(a)

\begin{align} {dy \over dx} & = 4 \\ \\ \text{Since } {dy \over dx} > 0 \text{ for all real} & \text{ values of } x, y \text{ is an increasing function} \end{align}

(b)

\begin{align} {dy \over dx} & = (2)(x) + (2)(1) \\ & = 2x + 2 \\ & = 2(x + 1) \\ \\ \text{For } & x > 0, \\ 2(x + 1) & > 0 \\ \therefore {dy \over dx} & > 0 \\ \\ \text{Thus, } y & \text{ is an increasing function} \end{align}

(c)

\begin{align} y & = \sqrt{x^3} - 1 \\ & = x^{3 \over 2} - 1 \\ \\ {dy \over dx} & = \left(3 \over 2\right)(x^{1 \over 2}) \\ & = {3 \over 2} \sqrt{x} \\ \\ \text{For } & x > 0, \\ \sqrt{x} & > 0 \\ {3 \over 2}\sqrt{x} & > 0 \\ \therefore {dy \over dx} & > 0 \\ \\ \text{Thus, } y & \text{ is an increasing function} \end{align}

(a)

\begin{align} f'(x) & = 0 - (1) \\ & = -1 \\ \\ \text{Since } f'(x) < 0 \text{ for all real} & \text{ values of } x, f(x) \text{ is a decreasing function} \end{align}

(b)

\begin{align} f(x) & = -(x^3 + 1) \\ & = -x^3 - 1 \\ \\ f'(x) & = -(3)(x^2) - 0 \\ & = -3x^2 \\ \\ \text{For all} & \text{ real values of }x, \\ x^2 & > 0 \\ -3x^2 & < 0 \\ \therefore f'(x) & < 0 \\ \\ \text{Thus, } f(x) & \text{ is a decreasing function} \end{align}

(c)

\begin{align} f'(x) & = (6)(x^5) - 5(1) \\ & = 6x^5 - 5 \\ \\ \text{For } & x < 0, \\ x^5 & < 0 \\ 6x^5 & < 0 \\ 6x^5 - 5 & < 0 \\ \therefore f'(x) & < 0 \\ \\ \text{Thus, } f(x) & \text{ is a decreasing function} \end{align}

(i)

\begin{align} {dy \over dx} & = (3)(x^2) + 3(1) \\ & = 3x^2 + 3 \\ & = 3(x^2 + 1) \end{align}

(ii)

\begin{align} \text{For all} & \text{ real values of } x, \\ x^2 & > 0 \\ x^2 + 1 & > 0 \\ 3(x^2 + 1) & > 0 \\ \therefore {dy \over dx} & > 0 \\ \\ \text{Thus, } y \text{ is an increasing } & \text{function and } y \text{ increases as } x \text{ increases} \end{align}

\begin{align} u & = 2x &&& v & = x + 1 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x + 1)(2) - (2x)(1) \over (x + 1)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {2(x + 1) - 2x \over (x + 1)^2} \\ & = {2x + 2 - 2x \over (x + 1)^2} \\ & = {2 \over (x + 1)^2} \\ \\ \text{For all real values} & \text{ of } x \text{ except } x = -1, \\ (x + 1)^2 & > 0 \\ {2 \over (x + 1)^2} & > 0 \\ \therefore {dy \over dx} & > 0 \\ \\ \text{Thus, } y \text{ is an increasing } & \text{function for all real values of } x \text{ except } x = -1 \end{align}

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

(a)

\begin{align} {dy \over dx} & = (2)(x) + 2(1) - 0 \\ & = 2x + 2 \\ \\ {dy \over dx} & > 0 \phantom{00000} [\text{Increasing function}] \\ 2x + 2 & > 0 \\ 2x & > - 2 \\ x & > {-2 \over 2} \\ x & > - 1 \end{align}

