Brace notation, evaluating at a point, graphing with open/closed circles, continuity at breakpoints, and writing piecewise rules from real-world scenarios.
A piecewise function is a function defined by different formulas on different subsets (pieces) of its domain. Instead of one rule that works everywhere, each piece has its own formula that applies only when x satisfies a specified condition.
General form:
f(x) = {
formula₁ if condition₁
formula₂ if condition₂
formula₃ if condition₃
}
The key idea: the function behaves differently depending on where x lives on the number line. Each piece is only active on its own restricted domain.
The large brace groups all the pieces together. Each line is a separate rule. Read each line as: "use [formula] when x satisfies [condition]."
Example function:
f(x) = {
x² + 1 if x < −1
3x if −1 ≤ x < 2
7 if x ≥ 2
}
Piece 1: x² + 1
x < −1
Active only for x strictly less than −1
Piece 2: 3x
−1 ≤ x < 2
Active for x from −1 (included) to 2 (excluded)
Piece 3: 7 (constant)
x ≥ 2
Active for all x at or above 2
Important rule
A well-defined piecewise function has conditions that do not overlap and together cover every x in the domain. Each input x belongs to exactly one piece.
Two-step process:
Worked Example — find f(3), f(0), and f(−2)
f(x) = {
x² + 1 if x < −1
3x if −1 ≤ x < 2
7 if x ≥ 2
}
f(3) — find which piece
x = 3. Is 3 < −1? No. Is −1 ≤ 3 < 2? No. Is 3 ≥ 2? Yes.
Use Piece 3: f(3) = 7
f(3) = 7
f(0) — find which piece
x = 0. Is 0 < −1? No. Is −1 ≤ 0 < 2? Yes.
Use Piece 2: f(0) = 3(0) = 0
f(0) = 0
f(−2) — find which piece
x = −2. Is −2 < −1? Yes.
Use Piece 1: f(−2) = (−2)² + 1 = 4 + 1 = 5
f(−2) = 5
Condition: x ≤ 2 or x ≥ 2
The endpoint x = 2 IS part of this piece. Draw a filled dot at that endpoint.
Condition: x < 2 or x > 2
The endpoint x = 2 is NOT part of this piece. Draw an empty dot at that endpoint.
Graphing Walkthrough
g(x) = {
2x + 3 if x < 1
5 if x ≥ 1
}
Piece 1 (2x + 3, x < 1): Graph the line y = 2x + 3 for all x to the LEFT of 1. Stop at x = 1. Since x < 1 (strict), put an open circle at the point (1, 5).
Piece 2 (5, x ≥ 1): Graph the horizontal line y = 5 for all x at or to the RIGHT of 1. Since x ≥ 1 (inclusive), put a closed circle at (1, 5).
Observation: both pieces land on the same y-value (5) at x = 1 — so this function is continuous at x = 1.
A piecewise function is continuous at a breakpoint x = a if all three of the following are equal:
limx→a⁻ f(x) = limx→a⁺ f(x) = f(a)
(left-hand limit) = (right-hand limit) = (function value)
The graph has no gap or jump at x = a.
Left piece and right piece meet at the same point.
h(x) = { 2x if x < 3 ; 6 if x ≥ 3 }
Left limit at 3: 2(3) = 6
Right limit at 3: 6
h(3) = 6
All equal → continuous
Left and right limits both exist but are different.
The graph visibly "jumps" at the breakpoint.
p(x) = { x + 1 if x < 2 ; 5 if x ≥ 2 }
Left limit at 2: 2 + 1 = 3
Right limit at 2: 5
3 ≠ 5 → jump discontinuity at x = 2
How to test continuity at x = a
The domain of a piecewise function is the union of all the pieces' domains. Every x that belongs to at least one condition is in the domain.
Example
f(x) = {
√x if 0 ≤ x < 4
x − 2 if x ≥ 4
}
Domain: [0, 4) ∪ [4, ∞) = [0, ∞)
If there is a gap between the pieces (for example, piece 1 covers x < 0 and piece 2 covers x > 1), then x-values in that gap are not in the domain of the function.
