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Piecewise Functions

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Piecewise Functions - Complete Interactive Lesson

Part 1: Anatomy & Notation

🧩 Piecewise Functions

Part 1 of 7 — Anatomy & Notation

Topics in This Part

Section
What Is a Piecewise Function?
Reading the Brace Notation
Which Rule Applies Where

🔑 Key Concept: A piecewise function is one function defined by different rules on different parts of its domain. Think of it as a recipe with conditions: "if the input is over here, use this rule; if it's over there, use that rule."

What Is a Piecewise Function?

Most functions you've met use one rule for every input — like $f(x) = 2x + 1$ . A piecewise function switches rules depending on where $x$ lands.

A classic real-world example is a parking garage:

Input ( $x$ = hours parked)	Rule (cost)
First 2 hours	flat $5
After 2 hours	$5 plus $3 per extra hour

The cost is still a function of hours — every input has exactly one output — but the formula you use depends on the input. That's the whole idea.

We collect the rules under a single brace:

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

💡 The function is still one function. The brace is just shorthand for "here are all my rules and the conditions that select them."

Reading the Brace Notation

Every row of a piecewise definition has two parts:

$\underbrace{2x + 1}_{\text{the rule}} \qquad \underbrace{x < 0}_{\text{the condition (a piece of the domain)}}$

Concept Check 🎯

Mind the Boundary 🎯

Match Input to Rule 🔽

For $f(x) = \begin{cases} 2x+1 & x<0 \\ x^2 & x\ge 0 \end{cases}$ , pick the rule each input triggers.

Recap

You now know how to read a piecewise function:

Each row is a rule paired with a condition.
To evaluate, find the one condition your input satisfies, then apply that row's rule.
The conditions tile the domain — no gaps, no overlaps — and the boundary value belongs to whichever piece uses $\le$ or $\ge$ (the one with the "equals" line).

In Part 2 we put this into action: plugging numbers in carefully, especially right at the boundaries where mistakes love to hide.

Part 2: Evaluating Piecewise Functions

🧩 Piecewise Functions

Part 2 of 7 — Evaluating Piecewise Functions

🔑 The One-Step Method: To find $f(a)$ , ignore the rules at first. Ask only: which condition does $a$ satisfy? Then plug $a$ into that — and only that — rule.

Evaluating, Step by Step

Use this function throughout:

f (x) = {\begin{cases}  \end{cases}

Part 3: Graphing & Open/Closed Dots

🧩 Piecewise Functions

Part 3 of 7 — Graphing & Open/Closed Dots

🔑 The Plan: Graph each piece as if it were a normal function, but only draw it over its own interval. The open and closed dots at the boundaries tell the whole story of what happens at the edges.

How to Graph a Piecewise Function

For each piece:

Graph the rule as usual (line, parabola, constant, etc.).
Keep only the part over that piece's interval — erase everything outside it.
Mark the endpoints:
- $\le$ or $\ge$ (endpoint included) → a closed (filled) dot ●
- $<$ or (endpoint ) → an ○

Part 4: Continuity at the Boundaries

🧩 Piecewise Functions

Part 4 of 7 — Continuity at the Boundaries

🔑 The Big Question: Does the graph connect at the seam between two pieces? A piecewise function is continuous at a boundary $x=c$ when the two pieces meet at the same point — no hole, no jump.

The Continuity Test at a Boundary

At a boundary $x = c$ , compute three things:

The value the left piece approaches as $x \to c$ (plug into the left rule).

Part 5: Domain & Range

🧩 Piecewise Functions

Part 5 of 7 — Domain & Range

🔑 Domain = all valid inputs; Range = all possible outputs. For piecewise functions, you find each by examining the pieces together, paying special attention to the boundaries and which dots are open vs. closed.

Finding the Domain

The domain is every $x$ for which some piece is defined. Just take the union of all the condition-intervals.

f(x) = \begin{cases} \sqrt{x} & 0 \le x \le 4 \\ x - 2 & x > 4 \end{cases}

Part 6: Absolute Value & Step Functions

🧩 Piecewise Functions

Part 6 of 7 — Absolute Value & Step Functions

🔑 Hidden Pieces: Two famous functions are secretly piecewise — the absolute value function and the greatest-integer (step) function. Recognizing them as piecewise unlocks how to evaluate and graph them.

