Solutions — Functions, Relations and Function Notation

← Back to Questions

1. Q1 — Function or relation

   (a) y=3x−5: Function — linear, every x gives one y.   (b) x²+y²=16: Relation (not a function) — e.g. x=0 gives y=±4. Fails vertical line test.   (c) y=x²: Function — every x gives exactly one y (even though two x-values give the same y, each x still gives only one y).   (d) x=4: Relation (not a function) — infinite y-values for x=4.   (e) y=√x: Function — domain x≥0, every x gives one non-negative y.

2. Q2 — Evaluating f(x) = 2x² − 3x + 1

   (a) f(0)=0−0+1=1   (b) f(2)=2(4)−3(2)+1=8−6+1=3   (c) f(−1)=2(1)−3(−1)+1=2+3+1=6   (d) f(3)=2(9)−9+1=10; f(1)=2−3+1=0; f(3)−f(1)=10

3. Q3 — Natural domain

   (a) f(x)=5x−3: All real numbers, x ∈ ℝ   (b) g(x)=√(2x−6): Need 2x−6≥0, x≥3. Domain: x ≥ 3   (c) h(x)=1/(x−2): Need x−2≠0. Domain: x ≠ 2 (or x ∈ ℝ \ {2})   (d) p(x)=√(9−x²): Need 9−x²≥0, x²≤9, −3≤x≤3. Domain: −3 ≤ x ≤ 3

4. Q4 — Domain and range

   (a) f(x)=x²−4: Domain all reals; vertex at (0,−4) so Range: y ≥ −4   (b) f(x)=√x+1: Domain x≥0; Range y≥1 (since √x≥0). Range: y ≥ 1   (c) f(x)=−x²+9: Maximum at x=0, f(0)=9. Opens downward. Range: y ≤ 9

5. Q5 — Piecewise function evaluation

   f(−3): x<0, use 2(−3)+1=−5   f(0): x=0 is in 0≤x≤3, use 0²=0   f(1): x=1 is in 0≤x≤3, use 1²=1   f(4): x>3, use 10−4=6

6. Q6 — Substituting expressions

   (a) f(2a)=4(2a)−1=8a−1   (b) g(x+1)=(x+1)²+(x+1)=x²+2x+1+x+1=x²+3x+2   (c) h(a+b)=2(a+b)+3=2a+2b+3. h(a)+h(b)−3=(2a+3)+(2b+3)−3=2a+2b+3 ✓

7. Q7 — Taxi piecewise function

   (a) C(d) = { 3.50,                  d ≤ 1
              3.50 + 2.10(d−1),   d > 1 }   (b) d=1: C(1)=$3.50. d=5: 3.50+2.10(4)=3.50+8.40=$11.90   (c) 3.50+2.10(d−1)=14.30 → 2.10(d−1)=10.80 → d−1=10.80/2.10=5.14... → d≈6.14 km (exact: 10.80/2.10 = 36/7 ≈ 5.143, d = 36/7 + 1 = 43/7 ≈ 6.14 km)

8. Q8 — Functions from tables

   (a) Inputs 1,2,3,4 each appear once → each input gives one output. It is a function. (The same output can repeat for different inputs — that is fine.)   (b) Input 2 appears twice with different outputs (6 and 8) → Not a function. One input gives two outputs.   (c) x=k is a vertical line. Every point on it has the same x-value but different y-values → fails the vertical line test → not a function.

9. Q9 — Electricity piecewise function

   (a) C(u) = { 0.22u,                         0 ≤ u ≤ 500
              110 + 0.35(u−500),   u > 500 }   (b) u=480: C(480)=0.22(480)=$105.60. u=750: 110+0.35(250)=110+87.50=$197.50   (c) Try first piece: 0.22u=225.50 → u=1025 > 500, so use second piece. 110+0.35(u−500)=225.50 → 0.35(u−500)=115.50 → u−500=330 → u=830 kWh

10. Q10 — Function composition

    (a) g(2)=4+3=7; f(7)=2(7)−1=13. So f(g(2))=13.   (b) g(f(x))=(2x−1)²+3=4x²−4x+1+3=4x²−4x+4   (c) f(g(2))=13 (from above). f(g(x))=2(x²+3)−1=2x²+5; f(g(2))=2(4)+5=13. g(f(2))=g(3)=9+3=12. Since 13≠12, f(g(x)) ≠ g(f(x)) in general — function composition is not commutative.