(a)

\begin{align} f(x) & = (x - 4)^2 \\ \\ f'(x) & = 2(x - 4). (1) \phantom{000000} [\text{Chain rule}] \\ & = 2(x - 4) \\ \\ \text{For increasing} & \text{ function, } f'(x) > 0 \\ \\ 2(x - 4) & > 0 \\ x - 4 & > 0 \\ x & > 4 \end{align}

(b)

\begin{align} f(x) & = (x - 4)^3 \\ \\ f'(x) & = 3(x - 4)^2 . (1) \phantom{000000} [\text{Chain rule}] \\ & = 3(x - 4)^2 \\ \\ \text{For all real} & \text{ values of } x \text{ except } x = 4, \\ 3(x - 4)^2 & > 0 \\ \\ \therefore f(x) \text{ is incre} &\text{asing for all real values of } x \text{ except } x = 4 \end{align}

(c)

\begin{align} f(x) & = (x - 4)^4 \\ \\ f'(x) & = 4(x - 4)^3 . (1) \phantom{000000} [\text{Chain rule}] \\ & = 4(x - 4)^3 \\ \\ \text{For } x > 4, & \phantom{0} 4(x - 4)^3 > 0 \\ \\ \therefore f(x) \text{ is incre} &\text{asing for } x > 4 \end{align}

(d)

\begin{align} f(x) & = {1 \over x - 4} \\ & = (x - 4)^{-1} \\ \\ f'(x) & = (-1)(x - 4)^{-2} . (1) \phantom{000000} [\text{Chain rule}] \\ & = - (x - 4)^{-2} \\ & = - {1 \over (x - 4)^2} \\ \\ \text{For all } & \text{real values of } x, \\ (x - 4)^2 & \ge 0 \\ -{1 \over (x - 4)^2} & < 0 \\ \\ \therefore \text{Not possible} & \text{ to find the values of } x \end{align}

(i)

\begin{align} u & = 2x - 5 &&& v & = 1 - 6x \\ {du \over dx} & = 2 &&& {dv \over dx} & = -6 \end{align} \begin{align} {dy \over dx} & = {(1 - 6x)(2) - (2x - 5)(-6) \over (1 - 6x)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {2(1 - 6x) + 6(2x - 5) \over (1 - 6x)^2} \\ & = {2 - 12x + 12x - 30 \over (1 - 6x)^2} \\ & = { - 28 \over (1 - 6x)^2 } \\ \\ \text{For } x > {1 \over 6}, & \phantom{0} (1 - 6x)^2 > 0 \\ \\ \therefore {-28 \over (1 - 6x)^2} & < 0 \\ \\ \therefore \text{For } x > {1 \over 6}, & {dy \over dx} < 0 \text{ and } y \text{ is a decreasing function} \end{align}

(ii)

\begin{align} {dy \over dx} & = { - 28 \over (1 - 6x)^2 } \\ & = -28(1 - 6x)^{-2} \\ \\ {d^2 y \over dx^2} & = -28 (-2) (1 - 6x)^{-3} . (-6) \phantom{000000} [\text{Chain rule}] \\ & = -336 (1 - 6x)^{-3} \\ \\ {d^3 y \over dx^3} & = -336 (-3) (1 - 6x)^{-4} . (-6) \\ & = -6048 (1 - 6x)^{-4} \\ & = -{6048 \over (1 - 6x)^4} \\ \\ \text{For } x > {1 \over 6}, & \phantom{0} (1 - 6x)^4 > 0 \\ \\ \therefore -{6048 \over (1 - 6x)^4} & < 0 \\ \\ \therefore \text{For } x > {1 \over 6}, & {d^3 y \over dx^3} < 0 \text{ and } {d^2 y \over dx^2} \text{ is a decreasing function} \end{align}

(i)

\begin{align} u & = x^2 &&& v & = (4 - x)^7 \\ {du \over dx} & = 2x &&& {dv \over dx} & = \underbrace{ 7(4 - x)^6 . (-1)}_\text{Chain rule} \\ & &&& & = -7 (4 - x)^6 \end{align} \begin{align} {dy \over dx} & = (x^2)[-7 (4 - x)^6 ] + (4 - x)^7 (2x) \phantom{000000} [\text{Product rule}] \\ & = -7x^2 (4 - x)^6 + 2x (4 - x)^7 \\ & = x(4 - x)^6 [ -7x + 2(4 - x)] \\ & = x(4 - x)^6 ( -7x + 8 - 2x ) \\ & = x(4 - x)^6 (8 - 9x) \\ & = (4 - x)^6 (8x - 9x^2) \end{align}