Absolute Value Is Piecewise

The absolute value $|x|$ keeps positives the same and flips negatives positive. As a piecewise rule:

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

Part 7: Modeling, Mixed Practice & Exit Quiz

🧩 Piecewise Functions

Part 7 of 7 — Modeling, Mixed Practice & Exit Quiz

You can read brace notation, evaluate carefully at boundaries, graph with open/closed dots, test continuity, find domain and range, and decode absolute-value and step functions. Now let's use it on a real model — then prove your mastery.

Real-World Modeling: A Tax Bracket

Piecewise functions are how the world handles "the rate changes after a threshold." Suppose a simplified tax on income $x$ (in dollars) is:

T(x) = \begin{cases} 0.10\,x & 0 \le x \le 10000 \\ 1000 + 0.20(x - 10000) & x > 10000 \end{cases}

the condition (a piece of the domain)x<0

To use the function, you ask one question: which condition does my input satisfy? That tells you which rule to apply.

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

An input like $x = -3$ satisfies $x < 0$ , so use the top rule: $2(-3)+1$ .
An input like $x = 4$ satisfies $x \ge 0$ , so use the bottom rule: $4^2$ .

⚠️ The conditions must not overlap — every input belongs to exactly one piece. If two conditions claimed the same $x$ , you'd get two outputs, and it wouldn't be a function. Notice $x<0$ and $x\ge 0$ together cover all real numbers with no overlap.

f (x) = ⎩ ⎨ ⎧ - x - 2 x^{2} 6 x < - 1 - 1 \leq x < 2 x \geq 2

Step 1 — pick the piece. Is $-4 < -1$ ? Yes. Use the top rule $-x-2$ . Step 2 — substitute. $f(-4) = -(-4) - 2 = 4 - 2 = 2$ .

Step 1 — pick the piece. Is $-1 \le 1 < 2$ ? Yes. Use the middle rule $x^2$ . Step 2 — substitute. $f(1) = 1^2 = 1$ .

Step 1 — pick the piece. Is $5 \ge 2$ ? Yes. Use the bottom rule, the constant $6$ . Step 2 — substitute. $f(5) = 6$ .

💡 The bottom piece is a constant: $f(x)=6$ for every $x \ge 2$ . The input doesn't appear in the rule, so the output never changes there.

The Boundary Is Where Points Are Lost

The danger zone is the boundary between pieces. Check the inequality signs carefully.

For the same $f$ , evaluate at the boundary $x = -1$ :

f(x) = \begin{cases} -x - 2 & x < -1 \\ x^2 & -1 \le x < 2 \\ 6 & x \ge 2 \end{cases}

Top condition $x < -1$ : is $-1 < -1$ ? No (strict).
Middle condition $-1 \le x < 2$ : is ? (the includes ).

So $x=-1$ belongs to the middle piece, and $f(-1) = (-1)^2 = 1$ .

⚠️ Don't grab the first rule whose number "looks close." Read the inequality: $<$ and $>$ exclude the endpoint; $\le$ and $\ge$ include it.

Evaluate It 🧮

Use $f(x) = \begin{cases} -x - 2 & x < -1 \\ x^2 & -1 \le x < 2 \\ 6 & x \ge 2 \end{cases}$

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

Which Piece? 🔽

For $f(x) = \begin{cases} -x - 2 & x < -1 \\ x^2 & -1 \le x < 2 \\ 6 & x \ge 2 \end{cases}$ , choose the piece each input uses.

Concept Check 🎯

open (hollow) dot

At a boundary $x = c$ , evaluate the rule on each side at $c$ :

The piece whose condition includes $c$ gets the closed dot (that's the actual function value).
The piece whose condition excludes $c$ gets an open dot.

