L41 Solutions — Domain, Range and Types of Functions

1. Natural domain. Fluency

   • (a) Linear — no restrictions. Domain: all real numbers (−∞, ∞).
   • (b) Need x − 5 ≥ 0. Domain: x ≥ 5, or [5, ∞).
   • (c) Need x + 3 ≠ 0. Domain: all reals except x = −3.
   • (d) Need 9 − x² ≥ 0 ⇒ x² ≤ 9 ⇒ −3 ≤ x ≤ 3. Domain: [−3, 3].

2. Range of functions. Fluency

   • (a) x² ≥ 0, so f(x) = x² − 4 ≥ −4. Range: y ≥ −4, or [−4, ∞).
   • (b) −x² ≤ 0, so g(x) = −x² + 1 ≤ 1. Range: y ≤ 1, or (−∞, 1].
   • (c) √x ≥ 0 for all valid x. Range: y ≥ 0, or [0, ∞).
   • (d) Linear function — unbounded in both directions. Range: all real numbers.

3. Function types. Fluency

   • (a) f(x) = 2x − 3: linear. g(x) = x² + 1: quadratic. h(x) = 3^x: exponential. k(x) = √(x+2): square root.
   • (b) k(x) = √(x+2) has a restricted domain: need x + 2 ≥ 0, so x ≥ −2.
   • (c) Restricted ranges: g(x) = x²+1 (y ≥ 1); h(x) = 3^x (y > 0); k(x) = √(x+2) (y ≥ 0). The linear function f has range = all reals.
   • (d) |2x−4| ≥ 0 always. Domain: all reals. Range: y ≥ 0. f(x) = 0 when 2x − 4 = 0 ⇒ x = 2.

4. Combined domain restrictions. Fluency

   • (a) Need x − 1 ≥ 0 (x ≥ 1) AND x ≠ 4. Domain: x ≥ 1, x ≠ 4, or [1, 4) ∪ (4, ∞).
   • (b) Need x + 2 > 0 (strictly, since also in denominator). Domain: x > −2, or (−2, ∞).
   • (c) f(−2) = 4, f(−1) = 1, f(0) = 0, f(1) = 1, f(2) = 4. Range: {0, 1, 4}.
   • (d) h(x) = √x > 4 ⇒ x > 16. Smallest integer in domain with h(x) > 4 is x = 17.

5. Domain and range from a graph. Understanding

   • (a) Curve A (y = √x): Domain [0, 3) (closed at 0, open at 3). Range [0, √3) ≈ [0, 1.73).
   • (b) Curve B (y = −x²+2): Domain all reals. Range: vertex at (0, 2), opening down. Range: y ≤ 2, or (−∞, 2].
   • (c) Set √x = −x²+2. Let t = √x ≥ 0: t = −t⁴+2 ⇒ t⁴+t−2 = 0 ⇒ (t−1)(t³+t²+t+2) = 0. Only non-negative root: t = 1, so x = 1. They intersect at (1, 1).
   • (d) −x²+2 ≤ 0 ⇒ x² ≥ 2 ⇒ x ≤ −√2 or x ≥ √2. Solution: x ≤ −√2 or x ≥ √2 (≈ x ≤ −1.41 or x ≥ 1.41).

6. Restricted domains. Understanding

   • (a) With x ≥ 0, x² ≥ 0, so range is still y ≥ 0. The change: the function is now one-to-one (only the right half of the parabola), so it has an inverse.
   • (b) Need 4 − x² ≥ 0 ⇒ −2 ≤ x ≤ 2. Domain: [−2, 2]. At x = 0: y = 2 (max); at x = ±2: y = 0. Range: [0, 2]. This is the upper semicircle of x² + y² = 4.
   • (c) f(−2) = 2(−2)+1 = −3; f(5) = 2(5)+1 = 11. Range: [−3, 11].
   • (d) Need x² − 4 ≠ 0 ⇒ x ≠ ±2. Domain: all reals except x = 2 and x = −2.

7. Real-world domain and range. Understanding

   • (a) Domain: 0 ≤ t ≤ 4 seconds — from the moment the ball is thrown to when it hits the ground.
   • (b) Vertex at t = 20/10 = 2 seconds. h(2) = −5(4) + 40 = 20 metres.
   • (c) h = 0 at t = 0 and t = 4; h = 20 at t = 2. Range: 0 ≤ h ≤ 20 metres, or [0, 20].
   • (d) −5t² + 20t ≥ 15 ⇒ t² − 4t + 3 ≤ 0 ⇒ (t−1)(t−3) ≤ 0. Solution: 1 ≤ t ≤ 3 seconds.

8. Comparing function types. Understanding

   • (a) x³ is an odd function — as x increases through all reals, so does x³ (no turning point, unbounded in both directions). x² always ≥ 0 because squaring removes sign, so its range is restricted to y ≥ 0.
   • (b) a > 0: parabola opens up, vertex is a minimum → range y ≥ k. a < 0: parabola opens down, vertex is a maximum → range y ≤ k.
   • (c) Reflecting in x-axis: y → −y. New range: [−7, 3].
   • (d) Shift right 4: √(x−4). Shift down 3: √(x−4) − 3. Domain: x − 4 ≥ 0 ⇒ x ≥ 4. Range: √(x−4) ≥ 0, so y ≥ −3. Range: y ≥ −3.

9. Finding range algebraically. Problem Solving

   • (a) y = (x+2)/(x−1). Solve: y(x−1) = x+2 ⇒ x(y−1) = y+2 ⇒ x = (y+2)/(y−1). Need y ≠ 1. Range: all reals except y = 1.
   • (b) g(x) = 3 − 2x². Maximum at x = 0: g(0) = 3. As |x| → ∞, g → −∞. Range: y ≤ 3, or (−∞, 3].
   • (c) Need x² − 4 ≥ 0 ⇒ x ≤ −2 or x ≥ 2. Then h(x) = √(x²−4) ≥ 0, and increases as |x| increases. Range: y ≥ 0, or [0, ∞).
   • (d) Opening upward (range y ≥ −5) with vertex at x = 3 and minimum −5. Example: f(x) = (x−3)² − 5 = x² − 6x + 4. (Any positive a gives a valid answer: f(x) = a(x−3)² − 5.)

10. Domain and range analysis. Problem Solving

    • (a) Domain: 0 ≤ n ≤ 50. Units produced must be non-negative and at most 50 per week.
    • (b) Vertex: n = 80/(2×2) = 20 units. P(20) = −2(400) + 80(20) − 400 = −800 + 1600 − 400 = $400 000.
    • (c) P > 0: −2n² + 80n − 400 > 0 ⇒ n² − 40n + 200 < 0. Roots: n = 20 ± 10√2 ≈ 5.86 and 34.14. Profit when 5.86 < n < 34.14, i.e. producing between 6 and 34 units (integer values).
    • (d) P(0) = −400, P(50) = −2(2500)+4000−400 = −1400, P(20) = 400. Minimum on [0,50] is P(50) = −1400. Range: [−1400, 400] (thousand dollars), i.e. −$1.4M to $400K.