(ii)

\begin{align} {dy \over dx} & = (4 - x)^6 (8x - 9x^2) \\ \\ \text{For all real} & \text{ values of } x, (4 - x)^6 > 0 \\ \\ \implies \text{For } {dy \over dx} > 0, & \phantom{0} 8x - 9x^2 > 0 \\ \\ 8x - 9x^2 & > 0 \\ 9x^2 - 8x & < 0 \\ x(9x - 8) & < 0 \end{align}

$$ 0 < x < {8 \over 9} $$

(i)

\begin{align} A & = A_0 (1 + \sqrt{0.7t}) \\ \\ \text{When } & t = 0 \text{ and } A = 50 \phantom{.} 000, \phantom{000000} [\text{initial investment of } \$ 50 \phantom{.} 000] \\ 50 \phantom{.} 000 & = A_0 (1 + \sqrt{0.7(0)}) \\ 50 \phantom{.} 000 & = A_0 (1) \\ 50 \phantom{.} 000 & = A_0 \end{align}

(ii)

\begin{align} A & = A_0 (1 + \sqrt{0.7t}) \\ A & = 50 \phantom{.} 000 (1 + \sqrt{0.7t}) \\ A & = 50 \phantom{.} 000 + 50 \phantom{.} 000 \sqrt{0.7t} \\ A & = 50 \phantom{.} 000 + 50 \phantom{.} 000 \sqrt{0.7} \sqrt{t} \\ A & = 50 \phantom{.} 000 + 50 \phantom{.} 000 \sqrt{0.7} t^{1 \over 2} \\ \\ {dA \over dt} & = 50 \phantom{.} 000 \sqrt{0.7} \left({1 \over 2}\right) t^{-{1 \over 2}} \\ & = 25 \phantom{.} 000 \sqrt{0.7} \left(1 \over \sqrt{t}\right) \\ & = { 25 \phantom{.} 000 \sqrt{0.7} \over \sqrt{t} } \\ \\ \text{For } & t > 0, \phantom{0} \sqrt{t} > 0 \\ \\ \therefore & { 25 \phantom{.} 000 \sqrt{0.7} \over \sqrt{t} } > 0 \\ \\ \text{For } t > 0, & {dA \over dt} > 0 \text{ and } A \text{ is an increasing function} \\ \\ \therefore \text{The } & \text{investor should invest in the portfolio} \end{align}

(i)

\begin{align} \text{For } y = {x - 1 \over 4x + 7} \text{ to be} & \text{ undefined}, \phantom{0} 4x + 7 = 0 \\ \\ 4x + 7 & = 0 \\ 4x & = -7 \\ x & = -{7 \over 4} \\ \\ \therefore k & = - {7 \over 4} \end{align}

(ii)

\begin{align} u & = x - 1 &&& v & = 4x + 7 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 4 \end{align} \begin{align} {dy \over dx} & = { (4x + 7)(1) - (x - 1)(4) \over (4x + 7)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = {4x + 7 - 4x + 4 \over (4x + 7)^2 } \\ & = {11 \over (4x + 7)^2} \\ \\ \text{For } & x \ne -{7 \over 4}, (4x + 7)^2 > 0 \\ \\ & \therefore {11 \over (4x + 7)^2 } > 0 \text{ and } {dy \over dx} > 0 \\ \\ \implies y & \text{ is an increasing function for all real values of } x \\ \\ \\ y & = {x - 1 \over 4x + 7} \times { {1 \over x} \over {1 \over x} } \\ y & = { 1 - {1 \over x} \over 4 + {7 \over x} } \\ \\ \text{As } x \rightarrow \infty, & \phantom{0} y \rightarrow {1 - 0 \over 4 + 0} = {1 \over 4} \\ \\ \text{Horizontal } & \text{asymptote: } y = {1 \over 4} \\ \\ \text{Vertical } & \text{asymptote: } x = -{7 \over 4} \end{align}