Piecewise-Devined Functions
Understand piecewise-defined functions for Queensland Year 11 Mathematical Methods (QCAA). A piecewise function joins several sub-functions into one, each defined on its own restricted domain, so the branch you use depends on the input.
You will learn to choose and evaluate the correct branch at a boundary, handle open and closed endpoints, state the domain and range, and judge whether the pieces join smoothly — continuity at a join.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1) a piecewise-defined function is built from several sub-functions, each on its own restricted domain. This page shows how to evaluate a piecewise function by choosing the correct branch (including exactly at a boundary), how open and closed endpoints work, how to state the domain and range, and how to test continuity at a join.
A piecewise-defined function (or hybrid function) applies a different rule on each part of its domain. Each rule holds only on its restricted domain, and together those pieces must not overlap.
To evaluate \(f(a)\), first decide which restricted domain contains \(a\), then use that branch. At a boundary the inequality decides: \(\le\) or \(\ge\) (inclusive) means the branch owns that point (a closed dot), while \(<\) or \(>\) (strict) means it does not (an open dot).
The function is continuous at a join if the two branches meet there — the value from the left equals the value from the right, with no jump. Choosing a parameter for continuity means setting those two values equal.
A two-branch piecewise function is written with a brace:
Continuity at the join \(x=a\) requires the two rules to agree there:
How to evaluate and analyse a piecewise function
- Locate: to find \(f(a)\), decide which restricted domain contains \(a\) by checking each inequality, watching the boundary carefully.
- Substitute: put \(a\) into that branch's rule only.
- Analyse: for domain, union the restricted domains; for range, find each branch's outputs (with open/closed ends); for continuity, test whether the branches agree at each join.
\(f(-2)\) — here \(-2<1\), so use \(2x+1\):
| \(f(-2)\) | \(=\) | \(2(-2)+1\) |
| \(=\) | \(-3\) |
\(f(1)\) — the boundary: \(1\ge1\) is inclusive, so use \(5-x\):
| \(f(1)\) | \(=\) | \(5-(1)\) |
| \(=\) | \(4\) |
\(f(3)\) — here \(3\ge1\), so use \(5-x\):
| \(f(3)\) | \(=\) | \(5-(3)\) |
| \(=\) | \(2\) |
\(f(-2)=-3\), \(f(1)=4\), \(f(3)=2\).
Value from the left branch at \(x=2\) (inclusive):
| \(g(2)\) | \(=\) | \((2)^2\) |
| \(=\) | \(4\) |
Value the right branch approaches as \(x\to2\):
| \(x+3\) | \(\to\) | \((2)+3\) |
| \(=\) | \(5\) |
The left gives \(4\) but the right approaches \(5\), so the graph jumps.
Not continuous at \(x=2\) — there is a jump from \(4\) to \(5\).
Domain — join the two restricted domains:
| \(\text{dom}\) | \(=\) | \([-4,0]\cup(0,3]\) |
| \(=\) | \([-4,\,3]\) |
Branch outputs — \(x+4\) on \([-4,0]\) rises from \(0\) to \(4\):
| \(h(-4)\) | \(=\) | \(-4+4=0\) |
| \(h(0)\) | \(=\) | \(0+4=4\) |
\(4-x\) on \((0,3]\) falls from \(4\) to \(1\):
| \(h(3)\) | \(=\) | \(4-3=1\) |
Combine the outputs:
| \(\text{ran}\) | \(=\) | \([0,\,4]\) |
Domain \([-4,3]\); range \([0,4]\).
Set the two branches equal at the join \(x=3\):
| \(2(3)+k\) | \(=\) | \((3)^2-4\) |
Solve for \(k\):
| \(6+k\) | \(=\) | \(9-4\) |
| \(6+k\) | \(=\) | \(5\) |
| \(k\) | \(=\) | \(-1\) |
\(k=-1\) makes \(f\) continuous at \(x=3\).
Common pitfalls
Frequently asked questions
What is a piecewise-defined function?
A function built from several sub-functions, each applying on its own restricted part of the domain, joined together with a brace.
How do you evaluate a piecewise function at a point?
Decide which restricted domain contains the input by checking the inequalities, then substitute the input into that branch's rule only.
Which branch do you use exactly at the boundary?
The one whose inequality is inclusive there: \(x\le a\) or \(x\ge a\) owns the boundary (closed dot), while \(x<a\) or \(x>a\) does not (open dot).
What do open and closed dots mean at a join?
A closed (filled) dot means that branch includes the boundary point; an open (hollow) dot means it excludes it. Exactly one branch owns each boundary.
How do you check if a piecewise function is continuous at a join?
Find the value each branch gives at the join. If they are equal the graph meets with no jump and it is continuous there; if not, it is discontinuous.
How do you find a value that makes a piecewise function continuous?
Set the two branch rules equal at the join and solve for the unknown, for example \(2(3)+k=(3)^2-4\) gives \(k=-1\).