽

Use $f(x) = \begin{cases} -2 & -4 \le x < 0 \\ x & 0 \le x \le 3 \end{cases}$

Plot the Endpoints 🧮

Use $f(x) = \begin{cases} x + 4 & x < -1 \\ x^2 & x \ge -1 \end{cases}$ . Find the $y$ -coordinate of each endpoint at the boundary $x = -1$ .

1) Left piece's endpoint height: $f$ -value from $x + 4$ at $x = -1$ is $\,?$ (this dot is open ○) 2) Right piece's endpoint height: $f$ -value from at is The size of the (closed height minus open height) is

When $x$ is negative, $-x$ makes it positive; when $x$ is non-negative, you leave it alone. That's why the graph of $y = |x|$ is a V: a downward ray on the left, an upward ray on the right, meeting at the origin.

Rewriting a Shifted Absolute Value

$y = |x - 3|$ becomes $y = \begin{cases} -(x - 3) & x < 3 \\ x - 3 & x \ge 3 \end{cases} = \begin{cases} -x + 3 & x < 3 \\ x - 3 & x \ge 3 \end{cases}$

The "corner" of the V sits where the inside equals zero: $x - 3 = 0 \Rightarrow x = 3$ .

💡 To convert $|expression|$ into pieces: find where the inside is $0$ (that's the boundary), keep the expression as-is where it's $\ge 0$ , and negate it where it's $< 0$ .

Concept Check 🎯

Step Functions

A step function is piecewise-constant: it's flat on each interval and "jumps" to a new height at the boundaries, producing a staircase.

The most famous is the greatest integer (floor) function, $\lfloor x \rfloor$ , which rounds down to the nearest integer:

$\lfloor 2.7 \rfloor = 2, \quad \lfloor 2 \rfloor = 2, \quad \lfloor -1.3 \rfloor = -2$

$x$	$\lfloor x \rfloor$
$1.9$	$1$
$3.0$

⚠️ For negative inputs the floor goes more negative: $\lfloor -1.3 \rfloor = -2$ , not $-1$ , because $-2$ is the greatest integer that is still $\le -1.3$ .

Real-world steps: a parking meter charging by the started hour, or postage that jumps at each ounce — the price is constant within a band, then leaps at the edge.

Floor Function Practice 🧮

Evaluate the greatest-integer (floor) function $\lfloor x \rfloor$ — round down to the nearest integer.

1) $\lfloor 4.9 \rfloor = \,?$ 2) $\lfloor 7 \rfloor = \,?$ 3) $\lfloor -2.1 \rfloor = \,?$

Continuous or a Jump?

A piecewise function is continuous at a boundary if the two pieces meet at the same height there — no gap, no jump. Test it by plugging the boundary into both neighboring formulas:

$f(x) = \begin{cases} x + 1 & x \le 2 \\ 3x - 3 & x > 2 \end{cases}$

Left piece at $x = 2$ : $2 + 1 = 3$ .
Right piece approaching $x = 2$ : $3(2) - 3 = 3$ .

Both give $3$ , so the graph connects — it's continuous at $x = 2$ . ✓

Compare with $f(x) = \begin{cases} x + 1 & x \le 2 \\ 3x & x > 2 \end{cases}$ : now the left gives but the right approaches . The heights differ, so there's a of .

🔑 Continuity test at a boundary $x=a$ : evaluate both pieces at $a$ . Equal heights ⇒ continuous. Different heights ⇒ jump (discontinuity).

Continuous or Jump? 🔽

C (g) = {30 30 + 10 (g - 2) 0 \leq g \leq 2 g > 2

Data used $g$	Which piece	Cost $C(g)$
$1.5$ GB	$0 \le g \le 2$	$30
$2$ GB	$0 \le g \le 2$	$30
$5$ GB	$g > 2$	$30 + 10(5-2) = 60$ , i.e. $60

💡 Notice the model is continuous at $g = 2$ : the flat piece gives $30 and the second piece gives $30 + 10(0) = 30$ , i.e. $30. Good pricing models usually avoid surprise jumps right at the threshold.

Apply the Model 🧮

A ride-share charges a $4 base for the first $3$ miles, then $2 per additional mile:

$F(m) = \begin{cases} 4 & 0 \le m \le 3 \\ 4 + 2(m - 3) & m > 3 \end{cases}$

1) Cost of a $2$ -mile ride: $F(2) = \,?$ (dollars) 2) Cost of a $3$ -mile ride: $F(3) = \,?$ Cost of an -mile ride:

Building a Piecewise Function From Words

Translating a scenario into notation is a top exam skill. Work in three steps:

Identify the thresholds — the inputs where the rule changes (these become your boundaries).
Write a formula for each interval — often a constant or a "base + rate × extra" expression.
Attach the correct inequalities so every input lands in exactly one piece.

Example: Movie-Theater Tickets

"Kids ( $12$ and under) pay $6; everyone else pays $11."

The threshold is age $12$ . Translating:

$P(a) = \begin{cases} 6 & a \le 12 \\ 11 & a > 12 \end{cases}$

The $\le$ on the first piece makes sure a $12$ -year-old gets the kids' price, and the strict $>$ keeps $13+$ at the adult price — no overlap, no gap.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

Range: the first piece is the single value

y = -2

. The second piece,

y = x

for

0 \le x \le 3

, sweeps from

0

up to

3

. So the outputs are

\{-2\}

together with

[0, 3]

(this dot is closed ●)

{−x+3x−3x<3x≥3