What is a piecewise function in simple words?

A function that uses different formulas for different ranges of x, like a fare that changes after a certain distance.

How do you know if a dot is open or closed?

Look at the inequality sign: means open (not included); ≤ or ≥ means closed (included).

What is the difference between floor and ceiling?

Floor ⌊x⌋ rounds down to the integer at or below x; ceiling ⌈x⌉ rounds up to the integer at or above x.

Piecewise Functions: Absolute Value and Step Functions

A piecewise function uses different rules on different parts of its domain. To evaluate it, first find which interval x is in, then use only that rule. On the graph, a closed dot ● means the end point is included (≤ or ≥) and an open dot ○ means it is not (< or >). The absolute value function y = |x| is piecewise: −x for x < 0 and x for x ≥ 0, giving a V with vertex (0, 0); y = a|x − h| + k moves the vertex to (h, k). Step functions such as ⌊x⌋, ⌈x⌉ or a parking fee are flat pieces that jump at whole numbers.

What is a piecewise function?

A piecewise function is one function built from two or more rules. Each rule works on its own interval (a part of the x-axis). The intervals must not overlap, so every x gets exactly one output.

We write it with a big brace:

f(x) = x + 2, if x < 1
f(x) = 5 − x, if x ≥ 1

The number where the rule changes (here x = 1) is called the break point or split point.

Evaluating a piecewise function

Two steps, always in this order:

Where? Look at the conditions and find the interval that contains x.
What? Put x into that rule only.

For the function above: f(−2): −2 < 1, so use x + 2 → 0. f(1): 1 ≥ 1, so use 5 − x → 4. f(3): 3 ≥ 1 → 2.

Be careful at the break point: check whether the sign is < or ≤.

Graphing: open and closed dots, jumps

Graph each rule only over its own interval, like cutting a piece from a full line. At each end point:

Closed dot ● – the end point belongs to this piece (≤ or ≥).
Open dot ○ – the end point does not belong (< or >).

If the pieces meet at the same height, the graph is continuous (no gap). If they end at different heights, the graph has a jump. Each x may have only one closed dot above it, otherwise it is not a function (vertical line test).

Domain and range

Read the domain from the intervals and the range from the heights the pieces reach, using the dots to decide if an end value is included.

Absolute value functions

The absolute value |x| is the distance of x from 0, so it is never negative: |−3| = 3, |3| = 3.

As a piecewise function: |x| = −x if x < 0, and x if x ≥ 0. The graph is a V with its corner (vertex) at (0, 0).

Transformations: y = a|x − h| + k

h moves the V right (h > 0) or left.
k moves it up (k > 0) or down.
a changes the steepness (slopes are a and −a); if a < 0 the V opens downward.

The vertex is (h, k). To solve |x − 2| = 3, split it: x − 2 = 3 or x − 2 = −3, so x = 5 or x = −1.

Step functions

A step function is a piecewise function whose pieces are flat (constant). Its graph looks like stairs.

Floor / greatest integer ⌊x⌋: the largest whole number ≤ x. ⌊2.7⌋ = 2, ⌊−1.2⌋ = −2.
Ceiling ⌈x⌉: the smallest whole number ≥ x. ⌈2.1⌉ = 3.

Real step functions: postage by weight band, parking per started hour, mobile plans by GB used. Parking at ₹20 per started hour is fee = 20⌈t⌉ for t > 0.

Try it: a fare table at home

Pick a real tariff, such as a taxi or auto fare (fixed charge for the first 2 km, then a rate per km) or your home electricity slab rates. Write it as a piecewise function. Make a table for 5 values, plot the points on squared paper and mark open and closed dots at the break points. Then check one bill with your rule. In the 3D, set the V in step 6 and predict f(x) before sliding x.

Key formulas and definitions

f(x) = rule 1 if x in interval 1; rule 2 if x in interval 2 …
|x| = −x if x < 0; x if x ≥ 0
y = a|x − h| + k: vertex (h, k), slopes ±a
⌊x⌋ = greatest integer ≤ x; ⌈x⌉ = smallest integer ≥ x
|x − h| = c (c ≥ 0) ⇒ x = h + c or x = h − c

Worked examples

1. f(x) = 3x if x < 2, and x + 4 if x ≥ 2. Find f(0), f(2) and f(5).

f(0): 0 < 2 → 3·0 = 0. f(2): 2 ≥ 2 → 2 + 4 = 6. f(5): 5 ≥ 2 → 5 + 4 = 9.

2. g(x) = 1 if x ≤ 0, and −x if x > 0. Find g(0) and g(0.5).

g(0): 0 ≤ 0 → 1. g(0.5): 0.5 > 0 → −0.5.

3. Find the vertex and the direction of y = −2|x + 1| + 3.

Write it as −2|x − (−1)| + 3, so h = −1, k = 3. Vertex (−1, 3). a = −2 < 0, so the V opens downward and (−1, 3) is the highest point.

4. Is f(x) = x + 1 (x < 2), 3 (x ≥ 2) continuous at x = 2?

Left piece ends at 2 + 1 = 3 (open dot). Right piece starts at 3 (closed dot). Same height, so there is no jump: it is continuous at x = 2.

5. Solve |2x − 1| = 7.

Split: 2x − 1 = 7 → x = 4, or 2x − 1 = −7 → x = −3. Check: |7| = 7, |−7| = 7. Answer x = 4 or x = −3.

6. Electricity costs ₹5 per unit for the first 100 units and ₹7 per unit after that. Write the bill B(u) and find B(160).

B(u) = 5u if 0 ≤ u ≤ 100; B(u) = 500 + 7(u − 100) if u > 100. B(160) = 500 + 7·60 = 500 + 420 = ₹920.

7. Parking costs ₹20 per started hour. Find the fee for 2 h 10 min and for exactly 3 h.

Fee = 20⌈t⌉. 2 h 10 min ≈ 2.17 h → ⌈2.17⌉ = 3 → ₹60. Exactly 3 h → ⌈3⌉ = 3 → ₹60.

Common mistakes

Putting x into every rule. Use only the rule whose condition x satisfies.
Getting the dot wrong at a break point: < and > give an open dot, ≤ and ≥ give a closed dot.
Reading y = |x + 3| as moving right. x + 3 = x − (−3), so the V moves 3 units LEFT.
Thinking ⌊−1.5⌋ = −1. The floor goes DOWN to the next integer: ⌊−1.5⌋ = −2.

Practice quiz

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Piecewise, Absolute Value and Step Functions

🎬 Step-by-step story

🤔 Common doubts, cleared

Why can't both pieces include the break point?

Do I put x into both rules and choose?

Is |x| really a piecewise function?

Why does |x − 2| move the V to the right, not the left?

Why is 1.5 hours of parking charged as 2 hours?

What changes when a is negative?