💡 A vertical line at any $x$ should cross your graph exactly once (the vertical line test) — that's the visual version of "one output per input."

Worked Example

h(x) = \begin{cases} x + 3 & x < 1 \\ -2x + 6 & x \ge 1 \end{cases}

Left piece $y = x+3$ , drawn for $x<1$ . As $x \to 1$ from the left, $y \to 1 + 3 = 4$ . Since , put an dot ○ at .

Right piece $y=-2x+6$ , drawn for $x \ge 1$ . At $x=1$ , $y =$ . Since , put a dot ● at .

Here both dots land at $(1,4)$ — the open one is "filled in" by the closed one, so the graph has no hole. We'll call that continuous in Part 4.

At $x = 1$	Left rule $x+3$	Right rule $-2x+6$
Value	(open ○)

⚠️ Always plug the boundary into both rules. If the two values differ, the graph jumps there, leaving a visible gap.

Concept Check 🎯

Find the Endpoints 🧮

For $h(x) = \begin{cases} x+3 & x<1 \\ -2x+6 & x\ge 1 \end{cases}$ , the boundary is $x = 1$ .

1) Plug $x=1$ into the left rule $x+3$ : the open dot is at height $y = \,?$ 2) Plug $x=1$ into the rule : the closed dot is at height The -intercept of the graph is

When the Graph Jumps

Not every piecewise graph connects. Consider:

j(x) = \begin{cases} x + 1 & x < 2 \\ x - 3 & x \ge 2 \end{cases}

At the boundary $x = 2$ :

Left rule: $2 + 1 = 3$ → open dot ○ at $(2, 3)$ .
Right rule: $2 - 3 = -1$ → dot ● at .

The two endpoints are at different heights ( $3$ vs. $-1$ ), so the graph jumps down by $4$ at $x=2$ . There's a visible break — a hole at $(2,3)$ and a solid point at .

💡 A jump is perfectly legal: it's still a function (one output per input — $j(2)=-1$ , the closed dot). It just isn't continuous, which is exactly the idea of Part 4.

Dots at the Jump 🔽

For $j(x) = \begin{cases} x+1 & x<2 \\ x-3 & x\ge 2 \end{cases}$ , describe the endpoints at $x=2$ .

The value the right piece approaches as

x \to c

(plug

c

into the right rule).

The actual value

f(c)

(use the piece whose condition includes

c

The function is continuous at $c$ exactly when all three are equal:

$\text{(left value)} = \text{(right value)} = f(c)$

If the left and right values differ, you get a jump discontinuity. If they agree with each other but the defined value $f(c)$ sits somewhere else, you get a removable discontinuity (a hole).

💡 Shortcut for "two-rule" boundaries: just set the two rules equal at $c$ . If the left rule and right rule give the same number at $x=c$ , the pieces connect.

Worked Example: Is It Continuous?

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

At the boundary $x = 2$ :

Left rule $x^2$ at $2$ : $2^2 = 4$ .
Right rule $3x-2$ at : .

All three equal $4$ , so $f$ is continuous at $x=2$ . The parabola hands off smoothly to the line. ✓

Now change one number

g(x) = \begin{cases} x^2 & x \le 2 \\ 3x - 5 & x > 2 \end{cases}

Left at $2$ : $2^2 = 4$ .
Right at $2$ : $3(2) - 5 = 1$ .

Now $4 \ne 1$ , so $g$ has a jump discontinuity at $x=2$ — the graph leaps from $4$ down to .

⚠️ "Continuous" is about the boundary values matching, not about whether the pieces have the same shape. A parabola and a line can absolutely meet continuously.

Concept Check 🎯

Solving for a Value That Makes It Continuous

A favorite AP-style question: find the constant that forces continuity.

f(x) = \begin{cases} 2x + k & x < 3 \\ x^2 & x \ge 3 \end{cases}

For continuity at $x=3$ , set the two rules equal at $3$ :

$2(3) + k = 3^2$ $6 + k = 9$ $k =$

So $k = 3$ makes the pieces meet at $(3, 9)$ , and the graph connects with no jump.

