Solutions — Average Rate of Change and Limits

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1. Q1 — Average rate of change over an interval

   (a) AROC = (f(5)−f(1))/(5−1) = (17−5)/4 = 12/4 = 3   (b) AROC = (25−4)/(5−2) = 21/3 = 7   (c) f(0)=0, f(4)=16−12=4. AROC = (4−0)/4 = 1   (d) f(1)=3, f(3)=19. AROC = (19−3)/2 = 16/2 = 8

2. Q2 — Expanding expressions

   (a) (x+h)² = x² + 2xh + h²   (b) (x+h)²−x² = x²+2xh+h²−x² = 2xh + h²   (c) (x+h)³ = x³+3x²h+3xh²+h³ = x³ + 3x²h + 3xh² + h³   (d) 2(x+h)²+3(x+h)−2x²−3x = 2x²+4xh+2h²+3x+3h−2x²−3x = 4xh + 2h² + 3h

3. Q3 — Difference quotient

   (a) (5(x+h)−5x)/h = 5h/h = 5   (b) ((x+h)²−x²)/h = (2xh+h²)/h = 2x + h   (c) (3(x+h)+7−3x−7)/h = 3h/h = 3   (d) ((x+h)²+4(x+h)−x²−4x)/h = (2xh+h²+4h)/h = h(2x+h+4)/h = 2x + h + 4

4. Q4 — Evaluating limits

   (a) lim_h→0(2x+h) = 2x   (b) lim_h→0(3x²+3xh+h²) = 3x²   (c) lim_h→0(4x+2h+1) = 4x + 1   (d) lim_h→0(6x²+6xh+2h²) = 6x²

5. Q5 — AROC geometrically

   (a) (f(3)−f(1))/(3−1) = (9−1)/2 = 8/2 = 4.   (b) x=1 to 2: (4−1)/1=3. x=1 to 1.5: (2.25−1)/0.5=2.5. x=1 to 1.1: (1.21−1)/0.1=2.1. These approach 2 (the instantaneous rate at x=1).   (c) Gradient=4, passes through (1,1): y−1=4(x−1) → y = 4x − 3.

6. Q6 — Difference quotient and limit

   (a) f(x)=x²+3x. f(x+h)=(x+h)²+3(x+h)=x²+2xh+h²+3x+3h. Numerator: 2xh+h²+3h=h(2x+h+3). Quotient: 2x+h+3. Limit: 2x + 3.   (b) f(x)=2x²−x+1. f(x+h)=2x²+4xh+2h²−x−h+1. Numerator: 4xh+2h²−h=h(4x+2h−1). Quotient: 4x+2h−1. Limit: 4x − 1.   (c) Each limit represents the gradient of the tangent to the curve at the point (x, f(x)) — the instantaneous rate of change.

7. Q7 — AROC in context

   (a) h(1)=−5+20=15, h(3)=−45+60=15. AROC=(15−15)/(3−1)=0 m/s. The ball has the same height at t=1 and t=3 — it went up then came back to the same height; average velocity is zero.   (b) h(1+h)=−5(1+h)²+20(1+h)=−5(1+2h+h²)+20+20h=15+10h−5h². h(1)=15. AROC=(10h−5h²)/h=10−5h. Limit: 10 m/s. This is the instantaneous velocity at t=1 s.

8. Q8 — AROC from a table

   (a) AROC over [0,2] = (6−0)/(2−0) = 3   (b) AROC over [1,3] = (24−0)/(3−1) = 24/2 = 12   (c) f(x)=x³−x. (a) (8−2−0)/2=6/2=3 ✓. (b) (27−3−(1−1))/2=24/2=12 ✓.

9. Q9 — Limit process for x³

   (a) (x+h)³=x³+3x²h+3xh²+h³. Numerator: 3x²h+3xh²+h³=h(3x²+3xh+h²). Quotient: 3x² + 3xh + h².   (b) lim_h→0(3x²+3xh+h²) = 3x². The instantaneous rate of change of x³ at any x is 3x².   (c) At x=2: 3(4)=12. At x=−1: 3(1)=3.   (d) g(x)=x³−3x. g(x+h)−g(x)=(x+h)³−3(x+h)−x³+3x=3x²h+3xh²+h³−3h. Quotient: 3x²+3xh+h²−3. Limit: g′(x)=3x²−3. Set 3x²−3=0 → x²=1 → x=1 or x=−1.

10. Q10 — Displacement modelling

    (a) s(3)=9+6=15. s(3+h)=(3+h)²+2(3+h)=9+6h+h²+6+2h=15+8h+h². Avg velocity=(8h+h²)/h=8 + h m/s.   (b) lim_h→0(8+h) = 8 m/s.   (c) Avg velocity over [0,t]: s(t)/t=(t²+2t)/t=t+2. Instantaneous velocity at t=1: s′(t)=2t+2, so at t=1: v=4. Set t+2=4 → t=2 s.