(a) To find the x-intercept, let y = 0

\begin{align} y & = {2 - x \over x^2} \\ \\ \text{Let } & y = 0, \\ 0 & = {2 - x \over x^2} \\ 0 & = 2 - x \\ x & = 2 \\ \implies x & \text{-intercept is 2} \\ \\ y & = {2 \over x^2} - {x \over x^2} \\ & = 2x^{-2} - {1 \over x} \\ & = 2x^{-2} - x^{-1} \\ \\ {dy \over dx} & = 2(-2)x^{-3} - (-1)x^{-2} \\ & = -4x^{-3} + x^{-2} \\ & = -{4 \over x^3} + {1 \over x^2} \\ \\ \text{Let } & x = 2, \\ {dy \over dx} & = -{4 \over (2)^3} + {1 \over (2)^2} \\ & = -{1 \over 4} \end{align}

Note: For the differentiation part, you can use quotient rule as well

(b)

\begin{align} y & = {16x^3 - 1 \over x^2} \\ & = {16x^3 \over x^2} - {1 \over x^2} \\ & = 16x - x^{-2} \\ \\ {dy \over dx} & = 16 - (-2)x^{-3} \\ & = 16 + 2x^{-3} \\ & = 16 + {2 \over x^3} \\ \\ \text{Let } & x = 1, \\ {dy \over dx} & = 16 + {2 \over (1)^3} \\ & = 18 \end{align}

Note: For the differentiation part, you can use quotient rule as well

(c)

\begin{align} y & = \sqrt{x} + {1 \over \sqrt{x}} \\ & = x^{1 \over 2} + x^{-{1 \over 2}} \\ \\ {dy \over dx} & = {1 \over 2}x^{-{1 \over 2}} + \left(-{1 \over 2}\right) x^{-{3 \over 2}} \\ & = {1 \over 2} \left( 1 \over \sqrt{x} \right) - {1 \over 2} \left(1 \over \sqrt{x^3}\right) \\ & = {1 \over 2\sqrt{x}} - {1 \over 2\sqrt{x^3}} \\ \\ \text{Let } & x = 4, \\ {dy \over dx} & = {1 \over 2\sqrt{4}} - {1 \over 2\sqrt{(4)^3}} \\ & = {3 \over 16} \end{align}

(d)

\begin{align} y & = x^2 - 4x \\ \\ {dy \over dx} & = 2x - 4 \\ \\ \text{Substitute } & y = -4 \text{ into eqn of curve,} \\ -4 & = x^2 - 4x \\ 0 & = x^2 - 4x + 4 \\ 0 & = x^2 - 2(x)(2) + 2^2 \\ 0 & = (x - 2)^2 \phantom{00000000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ 0 & = x - 2\\ 2 & = x \\ \implies \text{Coor} & \text{dinates of point is (2, -4)} \\ \\ \text{Substitute } & x = 2 \text{ into } {dy \over dx}, \\ {dy \over dx} & = 2(2) - 4 \\ & = 0 \end{align}

(e) At the y-intercept, x = 0

\begin{align} y & = 3 - 5x - 2x^2 \\ \\ {dy \over dx} & = -5 - 2(2)x \\ & = -5 - 4x \\ \\ \text{Let } & x = 0, \\ {dy \over dx} & = -5 - 4(0) \\ & = -5 \end{align}

(f)

\begin{align} y & = 2x^2 - 3x + 1 \\ \\ {dy \over dx} & = 2(2)x - 3 \\ & = 4x - 3 \\ \\ \text{Substitute } & y = 21 \text{ into eqn of curve,} \\ 21 & = 2x^2 - 3x + 1 \\ 0 & = 2x^2 - 3x - 20 \\ 0 & = (2x + 5)(x - 4) \end{align} \begin{align} 2x + 5 & = 0 && \text{ or } & x - 4 & = 0 \\ 2x & = -5 &&& x & = 4 \\ x & = -{5 \over 2} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = 4\left(-{5 \over 2}\right) - 3 &&& {dy \over dx} & = 4(4) - 3 \\ & = -13 &&& & =13 \end{align}

(a)

\begin{align} {d \over dx} (ax^2 + bx)^6 & = 6(ax^2 + bx)^5 . (2ax + b) \phantom{00000} [\text{Chain rule}] \end{align}