✅ Check: With $k=3$ , left value $=2(3)+3 = 9$ and right value $=3^2=9$ . Equal. ✓

Make It Continuous 🧮

Find the constant that makes each function continuous at its boundary.

1) $f(x) = \begin{cases} x + k & x < 4 \\ 3x & x \ge 4 \end{cases}$ continuous at $x=4$ : $k = \,?$ 2) $g(x) = \begin{cases} 5 & x \le 1 \\ ax + 2 & x > 1 \end{cases}$ continuous at $x=1$ : $a = \,?$

Continuous or Not? 🔽

For each pair of pieces, plug the boundary into both rules and decide.

Piece 1 covers $[0, 4]$ .
Piece 2 covers $(4, \infty)$ .

Union: $[0, \infty)$ . There's no rule for $x<0$ , so negatives are not in the domain.

💡 If the conditions leave a gap (say nothing is defined on $1<x<2$ ), that gap is excluded from the domain. Read the conditions as the menu of allowed inputs.

Finding the Range

The range is every $y$ the graph actually reaches. Find the output set of each piece over its own interval, then take the union.

f(x) = \begin{cases} x^2 & -2 \le x < 0 \\ x + 1 & 0 \le x \le 3 \end{cases}

Piece 1, $x^2$ on $[-2, 0)$ : As $x$ runs from $-2$ up to (but not including) $0$ , runs from down toward . At (included) we get ; as (excluded) we approach but don't reach it. Outputs: .

Piece 2, $x+1$ on $[0, 3]$ : Linear and increasing, from $0+1=1$ to $3+1=4$ . Outputs: .

Union: $(0,4] \cup [1,4] = (0, 4]$ . So the range is $(0, 4]$ .

⚠️ The range comes from the outputs, not the inputs. Always plug each interval's endpoints into that piece's rule, and watch which $y$ -values are actually attained.

Domain & Range 🔽

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

Concept Check 🎯

Range Endpoints 🧮

For the piece $x+1$ defined on $0 \le x \le 3$ :

1) Smallest output (at $x=0$ ): $y = \,?$ 2) Largest output (at $x=3$ ): $y = \,?$

For the piece $x^2$ defined on $-2 \le x < 0$ :

3) Largest output (at $x=-2$ ): $y = \,?$

For a negative input like $-5$ , the rule $-x$ gives $-(-5)=5$ . That's why $|{-5}|=5$ .

A shifted absolute value

f(x) = |x - 3| = \begin{cases} -(x-3) & x < 3 \\ x - 3 & x \ge 3 \end{cases} = \begin{cases} 3 - x & x < 3 \\ x - 3 & x \ge 3 \end{cases}

The "corner" (vertex) sits where the inside equals zero: $x - 3 = 0 \Rightarrow x = 3$ . The graph is a V with its point at $(3, 0)$ .

💡 To split any $|{\text{stuff}}|$ into pieces, find where stuff $=0$ . Left of that, the inside is negative (so you negate it); right of that, it's positive (so you keep it).

Concept Check 🎯

The Greatest-Integer (Step) Function

The greatest-integer function $\lfloor x \rfloor$ rounds down to the nearest integer — it returns the largest integer less than or equal to $x$ .

$x$	$\lfloor x \rfloor$	Why
$2.7$	$2$	largest integer $\le 2.7$
$2$	$2$	$2$ itself
$-1.3$	$-2$	largest integer $\le -1.3$ is $-2$ , not $-1$ !
$5$	$5$	$5$ itself

Its graph is a series of flat steps, each one unit wide, with a closed dot on the left (included) and an open dot on the right (excluded) — a piecewise function with infinitely many constant pieces.

⚠️ The trap is negatives. $\lfloor -1.3 \rfloor = -2$ , because $-2 \le -1.3$ but $-1 > -1.3$ . "Round down" always moves toward , not toward zero.

