L3 · Domain and range in context (set-builder notation)
“I can state the domain and range of a linear function using set-builder notation, and in a real-world context I can find the restrictions the situation imposes - including knowing when the domain is the integers (counted things) and when it is the real numbers (measured things).”
A candle is 15 cm tall and burns down steadily at 1.5 cm per hour, so its height after t hours is h(t) = 15 - 1.5t. Step 1: The EQUATION would accept any t. The SITUATION does not: the candle exists from t = 0 until it is gone. Gone when h = 0: 0 = 15 - 1.5t -> t = 10. Step 2: Domain: D: {t | 0 ≤ t ≤ 10, t ∈ R} - time is measured, so every real value between the endpoints is allowed. Step 3: Range: outputs run from the starting height down to nothing: h(0) = 15, h(10) = 0 -> R: {h | 0 ≤ h ≤ 15, h ∈ R}. Say it plainly: the domain comes from the WORLD (when does the situation start and stop?); the range is then read off the function at the domain's endpoints - for a linear function, the extremes of the output happen at the extremes of the input.
The equation h(t) = 15 - 1.5t gives h(20) = -15. The domain still stops at 10 because...
A hiker walks a trail 18 km long at a steady pace, finishing in 6 hours. Distance covered after t hours is modelled linearly. State D and R. -> D: {t | 0 ≤ t ≤ 6, t ∈ R}; R: {d | 0 ≤ d ≤ 18, d ∈ R} (she starts at 0 km and the situation ends at the trail's end).
"f(x) = 5x is used only for inputs from 0 to 4. D: {x | 0 ≤ x ≤ 4, x ∈ R}. Endpoints: f(0) = ___, f(4) = ___, so R: {y | ___ ≤ y ≤ ___, y ∈ R}" -> 0; 20; 0; 20
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.