Scenario:
Income tax is 10% on the first $10,000, 22% on income from $10,001 to $41,775, and 24% on income above $41,775. Write T(x), the tax owed on income x.
T(x) = {
0.10x if 0 ≤ x ≤ 10,000
1,000 + 0.22(x − 10,000) if 10,000 < x ≤ 41,775
7,990 + 0.24(x − 41,775) if x > 41,775
}
Each bracket uses the cumulative tax from lower brackets plus the marginal rate on the remaining income.
The floor function ⌊x⌋ (also called the greatest integer function) is a special piecewise function that rounds every real number DOWN to the nearest integer. It is a step function — its graph looks like a staircase.
⌊2.7⌋ = 2
Round down to 2
⌊3⌋ = 3
Integer stays the same
⌊−1.3⌋ = −2
Round down (more negative)
Each "step" of the staircase corresponds to one piece of the function. Every piece is a constant (horizontal segment) defined on a half-open interval [n, n+1). The left endpoint is included (closed circle) and the right endpoint is excluded (open circle).
Absolute value is a classic piecewise function — it is one of the first examples students encounter without realizing it.
|x| = {
x if x ≥ 0
−x if x < 0
}
For x ≥ 0:
|5| = 5 (use x itself)
For x < 0:
|−5| = −(−5) = 5 (negate x)
Understanding absolute value as piecewise explains why the graph of y = |x| is a V-shape: the left piece (slope −1) and right piece (slope +1) meet at the origin. See the full absolute value guide for equations, inequalities, and transformations.
f(x) = {
−x if x < 0
4 if 0 ≤ x ≤ 3
x² − 5 if x > 3
}
q(x) = {
3x − 1 if x ≤ 2
x + 4 if x > 2
}
Test continuity at the breakpoint x = 2.
Left-hand limit (approach x = 2 from below, use Piece 1):
limx→2⁻ (3x − 1) = 3(2) − 1 = 5
Right-hand limit (approach x = 2 from above, use Piece 2):
limx→2⁺ (x + 4) = 2 + 4 = 6
Function value (x = 2 satisfies x ≤ 2, so use Piece 1):
q(2) = 3(2) − 1 = 5
Left limit (5) ≠ Right limit (6) → jump discontinuity at x = 2. The function is NOT continuous there.
Scenario:
A parking garage charges $3 for the first hour (or any part of it), $2 for each additional full hour, with a maximum daily rate of $15. Write P(t), the cost for t hours (0 < t ≤ 12).
Piece 1: For 0 < t ≤ 1 — only the flat first-hour charge applies.
Piece 2: For 1 < t ≤ 7 — first-hour fee plus $2 per additional hour. At t = 7: 3 + 2(6) = $15 (cap reached).
Piece 3: For 7 < t ≤ 12 — maximum rate applies regardless of additional hours.
P(t) = {
3 if 0 < t ≤ 1
3 + 2⌊t⌋ if 1 < t ≤ 7
15 if 7 < t ≤ 12
}
Verification: P(2.5) = 3 + 2⌊2.5⌋ = 3 + 2(2) = $7. P(8) = $15 (daily max).
A piecewise function is a function defined by different formulas on different parts of its domain. You write it using brace notation: each line shows a formula on the left and a domain condition on the right. To use the function, you first identify which condition your input x satisfies, then apply that piece's formula.
Step 1: Look at your input value x and find which condition it satisfies. Step 2: Use only the formula associated with that condition — ignore all other pieces. Step 3: Substitute x into that formula and simplify. For example, if f(x) = {x² for x < 0; 2x + 1 for x ≥ 0}, then f(3): since 3 ≥ 0, use 2(3) + 1 = 7. And f(−2): since −2 < 0, use (−2)² = 4.
Graph each piece only on its restricted domain. At each endpoint, use a closed circle (filled dot) if that x value is included in the piece (≤ or ≥), or an open circle (empty dot) if excluded (< or >). After sketching all pieces, check continuity at each breakpoint: if the left piece and right piece meet at the same point with the same y-value, the function is continuous there. If they land at different y-values, it is a jump discontinuity.
Interactive problems with step-by-step solutions and private tutoring — free to try.
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