(b) Use product rule for the differentiation

\begin{align} u & = (2ax + b)^5 &&& v & = (5x^2 - ab)^6 \\ {du \over dx} & = 5(2ax + b)^4 . (2a) &&& {dv \over dx} & = 6(5x^2 - ab)^5 . (10x) \phantom{00000} [\text{Chain rule (for both)}] \\ & = 10a (2ax + b)^4 &&& & = 60x (5x^2 - ab)^5 \end{align} \begin{align} {d \over dx} (2ax + b)^5(5x^2 - ab)^6 & = (2ax + b)^5 [60x(5x^2 - ab)^5] + (5x^2 - ab)^6 [10a(2ax+b)^4] \\ & = 60x(2ax + b)^5 (5x^2 - ab)^5 + 10a(2ax + b)^4 (5x^2 - ab)^6 \\ & = (2ax + b)^4 (5x^2 - ab)^5 [60x(2ax + b) + 10a(5x^2 - ab)] \\ & = 10(2ax + b)^4 (5x^2 - ab)^5 [6x(2ax + b) + a(5x^2 - ab)] \\ & = 10(2ax + b)^4 (5x^2 - ab)^5 (12ax^2 + 6bx + 5ax^2 - a^2 b) \\ & = 10(2ax + b)^4 (5x^2 - ab)^5 (17ax^2 + 6bx - a^2 b) \end{align}

(c) Use quotient rule for the differentiation (product rule is possible as well)

\begin{align} u & = 2ax^2 + bx &&& v & = bx^3 - cx \\ {du \over dx} & = 4ax + b &&& {dv \over dx} & = 3bx^2 - c \end{align} \begin{align} {d \over dx} \left( 2ax^2 + bx \over bx^3 - cx \right) & = {(bx^3 - cx)(4ax + b) - (2ax^2 + bx)(3bx^2 - c) \over (bx^3 - cx)^2 } \\ & = { 4abx^4 + b^2 x^3 - 4acx^2 - bcx - (6abx^4 - 2acx^2 + 3b^2x^3 - bcx) \over (bx^3 - cx)^2} \\ & = { 4abx^4 + b^2 x^3 - 4acx^2 - bcx - 6abx^4 + 2acx^2 - 3b^2x^3 + bcx \over (bx^3 - cx)^2} \\ & = { -2abx^4 - 2b^2 x^3 - 2acx^2 \over (bx^3 - cx)^2} \\ & = { -(2abx^4 + 2b^2 x^3 + 2acx^2) \over (bx^3 - cx)^2} \\ & = - { 2abx^4 + 2b^2 x^3 + 2acx^2 \over (bx^3 - cx)^2} \end{align}

(d) Use quotient rule for the differentiation (product rule is possible as well)

\begin{align} u & = ax + b &&& v & = cx^2 - a^2 \\ {du \over dx} & = a &&& {dv \over dx} & = 2cx \end{align} \begin{align} {d \over dx} \left( {ax + b \over cx^2 - a^2} \right) & = {(cx^2 - a^2)(a) - (ax + b)(2cx) \over (cx^2 - a^2)^2 } \\ & = {acx^2 - a^3 - 2acx^2 - 2bcx \over (cx^2 - a^2)^2} \\ & = {-acx^2 - a^3 - 2bcx \over (cx^2 - a^2)^2} \\ & = {-(acx^2 + a^3 + 2bcx) \over (cx^2 - a^2)^2} \\ & = -{acx^2 + a^3 + 2bcx \over (cx^2 - a^2)^2} \end{align}

(a)

\begin{align} \text{Let } y & = (5x - 4)^5 \\ \\ {dy \over dx} & = 5(5x - 4)^4 . (5) \phantom{00000} [\text{Chain rule}] \\ & = 25(5x - 4)^4 \\ \\ {d^2 y \over d x^2} & = 25(4)(5x - 4)^3 . (5) \phantom{00000} [\text{Chain rule}] \\ & = 500(5x - 4)^3 \end{align}

(b)

\begin{align} \text{Let } y & = \sqrt{2x^3 + 5} \\ & = (2x^3 + 5)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(2x^3 + 5)^{-{1 \over 2}}. (6x^2) \phantom{00000} [\text{Chain rule}] \\ & = 3x^2 (2x^3 + 5)^{-{1 \over 2}} \\ & = {3x^2 \over \sqrt{2x^3 +5}} \end{align}
\begin{align} u & = 3x^2 &&& v & = \sqrt{2x^3 + 5} \\ & &&& & = (2x^3 + 5)^{1 \over 2} \\ {du \over dx} & = 6x &&& {dv \over dx} & = {1 \over 2} (2x^3 + 5)^{-{1 \over 2}} . (6x^2) \\ & &&& & = (3x^2) (2x^3 + 5)^{-{1 \over 2}} \\ & &&& & = {3x^2 \over \sqrt{2x^3 + 5}} \end{align} \begin{align} {d^2 y \over dx^2} & = {(\sqrt{2x^3 + 5})(6x) - (3x^2) \left( {3x^2 \over \sqrt{2x^3 + 5}} \right) \over (\sqrt{2x^3 + 5})^2} \phantom{00000} [\text{Quotient rule}] \\ & = { {6x\sqrt{2x^3 +5} \over 1} - {9x^4 \over \sqrt{2x^3 + 5}} \over 2x^3 + 5} \\ & = { {6x\sqrt{2x^3 +5}(\sqrt{2x^3 + 5}) \over \sqrt{2x^3 + 5}} - {9x^4 \over \sqrt{2x^3 + 5}} \over 2x^3 + 5} \\ & = { {6x(2x^3 + 5) - 9x^4 \over \sqrt{2x^3 + 5}} \over 2x^3 + 5} \\ & = { 6x(2x^3 + 5) - 9x^4 \over (\sqrt{2x^3 + 5})(2x^3 + 5)} \\ & = { 12x^4 + 30x - 9x^4 \over (2x^3 + 5)^{1 \over 2} (2x^3 + 5) } \\ & = {3x^4 + 30x \over (2x^3 + 5)^{3 \over 2} } \\ & = {3x^4 + 30x \over \sqrt{(2x^3 + 5)^3} } \end{align}

