Chapter 3 · Functions
Choose the branch containing an input and check coverage and output consistency at boundaries.
Drag a handle or select an object, then tap its destination.
Give your reason before checking.
| t | Applicable branch values | Distinct outputs | Count |
|---|---|---|---|
| 1 | 6 | {6} | 1 |
Tab to an object. Arrow keys move handles; Enter selects a card or places it at a destination.
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Chapter 3 · Functions
Class ______ Name __________________ Date __________
For the rental rule, predict the charge at exactly 1 hour before testing values just below and above 1. Decide which branch contains the boundary.
For 0 < t ≤ 1 charge 6; for 1 < t ≤ 4 charge 6 + 4(t − 1). Record the branch and charge at t = 0.8, 1, 1.01 and 2.5.
Change both inequalities at t = 1 to strict inequalities. Record which input loses its output and restore the original coverage.
Let both rental branches include t = 1, then do the same for h(x) = x + 2 for x < 0 and x² for x ≥ 0. Compare the two formulas at each overlap.
A rule uses x + 1 for x ≤ 2 and 3 for x ≥ 2. Does the overlap invalidate the function? What changes if the second formula becomes 4?
A rule gives x + 1 for x ≤ 2 and 3 for x ≥ 2. Is it a function, and what is its value at 2?
Chapter 3 · Functions · Teacher reference
3.1.2(第6節) · 06_分段函數與臨界點.docx
5 min predict the one-hour charge; 12 min change endpoints; 8 min distinguish overlap from output conflict; 12 min paper evaluation; 3 min exit check.
For the rental rule, predict the charge at exactly 1 hour before testing values just below and above 1. Decide which branch contains the boundary.
Evaluate an input using a branch whose condition contains it. A missing boundary produces a gap only if that input is intended to be in the domain. Overlap is compatible with a function when all applicable branches give the same value: both rental formulas give 6 at t = 1. If overlapping branches give different values, uniqueness fails; at x = 0 the formulas x + 2 and x² give 2 and 0. A jump in a graph is allowed in a function; two different outputs for one input are not.
A rule uses x + 1 for x ≤ 2 and 3 for x ≥ 2. Does the overlap invalidate the function? What changes if the second formula becomes 4?
A rule gives x + 1 for x ≤ 2 and 3 for x ≥ 2. Is it a function, and what is its value at 2?
Yes. At the only overlap x = 2, both branches give 3, so there is one distinct output; f(2) = 3.
