Function definitions, domains and piecewise rules
Read a rule's domain and choose the appropriate branch before evaluating or solving.
What you will tackle
- Define a function and interpret domain/range
- Move between function notation, tables, graphs and contextual rules
- Evaluate piecewise functions and identify boundary inclusion
- Interpret transformations and solve inverse input questions
A function gives one output for each permitted input
A relation is a function if no input has two different outputs. The domain lists permitted inputs and the range the outputs produced. A vertical-line test expresses this on a graph. Restrictions such as nonzero denominators and nonnegative real square-root radicands must be retained.
Piecewise rules specify separate input regions
For f(x) = x + 2 when x < 1, and f(x) = 2x − 1 when x ≥ 1, the input 1 uses the second branch. Open or closed endpoints on a graph reflect these inequalities. Each branch must be evaluated only where it applies; applying both at a boundary would misread the definition.
Build your notes, then practise
Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.
Guided practice
Guided 1
For the piecewise rule above find f(0), f(1) and f(3). Is the relation {(1,2),(1,3),(2,4)} a function?
Show a starting hint
Select the branch using the input, not the resulting output.
Compare with the explained answer
- 0 < 1 uses x + 2, so f(0) = 2.
- 1 and 3 use 2x − 1, giving 1 and 5.
- The relation gives two outputs for input 1, so it is not a function.
Answer / response criteria: 2; 1; 5; the relation is not a function.
Independent practice
Independent 1
Find the real domain and range of g(x) = √(x − 4). Find the excluded input for r(x) = 3/(x + 2), and solve r(x) = 1.
Compare with the explained answer
- x − 4 ≥ 0 gives domain x ≥ 4; principal square roots give range y ≥ 0.
- The denominator excludes x = −2.
- 3/(x + 2) = 1 gives x + 2 = 3 and x = 1.
Answer / response criteria: g: domain [4,∞), range [0,∞); r excludes −2; r(x) = 1 at x = 1.
Application practice
Application 1
A delivery service charges $5 for parcels up to and including 2 kg, then $5 plus $3 per additional kilogram above 2 kg. Write a piecewise cost function for m > 0, find the cost at 2 kg and 4.5 kg, and interpret the boundary.
Compare with the explained answer
- Use C(m) = 5 for 0 < m ≤ 2, and C(m) = 5 + 3(m − 2) for m > 2.
- C(2) = 5 and C(4.5) = 5 + 7.5 = 12.5.
- The first branch includes 2 kg; the rule assumes continuous rather than rounded-up additional weight.
Answer / response criteria: $5 at 2 kg and $12.50 at 4.5 kg, under the stated rule.
What can you explain now?
A delivery service charges $5 for parcels up to and including 2 kg, then $5 plus $3 per additional kilogram above 2 kg. Write a piecewise cost function for m > 0, find the cost at 2 kg and 4.5 kg, and interpret the boundary.
Compare with the explanation
$5 at 2 kg and $12.50 at 4.5 kg, under the stated rule.
Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.
After trying it yourself, choose your next review. This is your self-assessment.
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Source and review status
Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.
Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.
Official source specification / guide