L40 Solutions — Functions and Function Notation

1. Relations and functions. Fluency

   • (a) Function. Each x-value (2, 3, 4, 5) appears exactly once — every input has exactly one output.
   • (b) Function. Each x-value (1, 2, 3, 4) appears exactly once. This is a constant function — many-to-one is allowed.
   • (c) Not a function. x = 1 appears twice, mapping to both 2 and 3. One input has two outputs.
   • (d) Function. Each x-value (−2, 0, 2) appears exactly once. Many-to-one is fine: −2 and 2 both give 4.

2. Evaluating functions. Fluency

   • (a) f(2) = 3(2) − 4 = 6 − 4 = 2.
   • (b) f(0) = 3(0) − 4 = −4.
   • (c) f(−3) = 3(−3) − 4 = −9 − 4 = −13.
   • (d) f(a+1) = 3(a+1) − 4 = 3a + 3 − 4 = 3a − 1.

3. Quadratic function evaluation. Fluency

   • (a) g(1) = 1 + 2 − 3 = 0. (x = 1 is an x-intercept.)
   • (b) g(−2) = 4 − 4 − 3 = −3.
   • (c) g(0) = 0 + 0 − 3 = −3. (The y-intercept is −3.)
   • (d) x² + 2x − 3 = 5 ⇒ x² + 2x − 8 = 0 ⇒ (x+4)(x−2) = 0. x = −4 or x = 2.

4. Vertical line test and function notation. Fluency

   • (a) Not a function. A vertical line at x = 0 crosses the circle at (0, 4) and (0, −4) — two y-values for one x-value. Fails the vertical line test.
   • (b) Constant function (horizontal line). h(100) = 5 — the output is always 5 regardless of input.
   • (c) 4x + 1 = 17 ⇒ 4x = 16 ⇒ x = 4.
   • (d) x² = 4x − 4 ⇒ x² − 4x + 4 = 0 ⇒ (x−2)² = 0 ⇒ x = 2 (a double root — the graphs touch at one point).

5. Reading a function graph. Understanding

   • (a) x-intercepts: (1, 0) and (3, 0). The parabola crosses the x-axis at x = 1 and x = 3.
   • (b) y-intercept: (0, 3). Substitute x = 0: f(0) = 0 − 0 + 3 = 3.
   • (c) Turning point (vertex): (2, −1). Vertex at x = 4/(2) = 2; f(2) = 4 − 8 + 3 = −1.
   • (d) f(x) < 0 between the x-intercepts: 1 < x < 3. The parabola is below the x-axis in this region.

6. Function operations. Understanding

   • (a) f(x) + g(x) = (2x+3) + (x²−1) = x² + 2x + 2.
   • (b) f(x) × g(x) = (2x+3)(x²−1) = 2x³ − 2x + 3x² − 3 = 2x³ + 3x² − 2x − 3.
   • (c) g(2) = 4 − 1 = 3. Then f(3) = 2(3) + 3 = 9. f(g(2)) = 9.
   • (d) 2x + 3 = x² − 1 ⇒ x² − 2x − 4 = 0. x = (2 ± √20) / 2 = 1 ± √5 (≈ 3.24 or ≈ −1.24).

7. Function notation in context. Understanding

   • (a) C(3) = 80(3) + 120 = 240 + 120 = $360. This is the total cost of hiring the plumber for 3 hours.
   • (b) 80h + 120 = 440 ⇒ 80h = 320 ⇒ h = 4 hours. A $440 bill means the plumber worked for 4 hours.
   • (c) C(0) = 120. This is the call-out fee — the fixed cost just to have the plumber attend, before any work is done.
   • (d) C(h+1) − C(h) = [80(h+1)+120] − [80h+120] = 80. This is the hourly rate — every additional hour adds $80 to the bill, regardless of how many hours have already been worked.

8. Identifying functions from graphs. Understanding

   • (a) Yes, it is a function. A horizontal line passes the vertical line test — each x gives exactly one y (= 5). f(7) = 5.
   • (b) Not a function. A vertical line x = 3 fails the vertical line test: for x = 3, every y-value is in the relation — infinitely many outputs for one input.
   • (c) Yes, y = |x| is a function. Each x gives exactly one output. f(−6) = |−6| = 6.
   • (d) Example: the cost of petrol as a function of litres filled. Input = litres, output = cost ($). For every quantity of petrol, there is exactly one total cost (at a fixed price per litre).

9. Piecewise functions. Problem Solving

   • (a) C(d) = 3.50 + 2.10d for 0 ≤ d ≤ 5; and C(d) = 3.50 + 2.10(5) + 1.60(d−5) = 6 + 1.60d for d > 5.
   • (b) C(3) = 3.50 + 2.10(3) = $9.80. C(5) = 3.50 + 2.10(5) = $14.00. C(8) = 6 + 1.60(8) = $18.80.
   • (c) For d > 5: 6 + 1.60d = 22.10 ⇒ 1.60d = 16.10 ⇒ d = 10.06 km (approximately).
   • (d) Competitor cost: 1.80d. Taxi cheaper when C(d) < 1.80d. For d > 5: 6 + 1.60d < 1.80d ⇒ 6 < 0.20d ⇒ d > 30. The taxi is cheaper for trips longer than 30 km. (For trips under 30 km, the competitor is cheaper due to no flag fall.)

10. Function relationships. Problem Solving

    • (a) T(0) = −5°C (temperature at midnight). T(12) = −0.5(144) + 72 − 5 = −72 + 72 − 5 = −5°C (temperature 12 hours after midnight, i.e. at noon).
    • (b) Vertex: t = −6 / (2 × −0.5) = 6 hours after midnight. T(6) = −0.5(36) + 36 − 5 = 13°C at 6:00 am.
    • (c) −0.5t² + 6t − 5 = 0. Multiply by −2: t² − 12t + 10 = 0. t = (12 ± √(144−40)) / 2 = 6 ± √26 ≈ 0.9 hours and 11.1 hours after midnight.
    • (d) T(t) > 0 for 6 − √26 < t < 6 + √26 (approximately 0.9 < t < 11.1). The temperature is above freezing for about 10.2 hours — from roughly 12:54 am to 11:06 am.