Piecewise & Absolute Value Functions
Ever noticed how a ride-share app charges one rate for the first few kilometres, then a different rate after that? The rule changes depending on how far you travel. That is exactly what a piecewise function does, and it is one of the most useful tools in applied mathematics.
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
A delivery app charges $5 for any trip under 3 km, then adds $1.50 for every kilometre beyond 3 km. How much would a 2 km trip cost? How much would a 5 km trip cost? Why can't we describe both trips with a single simple formula like $C = 5 + 1.5d$?
Piecewise
Key insight: The absolute value of a negative number is positive because the negative sign flips the sign. For example, $|-7| = -(-7) = 7$.
Key facts
- The definition of a piecewise function
- The piecewise definition of absolute value
- How to evaluate piecewise functions at given inputs
Concepts
- Why real-world pricing models often need piecewise rules
- Why $|x|$ is always non-negative
- How the graph of a piecewise function can change direction at a boundary point
Skills
- Evaluate piecewise functions for numerical inputs
- Write and interpret piecewise rules in context
- Solve basic absolute value equations
- Sketch piecewise linear functions
A piecewise function is a function defined by different rules for different parts of its domain. Instead of one formula that works everywhere, we use multiple formulas, each with its own condition.
For example:
$$f(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ 10 - x & \text{if } x \geq 3 \end{cases}$$
To evaluate this function, first check which condition the input satisfies, then apply the matching rule.
How to evaluate a piecewise function
- Look at the input value.
- Find the condition that matches the input.
- Apply only that rule.
- Simplify.
For the function above:
- $f(2)$: since $2 < 3$, use $2x + 1 \Rightarrow f(2) = 2(2) + 1 = 5$
- $f(3)$: since $3 \geq 3$, use $10 - x \Rightarrow f(3) = 10 - 3 = 7$
- $f(5)$: since $5 \geq 3$, use $10 - x \Rightarrow f(5) = 10 - 5 = 5$
Misconceptions to fix
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
A piecewise function uses different rules for different parts of the domain; Always check the condition before choosing which rule to apply
Pause, copy the piecewise definition (different rules for different parts of the domain) and the two-step evaluation procedure (check the condition first, then apply the matching rule) into your book.
Did you get this? True or false: for the piecewise function $f(x) = \begin{cases} 2x+1 & x < 3 \\ 10-x & x \geq 3 \end{cases}$, we evaluate $f(3)$ using the rule $2x + 1$.
Quick check: For $f(x) = \begin{cases} x^2+1 & x \leq 2 \\ 5x-3 & x > 2 \end{cases}$, what is $f(2)$?
We just saw that a piecewise function uses a different rule depending on which part of the domain the input falls into. That raises a question: is there a common function that already has a built-in two-piece rule baked into its definition? This card answers it → the absolute value function $|x|$ is itself piecewise: $x$ when $x \geq 0$, and $-x$ when $x < 0$.
The absolute value of a number is its distance from zero on the number line. Distance is always non-negative, so absolute value always produces a positive result or zero.
We can define absolute value as a piecewise function:
$$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$
This definition explains why $|-7| = 7$: because $-7 < 0$, we use the second piece and get $-(-7) = 7$.
Solving absolute value equations
If $|A| = B$ where $B \geq 0$, then there are two possibilities:
- $A = B$
- $A = -B$
For example, to solve $|2x - 4| = 6$:
- Case 1: $2x - 4 = 6 \Rightarrow 2x = 10 \Rightarrow x = 5$
- Case 2: $2x - 4 = -6 \Rightarrow 2x = -2 \Rightarrow x = -1$
Both solutions should be checked in the original equation.
$|x|$ = distance from zero; always $\geq 0$; $|x| = x$ if $x \geq 0$; $|x| = -x$ if $x < 0$
Pause, copy the absolute value piecewise definition ($|x| = x$ if $x \geq 0$; $|x| = -x$ if $x < 0$) and the distance-from-zero interpretation into your book.
Fill the blanks: drag each token into the matching blank.
The absolute value of a ___ number is ___ because $-x$ means the ___ of $x$. An absolute value equation $|A| = B$ always has ___ cases to solve.
Worked examples · reveal as you go
Evaluate $f(-1)$, $f(2)$, and $f(4)$ for $\displaystyle f(x) = \begin{cases} x^2 + 1 & \text{if } x \leq 2 \\ 5x - 3 & \text{if } x > 2 \end{cases}$
Solve $|3x - 6| = 9$.
A courier company charges $\$5$ for deliveries up to and including $2$ km, and $\$5$ plus $\$1.50$ for each kilometre beyond $2$ km for longer deliveries. Write a piecewise function $C(d)$ for the cost in dollars of a delivery of $d$ km.
Common mistakes · the 4 traps that cost marks
Using the wrong piece at the boundary
At the boundary value, always check whether the condition includes $\leq$, $\geq$, $<$, or $>$. If the condition says $x \leq 3$, then $x = 3$ belongs to that piece. If it says $x < 3$, then $x = 3$ does not.
✓ Fix: Before substituting, write the condition next to the input and tick the one that matches.
Thinking $|x|$ is always positive, so $|x| = x$ for all $x$
Many students incorrectly write $|-5| = -5$ because they forget the piecewise definition. The negative piece $|x| = -x$ is what turns negative inputs into positive outputs.
✓ Fix: Always ask: is the expression inside the absolute value positive or negative? If negative, multiply it by $-1$.
