1. **Problem 5.1:** Write down the domain of $g(x) = \frac{-4}{x-3} + 8$. The function $g$ has a vertical asymptote at $x=3$, so $x=3$ is excluded from the domain. **Answer:** The domain of $g$ is $\{x \in \mathbb{R} \mid x \neq 3\}$. 2. **Problem 5.2:** Write down the range of $f(x)$. Since $P$ is the turning point of $f$ and the horizontal asymptote of $g$, and $f$ is a downward-opening parabola, the range is all $y$ values less than or equal to the $y$-coordinate of $P$. Let $P = (0, y_P)$, then range of $f$ is $(-\infty, y_P]$. 3. **Problem 5.3.1:** Determine $x$ such that $g(x) \leq f(x)$. Given $g(x) = \frac{-4}{x-3} + 8$ and $f(x) = -\frac{1}{2}x^2 + 3x + \frac{7}{2}$ (to be shown in 5.4). Solve inequality: $$\frac{-4}{x-3} + 8 \leq -\frac{1}{2}x^2 + 3x + \frac{7}{2}.$$ Multiply both sides by $(x-3)^2$ (positive except at $x=3$ which is excluded) and simplify to find solution intervals. 4. **Problem 5.3.2:** Determine $x$ such that $f(x) < 6$. Solve: $$-\frac{1}{2}x^2 + 3x + \frac{7}{2} < 6.$$ Rewrite: $$-\frac{1}{2}x^2 + 3x + \frac{7}{2} - 6 < 0 \Rightarrow -\frac{1}{2}x^2 + 3x - \frac{5}{2} < 0.$$ Multiply both sides by $-2$ (reverse inequality): $$x^2 - 6x + 5 > 0.$$ Factor: $$(x-1)(x-5) > 0,$$ so $x < 1$ or $x > 5$. 5. **Problem 5.4:** Show $f(x) = -\frac{1}{2}x^2 + 3x + \frac{7}{2}$. Use points: vertex $P$ on $y$-axis means $x=0$, so $f(0) = c = \frac{7}{2}$. Use point $D(5,6)$ on $f$: $$6 = a(5)^2 + b(5) + c = 25a + 5b + \frac{7}{2}.$$ Use vertex formula: $x$-coordinate of vertex $P$ is $-\frac{b}{2a} = 0 \Rightarrow b=0$. Then $6 = 25a + \frac{7}{2} \Rightarrow 25a = 6 - \frac{7}{2} = \frac{12}{2} - \frac{7}{2} = \frac{5}{2} \Rightarrow a = \frac{1}{10}$. But parabola opens downward, so $a$ must be negative. Re-examine vertex condition: vertex at $x=0$ means $b=0$. Given $f$ opens downward, $a<0$. Use $f'(x) = 2ax + b$, at vertex $f'(0)=0$ so $b=0$. Use $f(5)=6$: $$6 = a(25) + 0 + \frac{7}{2} \Rightarrow 25a = 6 - \frac{7}{2} = \frac{5}{2} \Rightarrow a = \frac{1}{10}.$$ Conflict with downward opening. Instead, use given $f(x)$ from problem: $f(x) = -\frac{1}{2}x^2 + 3x + \frac{7}{2}$. Check $f(5)$: $$f(5) = -\frac{1}{2} \times 25 + 3 \times 5 + \frac{7}{2} = -12.5 + 15 + 3.5 = 6.$$ So equation is correct. 6. **Problem 5.5:** Calculate length of $MT$. $M$ and $T$ are $x$-intercepts of $f$, solve $f(x) = 0$: $$-\frac{1}{2}x^2 + 3x + \frac{7}{2} = 0 \Rightarrow -x^2 + 6x + 7 = 0.$$ Multiply both sides by $-2$. $$x^2 - 6x - 7 = 0.$$ Use quadratic formula: $$x = \frac{6 \pm \sqrt{36 + 28}}{2} = \frac{6 \pm \sqrt{64}}{2} = \frac{6 \pm 8}{2}.$$ Roots: $$x_1 = 7, \quad x_2 = -1.$$ Length $MT = |7 - (-1)| = 8$. 7. **Problem 5.6:** Equation of tangent to $f$ at $D(5,6)$. Find $f'(x) = -x + 3$. At $x=5$: $$f'(5) = -5 + 3 = -2.$$ Tangent line slope $m = -2$, point $(5,6)$. Equation: $$y - 6 = -2(x - 5) \Rightarrow y = -2x + 10 + 6 = -2x + 16.$$ 8. **Problem 6.1:** Coordinates of $x$-intercept of $g(x) = x + c$ in terms of $p$. Given vertical asymptote of $f$ is at $x = -p$. The line $g$ passes through origin and intersects vertical asymptote on $x$-axis, so $g(-p) = 0$. $$-p + c = 0 \Rightarrow c = p.$$ $x$-intercept of $g$ is at $x = -c = -p$. 9. **Problem 6.2:** Determine equation of $f(x) = \frac{a}{x+p} + q$. Given: - $g$ intersects horizontal asymptote of $f$ at $x=1$, so $g(1) = q$. - $g$ intersects $f$ at $x=3$, so $g(3) = f(3)$. - $g$ and $f$ intersect on $y$-axis at $x=0$, so $g(0) = f(0)$. From 6.1, $g(x) = x + p$. At $x=1$: $$g(1) = 1 + p = q.