T4T1 Review Solutions — Functions and Relations

1. Functions and notation. Fluency

   • (a) f(5) = 4(5)−3 = 17.
   • (b) f(−2) = 4(−2)−3 = −11.
   • (c) f(a−1) = 4(a−1)−3 = 4a − 7.
   • (d) 4x−3=21 ⇒ 4x=24 ⇒ x = 6.

2. Function or not? Fluency

   • (a) No — x = 1 maps to both 3 and 9 (one-to-many).
   • (b) Yes — constant function; every x gives exactly one output (7).
   • (c) No — vertical line fails the vertical line test; one x-value has infinitely many y-values.
   • (d) Yes — every x gives exactly one y; passes the vertical line test.

3. Natural domain. Fluency

   • (a) Need 2x+8 ≥ 0 ⇒ x ≥ −4. Domain: x ≥ −4.
   • (b) Need x ≠ 2. Domain: all reals except x = 2.
   • (c) Need x² ≤ 16 ⇒ −4 ≤ x ≤ 4. Domain: [−4, 4].
   • (d) Need x+5 > 0 (strictly, as also under square root and in denominator). Domain: x > −5.

4. Finding inverses. Fluency

   • (a) x=6y+2 ⇒ 6y=x−2 ⇒ f⁻¹(x) = (x−2)/6.
   • (b) x=(y+4)/3 ⇒ 3x=y+4 ⇒ g⁻¹(x) = 3x − 4.
   • (c) x=y³−2 ⇒ y³=x+2 ⇒ h⁻¹(x) = &cbrk;(x+2) (cube root).
   • (d) The inverse reverses each mapping: f⁻¹(10) = 3 and f⁻¹(4) = 7.

5. Graph reading — function properties. Understanding

   • (a) Domain: all reals. f(x) ≤ 4 (maximum at vertex (2,4)). Range: y ≤ 4.
   • (b) x-intercepts: (0, 0) and (4, 0). y-intercept: f(0) = 0, so (0, 0) (same point).
   • (c) The parabola is ≥ 0 between its x-intercepts: 0 ≤ x ≤ 4.
   • (d) No — f is not one-to-one (parabola fails horizontal line test). Restrict domain to x ≥ 2 (or x ≤ 2) to obtain an invertible function.

6. Composite functions. Understanding

   • (a) f(g(x)) = x²+5. g(f(x)) = (x+5)² = x²+10x+25.
   • (b) g(3)=9; f(9)=14. f(g(3))=14. f(3)=8; g(8)=64. g(f(3))=64.
   • (c) x²+5=x²+10x+25 ⇒ 5=10x+25 ⇒ 10x=−20 ⇒ x = −2.
   • (d) x²+5=14 ⇒ x²=9 ⇒ x = 3 or x = −3.

7. Range and inverse. Understanding

   • (a) Range: y ≥ −5 (vertex at (0,−5)). With x≥0: swap gives x=3y²−5 ⇒ y²=(x+5)/3 ⇒ f⁻¹(x) = √((x+5)/3), domain x ≥ −5.
   • (b) f(−3)=−7, f(6)=11. Range of f: [−7, 11]. Domain of f⁻¹: [−7, 11]. Range of f⁻¹: [−3, 6].
   • (c) Domain: x ≠ −2. Solving for inverse: y(x+2)=3x−1 ⇒ x(y−3)=−1−2y ⇒ f⁻¹(x) = (−2x−1)/(x−3). y = 3 is excluded from the range of f.
   • (d) Let u=(−2x−1)/(x−3). f(u)=(3u−1)/(u+2). Numerator: 3(−2x−1)/(x−3)−1=(−6x−3−x+3)/(x−3)=−7x/(x−3). Denominator: (−2x−1)/(x−3)+2=(−2x−1+2x−6)/(x−3)=−7/(x−3). Ratio: (−7x)/(−7)=x ✓.