(b)

\begin{align} {dy \over dx} & = 2(3)(x^2) - 3(2)(x) + 0 \\ & = 6x^2 - 6x \\ \\ {dy \over dx} & > 0 \phantom{00000} [\text{Increasing function}] \\ 6x^2 - 6x & > 0 \\ x^2 - x & > 0 \\ x(x - 1) & > 0 \end{align}

$$ x < 0 \text{ or } x > 1 $$

(a)

\begin{align} {dy \over dx} & = 3(2)(x) + 4(1) - 0 \\ & = 6x + 4 \\ \\ {dy \over dx} & < 0 \phantom{00000} [\text{Decreasing function}] \\ 6x + 4 & < 0 \\ 6x & < -4 \\ x & < {-4 \over 6} \\ x & < -{2 \over 3} \end{align}

(b)

\begin{align} {dy \over dx} & = 2(3)(x^2) - 9(2)(x) + 12(1) - 0 \\ & = 6x^2 - 18x + 12 \\ \\ {dy \over dx} & < 0 \phantom{00000} [\text{Decreasing function}] \\ 6x^2 - 18x + 12 & < 0 \\ x^2 - 3x + 2 & < 0 \\ (x - 2)(x - 1) & < 0 \end{align}

$$ 1 < x < 2 $$

(i)

\begin{align} {dy \over dx} & = (4)(x + 1)^3 . (1) \phantom{000000} [\text{Chain rule}] \\ & = 4(x + 1)^3 \\ \\ {dy \over dx} & > 0 \phantom{00000} [\text{Increasing function}] \\ 4(x + 1)^3 & > 0 \\ (x + 1)^3 & > 0 \\ x + 1 & > \sqrt[3]{0} \\ x + 1 & > 0 \\ x & > - 1 \end{align}

(ii)

\begin{align} {dy \over dx} & < 0 \phantom{00000} [\text{Decreasing function}] \\ 4(x + 1)^3 & < 0 \\ (x + 1)^3 & < 0 \\ x + 1 & < \sqrt[3]{0} \\ x + 1 & < 0 \\ x & < - 1 \end{align}

(i)

\begin{align} u & = x^2 - 1 &&& v & = x^2 + 1 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2 + 1)(2x) - (x^2 - 1)(2x) \over (x^2 + 1)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {2x(x^2 + 1) - 2x(x^2 - 1) \over (x^2 + 1)^2} \\ & = {2x^3 + 2x - 2x^3 + 2x \over (x^2 + 1)^2} \\ & = {4x \over (x^2 + 1)^2} \\ \\ {dy \over dx} & > 0 \phantom{00000} [\text{Increasing function}] \\ {4x \over (x^2 + 1)^2} & > 0 \\ \\ \text{Since } (x^2 + 1)^2 > 0 & \text{ for all real values of } x, \\ 4x & > 0 \\ x & > 0 \end{align}

(ii)

\begin{align} {dy \over dx} & < 0 \phantom{00000} [\text{Decreasing function}] \\ {4x \over (x^2 + 1)^2} & < 0 \\ \\ \text{Since } (x^2 + 1)^2 > 0 & \text{ for all real values of } x, \\ 4x & < 0 \\ x & < 0 \end{align}

When the metal starts cooling and temperature falls, the function should behaves as a decreasing function.

\begin{align} {dT \over dx} & = -5(2)(x) + 90(1) + 0 \\ & = -10x + 90 \\ \\ {dT \over dx} & < 0 \phantom{00000} [\text{Decreasing function}]\\ -10x + 90 & < 0 \\ -10x & < -90 \\ 10x & > 90 \\ x & > {90 \over 10} \\ x & > 9 \\ \\ \therefore \text{Metal} & \text{ starts cooling after 9 seconds} \end{align}

(i)

(ii)

The converse is true only if the gradient function ${dy \over dx}$ is defined for all real values of $x$