Forgetting the second case in absolute value equations
Equations like $|2x - 1| = 7$ almost always have two solutions. Students frequently stop after finding the first one.
✓ Fix: Every time you see $|A| = B$, immediately write $A = B$ and $A = -B$ side by side.
Writing the extra-distance piece incorrectly in context problems
In a ride-share problem, students sometimes write $1.5d$ for the entire trip when only the distance beyond the threshold should be charged at the higher or lower rate.
✓ Fix: Use $d - \text{threshold}$ for the variable part, not $d$ itself. The fixed fee covers the first part of the trip.
Activity 1, Evaluate the piecewise function
Consider the function $\displaystyle f(x) = \begin{cases} 2x + 3 & \text{if } x < 1 \\ x^2 & \text{if } 1 \leq x \leq 3 \\ 12 - x & \text{if } x > 3 \end{cases}$
Find $f(0)$.
Find $f(1)$.
Find $f(3)$.
Find $f(5)$.
Quick check: For the piecewise function above, which rule is used to evaluate $f(3)$?
Quick-fire practice · 5 reps +2 XP per reveal
For $f(x) = \begin{cases} 3x-1 & x < 2 \\ x^2-1 & x \geq 2 \end{cases}$, find $f(0)$.
Evaluate $|-8|$.
Solve $|x - 2| = 5$.
A shop charges $\$20$ flat for orders under $\$100$, and $\$20$ plus $5\%$ of the amount over $\$100$ for larger orders. What is $C(150)$?
Why does $|x|$ always give a non-negative result?
Earlier you were asked: A delivery app charges $5 for any trip under 3 km, then adds $1.50 for every kilometre beyond 3 km. How much would a 2 km trip cost? How much would a 5 km trip cost? Why can't we describe both trips with a single simple formula?
A 2 km trip costs $5 because it falls under the flat-rate condition ($d < 3$). A 5 km trip costs $5 + 1.50(5-3) = \$8$ because it exceeds the threshold. A single simple formula like $C = 5 + 1.5d$ would charge $5 + 1.5(2) = \$8$ for the 2 km trip, which is wrong. The rule genuinely changes at the 3 km boundary, so a piecewise function is required to model the pricing accurately.
Pick your answer, then rate your confidencethat tells the system what to drill next.
Q1. Consider $\displaystyle f(x) = \begin{cases} 3x - 1 & \text{if } x < 2 \\ 5 & \text{if } x = 2 \\ x^2 - 1 & \text{if } x > 2 \end{cases}$. Find $f(0)$, $f(2)$, and $f(3)$. Show which piece you used for each. (3 marks)
Q2. A theme park charges $\$40$ for entry if you are under 16 years old, and $\$60$ if you are 16 or older. Let $A(x)$ be the admission cost in dollars for a person of age $x$. (a) Write $A(x)$ as a piecewise function. (b) Find $A(15)$ and $A(16)$. (c) Explain why a single linear formula cannot model this pricing structure. (4 marks)
Q3. A student solves $|2x - 4| = 8$ and writes: "$2x - 4 = 8$, so $x = 6$." Evaluate whether this answer is complete. If it is not, find the missing solution and explain why it must be included. (3 marks)
Comprehensive answers (click to reveal)
Multiple choice, drill bank
MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.
1. B$x = 2$ satisfies $x \leq 2$, so $f(2) = 3(2) + 1 = 7$.
2. B$|-5| = 5$.
3. A$|x| = x$ if $x \geq 0$, and $-x$ if $x < 0$.
4. B$4 \geq 3$, so $f(4) = 2(4) + 1 = 9$.
5. B$y = |x - 2|$ has vertex at $(2, 0)$.
Activity 1, Evaluate the piecewise function model answers
1. $f(0) = 2(0) + 3 = 3$ (first piece, since $0 < 1$)
2. $f(1) = (1)^2 = 1$ (second piece, since $1 \leq 1 \leq 3$)
3. $f(3) = (3)^2 = 9$ (second piece, since $1 \leq 3 \leq 3$)
4. $f(5) = 12 - 5 = 7$ (third piece, since $5 > 3$)
Short answer model answers
Q1 (3 marks): $f(0) = 3(0) - 1 = -1$ using $3x - 1$ because $0 < 2$ [1]. $f(2) = 5$ using the middle piece because $x = 2$ [1]. $f(3) = (3)^2 - 1 = 8$ using $x^2 - 1$ because $3 > 2$ [1].
Q2 (4 marks):
(a) $A(x) = \begin{cases} 40 & \text{if } x < 16 \\ 60 & \text{if } x \geq 16 \end{cases}$ [1]
(b) $A(15) = \$40$ and $A(16) = \$60$ [1].
(c) A single linear formula would produce a gradual increase in cost as age increases, but the actual pricing jumps from $\$40$ to $\$60$ at exactly age 16 [1]. A piecewise function is needed because the rate of change is not constant across all ages [1].
Q3 (3 marks): The student's answer is incomplete [1]. They missed the second case: $2x - 4 = -8 \Rightarrow 2x = -4 \Rightarrow x = -2$ [1]. The missing solution must be included because absolute value represents distance from zero, so there are two values of $x$ that give an expression inside the absolute value with magnitude 8 [1].
Five timed questions on piecewise and absolute value functions. Beat the boss to bank a tier, gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaClimb platforms, hit checkpoints, and answer piecewise function questions. Quick recall, lighter than the boss.
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