8. Real-world functions. Understanding

   • (a) c(f(d)) = 2.20 × d/12 = 11d/60 ≈ 0.183d. This gives the petrol cost (dollars) for driving d km.
   • (b) c(f(300)) = 2.20 × 300/12 = 2.20 × 25 = $55.00.
   • (c) Let h(d)=11d/60. x=11d/60 ⇒ d=60x/11. h⁻¹(c) = 60c/11. This gives the distance you can drive on a petrol budget of $c.
   • (d) h⁻¹(40) = 60(40)/11 = 2400/11 ≈ 218 km.

9. Decomposition and domain. Problem Solving

   • (a) g(x)=x+3, f(x)=1/√x. Domain: need x+3 > 0. x > −3.
   • (b) f(g(x))=√(4−x²). Need 4−x² ≥ 0 ⇒ x² ≤ 4. Domain: −2 ≤ x ≤ 2.
   • (c) f(3x−2)=9x²−12x+5=(3x−2)²+1. Let u=3x−2: f(u)=u²+1. So f(x) = x² + 1.
   • (d) If f(g(a))=f(g(b)): since f is one-to-one, g(a)=g(b); since g is one-to-one, a=b. Therefore f∘g is one-to-one. ✓

10. Modelling with functions. Problem Solving

    • (a) Domain: 10 ≤ r ≤ 30. Y(10) = −20+80−60 = 0 tonnes. Y(30) = −180+240−60 = 0 tonnes.
    • (b) Vertex: r = −8/(2×−0.2) = 20 mm. Y(20) = −0.2(400)+160−60 = 20 tonnes.
    • (c) Min at endpoints Y=0, max at vertex Y=20. Range: 0 ≤ Y ≤ 20 tonnes.
    • (d) Complete the square: Y = −0.2(r−20)²+20. Swap: x=−0.2(r−20)²+20 ⇒ (r−20)²=5(20−x) ⇒ r=20+√(100−5x). Y⁻¹(x) = 20+√(100−5x). Represents: rainfall ≥ 20 mm needed to achieve yield x. Y⁻¹(12) = 20+√40 = 20+2√10 ≈ 26.3 mm.

11. Multi-step function problem. Problem Solving

    • (a) 2a+b=7, 5a+b=13 ⇒ 3a=6 ⇒ a=2, b=3. f(x)=2x+3. f⁻¹(x) = (x−3)/2.
    • (b) f(g(x))=2(x−1)+3=2x+1. g(f(x))=(2x+3)−1=2x+2. Setting equal: 2x+1=2x+2 ⇒ 1=2. No solution — the composites are parallel (never intersect).
    • (c) h(x)=f(g(f(x))): f(x)=2x+3 ⇒ g(f(x))=2x+2 ⇒ f(g(f(x)))=2(2x+2)+3=4x+7. h(x)=4x+7. h⁻¹(x) = (x−7)/4.
    • (d) h⁻¹(h(x)) = h⁻¹(4x+7) = (4x+7−7)/4 = 4x/4 = x ✓.

12. Extended problem. Problem Solving

    • (a) K(F(c)) = (9c/5)+32+273.15 = (9c/5) + 305.15 (or 1.8c + 305.15).
    • (b) 1.8(100)+305.15 = 180+305.15 = 485.15 K. (This reveals an error in the K(f) definition — see (d).)
    • (c) f=(9c/5)+32 ⇒ f−32=9c/5 ⇒ c=5(f−32)/9. F⁻¹(f) = 5(f−32)/9.
    • (d) The direct formula K = C+273.15 gives 100°C = 373.15 K. The composite C→F→K using K(f)=f+273.15 gives 485.15 K — these disagree. The issue: K(f)=f+273.15 is wrong because Fahrenheit and Kelvin are not related by a simple shift. The correct direct path (C → K) is K=C+273.15. This demonstrates that incorrect intermediate formulas propagate errors through composition.