(i) I think's a typo in the answer for ${dy \over dx}$

\begin{align} u & = 3x + 4 &&& v & = \sqrt{2x - 1} \\ & &&& & = (2x - 1)^{1 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = {1 \over 2}(2x - 1)^{-{1 \over 2}} . (2) \phantom{00000} [\text{Chain rule}] \\ & &&& & = (2x - 1)^{-{1 \over 2}} \\ & &&& & = {1 \over \sqrt{2x - 1}} \end{align} \begin{align} {dy \over dx} & = {(\sqrt{2x - 1})(3) - (3x + 4)\left(1 \over \sqrt{2x - 1}\right) \over (\sqrt{2x - 1})^2} \phantom{00000} [\text{Quotient rule}] \\ & = {3\sqrt{2x - 1} - {3x + 4 \over \sqrt{2x - 1}} \over 2x - 1} \\ & = {3\sqrt{2x - 1} - {3x + 4 \over \sqrt{2x - 1}} \over 2x - 1} \times {\sqrt{2x - 1} \over \sqrt{2x - 1}} \\ & = {3\sqrt{2x - 1}(\sqrt{2x - 1}) - (3x + 4) \over (2x - 1)\sqrt{2x - 1}} \\ & = {3(2x - 1) - 3x - 4 \over (2x - 1)(2x - 1)^{1 \over 2}} \\ & = {6x - 3 - 3x - 4 \over (2x - 1)^{3 \over 2}} \\ & = {3x - 7 \over \sqrt{(2x - 1)^3}} \\ \\ {dy \over dx} & > 0 \phantom{00000} [\text{Increasing function}] \\ {3x - 7 \over \sqrt{(2x - 1)^3}} & > 0 \\ \\ \text{For } x > {1 \over 2}, & \sqrt{(2x - 1)^3} > 0 \\ 3x - 7 & > 0 \\ 3x & > 7 \\ x & > {7 \over 3} \end{align}

(ii)

\begin{align} {dy \over dx} & > 0 \phantom{00000} [\text{Decreasing function}] \\ {3x - 7 \over \sqrt{(2x - 1)^3}} & < 0 \\ \\ \text{For } x > {1 \over 2}, & \sqrt{(2x - 1)^3} > 0 \\ \\ \therefore 3x - 7 & < 0 \\ 3x & < 7 \\ x & < {7 \over 3} \\ \\ \text{Since } y = {3x + 4 \over \sqrt{2x - 1}} & \text{ is only defined for } x > {1 \over 2}, \\ {1 \over 2} < & \phantom{.} x < {7 \over 3} \end{align}

(i)

\begin{align} u & = x^2 - 2ax - 2 &&& v & = x - a \\ {du \over dx} & = 2x - 2a &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x - a)(2x - 2a) - (x^2 - 2ax - 2)(1) \over (x - a)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {2x^2 - 2ax - 2ax + 2a^2 - (x^2 - 2ax - 2) \over (x - a)^2} \\ & = {2x^2 - 4ax + 2a^2 - x^2 + 2ax + 2 \over (x - a)^2 } \\ & = {x^2 - 2ax + 2a^2 + 2 \over (x - a)^2} \\ & = {x^2 - 2ax + a^2 + a^2 + 2 \over (x - a)^2} \\ & = {x^2 - 2ax + a^2 \over (x - a)^2} + {a^2 + 2 \over (x - a)^2} \\ & = {(x - a)^2 \over (x - a)^2} + {a^2 + 2 \over (x - a)^2} \\ & = 1 + {a^2 + 2 \over (x - a)^2} \end{align}

(ii)

\begin{align} {dy \over dx} & = 1 + {a^2 + 2 \over (x - a)^2} \\ \\ \text{For all } & \text{real values of } a, \\ a^2 > 0 & \text{ and } (x - a)^2 \ge 0 \\ \\ \therefore 1 + {a^2 + 2 \over (x - a)^2} & > 0 \\ {dy \over dx} & > 0 \\ \\ \therefore \text{For all } & \text{real values of } a \end{align}