| Level | Description | % |
|---|---|---|
| 4 | Thorough, insightful | 80–100 |
| 3 | Considerable (provincial standard) | 70–79 |
| 2 | Some effectiveness | 60–69 |
| 1 | Limited effectiveness | 50–59 |
| R | Insufficient | <50 |
1. \( \frac{(x-4)(x+4)}{x-4}=x+4\to 8\).
2. Rationalize: \(\frac{1}{\sqrt{x+4}+2}\to 1/4\).
3. Same degree → 4/2 = 2 → answer (b).
4. \(\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}=6\), so \(k=6\).
5. \(f'(x)=4x-5\); \(f'(3)=7\). Limit work: \(\frac{2(x+h)^2-5(x+h)+1-(2x^2-5x+1)}{h}=\frac{4xh+2h^2-5h}{h}=4x+2h-5\to 4x-5\).
6. \(f'(x)=-\frac{1}{(x+2)^2}\). Common denominator: \(\frac{(x+2)-(x+h+2)}{h(x+h+2)(x+2)}=\frac{-1}{(x+h+2)(x+2)}\to -1/(x+2)^2\). Power rule (\(f=(x+2)^{-1}\)): \(-(x+2)^{-2}\). ✓
7. \(\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\frac{1}{\sqrt{4+h}+2}\to 1/4\).
8. Limit at 1 doesn't exist (left ≠ right). Continuity fails because 2-sided limit doesn't exist; redefine \(g(1)\) doesn't fix that. To make it continuous, you must change the definitions so that the one-sided limits agree — e.g. redefine the right-side rule.
9. \(f(x+h)-f(x)\) = rise; \(h\) = run; ratio = secant slope; limit \(h\to 0\) = tangent slope = instantaneous rate of change.
10. Removable: hole, e.g. \((x^2-1)/(x-1)\) at 1. Jump: piecewise with different one-sided limits. Infinite: vertical asymptote, e.g. \(1/x\) at 0.
11. Error: \(0/0\) is indeterminate, not "DNE." Correct: factor → \(x+2\to 4\).
12. (i) factor & cancel; (ii) rationalize numerator/denominator; (iii) common denominator (for complex fractions); (iv) divide by highest power (limits at infinity).
13. a) \(s(3)=35,s(1)=75\); ARoC = \(-20\) m/s. b) \(s'(t)=-10t\); \(s'(2)=-20\) m/s. c) \(0=80-5t^2\) → \(t=4\) s.
14. a) \(V(0)=200\) L (full); \(V(20)=0\) (empty). b) \(V(10)=50\); ARoC = \((50-200)/10=-15\) L/min. c) \(V'(t)=200\cdot 2(1-t/20)(-1/20)=-20(1-t/20)\); at 10: \(-10\) L/min (≈ matches \(h=0.01\) estimate).
15. a) Degree 1 over 2 → 0 mg/L (drug clears). b) \(C(2)=12/8=1.5\) mg/L. c) Quotient rule (or numerical): \(C'(t)=\frac{6(t^2+4)-6t(2t)}{(t^2+4)^2}=\frac{24-6t^2}{(t^2+4)^2}\); at 1: \(18/25=0.72\) mg/L/h.