Step Function Values 🧮

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

1) $\lfloor 4.9 \rfloor = \,?$ 2) $\lfloor 7 \rfloor = \,?$ 3) $\lfloor -2.5 \rfloor = \,?$

Special Functions 🔽

{0.10x1000+0.20(x−10000)0≤x≤10000x>10000

The first $10,000 is taxed at $10\%$ : that's at most $0.10 \times 10000 = 1000$ , i.e. $1000.
Every dollar above $10,000 is taxed at $20\%$ , and the constant $1000$ carries over the tax already owed on the first bracket.

Since $25000 > 10000$ , use the bottom rule:

$T(25000) = 1000 + 0.20(25000 - 10000) = 1000 + 0.20(15000) = 1000 + 3000 = 4000$

So the tax is $4000.

💡 Continuity matters here too. At $x=10000$ : top gives $0.10(10000)=1000$ ; bottom gives $1000+0.20(0)=1000$ . They match, so the tax has no sudden jump at the bracket edge — exactly how real brackets are designed.

Use the Model 🧮

Use $T(x) = \begin{cases} 0.10\,x & 0 \le x \le 10000 \\ 1000 + 0.20(x - 10000) & x > 10000 \end{cases}$

1) Tax on an income of $8000: $T(8000) = \,?$ 2) Tax on an income of $10000: $T(10000) = \,?$ 3) Tax on an income of $30000: $T(30000) = \,?$

Mixed Practice 🎯

Quick Reference

Task	What to do
Evaluate $f(a)$	Find the one condition $a$ satisfies; plug into that rule
Boundary value	$\le, \ge$ include the endpoint; $<, >$ exclude it
Graph a piece	Draw the rule, keep it over its interval, ● if included / ○ if excluded
Continuous at $c$	Left value $=$ right value $= f(c)$ (set the two rules equal at $c$ )
Domain	Union of all condition-intervals (allowed inputs)
Range	Union of each piece's output set over its interval
$\lfloor x \rfloor$	Round down toward $-\infty$ (watch negatives!)

🔑 The whole topic reduces to one habit: always read the inequality at the boundary, then plug into the right rule.

Exit Quiz ✅

Answer all three to finish the lesson.

x<−1−1≤x<2x≥2

y = - 2 (1) + 6 = 4

(use the correct piece)

Defined value:

x \le 2

includes

2

, so

f(2) = 2^2 = 4

Piecewise Functions

Find $f(-4)$

Find $f(1)$

Find $f(5)$

The Boundary Is Where Points Are Lost

The Endpoint Test

Worked Example

When the Graph Jumps

Worked Example: Is It Continuous?

Now change one number

Solving for a Value That Makes It Continuous

Finding the Range

A shifted absolute value

The Greatest-Integer (Step) Function

Tax on $25,000

Quick Reference

Piecewise Functions

Piecewise Functions - Complete Interactive Lesson

Part 1: Anatomy & Notation

🧩 Piecewise Functions

Topics in This Part

What Is a Piecewise Function?

Reading the Brace Notation

Recap

Part 2: Evaluating Piecewise Functions

🧩 Piecewise Functions

Evaluating, Step by Step

Part 3: Graphing & Open/Closed Dots

🧩 Piecewise Functions

How to Graph a Piecewise Function

Part 4: Continuity at the Boundaries

🧩 Piecewise Functions

The Continuity Test at a Boundary

Part 5: Domain & Range

🧩 Piecewise Functions

Finding the Domain

Part 6: Absolute Value & Step Functions

🧩 Piecewise Functions

Absolute Value Is Piecewise

Part 7: Modeling, Mixed Practice & Exit Quiz

🧩 Piecewise Functions

Real-World Modeling: A Tax Bracket

Find f(−4)f(-4)f(−4)

Find f(1)f(1)f(1)

Find f(5)f(5)f(5)

The Boundary Is Where Points Are Lost

The Endpoint Test

Worked Example

When the Graph Jumps

Worked Example: Is It Continuous?

Now change one number

Solving for a Value That Makes It Continuous

Finding the Range

A shifted absolute value

The Greatest-Integer (Step) Function

Tax on $25,000

Quick Reference

Find $f(-4)$

Find $f(1)$

Find $f(5)$