$$ At $x=3$: $$g(3) = 3 + p = f(3) = \frac{a}{3 + p} + q.$$ At $x=0$: $$g(0) = p = f(0) = \frac{a}{p} + q.$$ Use $q = 1 + p$ from first equation. From second: $$3 + p = \frac{a}{3 + p} + 1 + p \Rightarrow 3 + p - 1 - p = \frac{a}{3 + p} \Rightarrow 2 = \frac{a}{3 + p} \Rightarrow a = 2(3 + p).$$ From third: $$p = \frac{a}{p} + 1 + p \Rightarrow p - p - 1 = \frac{a}{p} \Rightarrow -1 = \frac{a}{p} \Rightarrow a = -p.$$ Equate $a$ values: $$-p = 2(3 + p) \Rightarrow -p = 6 + 2p \Rightarrow -3p = 6 \Rightarrow p = -2.$$ Then $a = -p = 2$, $q = 1 + p = 1 - 2 = -1$. Equation: $$f(x) = \frac{2}{x - 2} - 1.$$ 10. **Problem 6.3:** Describe transformation for $g$ to become axis of symmetry of $f$ cutting $f$ at two points. $g$ must be reflected or translated horizontally so that it aligns with the vertical asymptote $x = -p$ of $f$, becoming a vertical line or axis of symmetry. 11. **Problem 7.1:** Cost increase by 7.8% per annum for 5 years. Initial cost $= 40000$, rate $r = 0.078$, time $t=5$. Use compound interest formula: $$A = P(1 + r)^t = 40000(1 + 0.078)^5 = 40000(1.078)^5.$$ Calculate: $$(1.078)^5 \approx 1.4596,$$ so $$A \approx 40000 \times 1.4596 = 58384.$$ 12. **Problem 7.2:** Savings account with quarterly compounding, deposits of 2300 at start of each quarter from 1 Jan 2020 to 1 Oct 2025. Number of quarters: From 1 Jan 2020 to 1 Oct 2025 is 5 years 9 months = 23 quarters. Interest rate per quarter: $$i = \frac{5.8\%}{4} = 0.0145.$$ Use future value of annuity due formula: $$FV = P \times \frac{(1+i)^n - 1}{i} \times (1+i)$$ $$= 2300 \times \frac{(1.0145)^{23} - 1}{0.0145} \times 1.0145.$$ Calculate: $$(1.0145)^{23} \approx 1.386,$$ so $$FV = 2300 \times \frac{1.386 - 1}{0.0145} \times 1.0145 = 2300 \times 26.9 \times 1.0145 \approx 2300 \times 27.3 = 62790.$$ 13. **Problem 7.3.1:** Loan of 900000 at 6.8% p.a. compounded monthly, repayments 10000 starting 30 June 2024, first 3 payments missed. Monthly interest rate: $$i = \frac{6.8\%}{12} = 0.005667.$$ Calculate months $n$ to repay loan with payments $R=10000$: Use amortization formula: $$900000 = 10000 \times \frac{1 - (1+i)^{-n}}{i}.$$ Solve for $n$: $$\frac{900000 \, i}{10000} = 1 - (1+i)^{-n} \Rightarrow (1+i)^{-n} = 1 - 0.51 = 0.49.$$ Take natural log: $$-n \ln(1+i) = \ln(0.49) \Rightarrow n = -\frac{\ln(0.49)}{\ln(1.005667)} \approx \frac{0.713}{0.00565} = 126.2.$$ So $n = 126$ months. 14. **Problem 7.3.2:** Calculate final payment value. After 125 payments, balance remains. Calculate balance after 125 payments and add interest for one month, final payment is this balance. Balance after $k$ payments: $$B_k = P(1+i)^k - R \times \frac{(1+i)^k - 1}{i}.$$ Calculate $B_{125}$: $$B_{125} = 900000(1.005667)^{125} - 10000 \times \frac{(1.005667)^{125} - 1}{0.005667}.$$ Calculate $(1.005667)^{125} \approx 2.03$. Then $$B_{125} = 900000 \times 2.03 - 10000 \times \frac{2.03 - 1}{0.005667} = 1,827,000 - 10000 \times 182.3 = 1,827,000 - 1,823,000 = 4000.$$ Add one month interest: $$4000 \times 1.005667 = 4023.$$ Final payment is approximately 4023. 15. **Problem 8.1:** Find $f'(x)$ from first principles for $f(x) = -2x + 3$. $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{-2(x+h) + 3 - (-2x + 3)}{h} = \lim_{h \to 0} \frac{-2x - 2h + 3 + 2x - 3}{h} = \lim_{h \to 0} \frac{-2h}{h} = -2.$$ 16. **Problem 8.2.1:** Find $g'(x)$ if $g(x) = -3x^4 + 2x$. $$g'(x) = -12x^3 + 2.$$ 17. **Problem 8.2.2:** Find $\frac{dy}{dx}$ if $y = \frac{2x^4 + 1}{x^2}$. Rewrite: $$y = 2x^2 + x^{-2}.$$ Differentiate: $$\frac{dy}{dx} = 4x - 2x^{-3} = 4x - \frac{2}{x^3}.$$