(c)

\begin{align} \text{Let } y & = (5x^2 + 3)^{3 \over 2} \\ \\ {dy \over dx} & = {3 \over 2}(5x^2 + 3)^{1 \over 2} . (10x) \phantom{00000} [\text{Chain rule}] \\ & = 15x (5x^2 + 3)^{1 \over 2} \\ & = 15x \sqrt{5x^2 + 3} \end{align}
\begin{align} u & = 15x &&& v & = \sqrt{5x^2 + 3} \\ & &&& &= (5x^2 + 3)^{1 \over 2} \\ {du \over dx} & = 15 &&& {dv \over dx} & = {1 \over 2} (5x^2 + 3)^{-{1 \over 2}} . (10x) \\ & &&& & = 5x (5x^2 + 3)^{-{1 \over 2}} \\ & &&& & = {5x \over \sqrt{5x^2 + 3}} \end{align} \begin{align} {d^2 y \over dx^2} & = (15x) \left({5x \over \sqrt{5x^2 + 3}}\right) + (\sqrt{5x^2 + 3})(15) \phantom{00000} [\text{Product rule}] \\ & = {75x^2 \over \sqrt{5x^2 + 3}} + {15 \sqrt{5x^2 + 3} \over 1} \\ & = {75x^2 \over \sqrt{5x^2 + 3}} + {15 \sqrt{5x^2 + 3} (\sqrt{5x^2 + 3}) \over \sqrt{5x^2 + 3}} \\ & = {75x^2 + 15(5x^2 + 3) \over \sqrt{5x^2 + 3}} \\ & = {75x^2 + 75x^2 + 45 \over \sqrt{5x^2 + 3}} \\ & = {150x^2 + 45 \over \sqrt{5x^2 + 3}} \end{align}

(d)

\begin{align} \text{Let } y & = (2 + 9x^2)^{1 \over 3} \\ \\ {dy \over dx} & = {1 \over 3}(2 + 9x^2)^{-{2 \over 3}} . (18x) \phantom{00000} [\text{Chain rule}] \\ & = 6x (2 + 9x^2)^{-{2 \over 3}} \\ & = {6x \over \sqrt[3]{(2 + 9x^2)^2}} \end{align}
\begin{align} u & = 6x &&& v & = (2 + 9x^2)^{-{2 \over 3}} \\ {du \over dx} & = 6 &&& {dv \over dx} & = -{2 \over 3} (2 + 9x^2)^{-{5 \over 3}} . (18x) \\ & &&& & = -12x (2 + 9x^2)^{-{5 \over 3}} \end{align} \begin{align} {d^2 y \over dx^2} & = (2 + 9x^2)^{-{2 \over 3}}(6) + (6x)[-12x (2 + 9x^2)^{-{5 \over 3}}] \phantom{00000} [\text{Product rule}] \\ & = 6(2 + 9x^2)^{-{2 \over 3}} - 72x^2 (2 + 9x^2)^{-{5 \over 3}} \\ & = 6(2 + 9x^2)^{-{2 \over 3}} \left[ 1 - 12x^2 (2 + 9x^2)^{-1} \right] \\ & = {6 \over (2 + 9x^2)^{2 \over 3}} \left( 1 - {12x^2 \over 2 + 9x^2} \right) \\ & = {6 \over (2 + 9x^2)^{2 \over 3}} \left( {2 + 9x^2 \over 2 + 9x^2} - {12x^2 \over 2 + 9x^2} \right) \\ & = {6 \over (2 + 9x^2)^{2 \over 3}} \left( {2 - 3x^2 \over 2 + 9x^2} \right) \\ & = {6(2 - 3x^2) \over (2 + 9x^2)^{5 \over 3}} \\ & = {12 - 18x^2 \over \sqrt[3]{(2 + 9x^2)^5}} \end{align}

(e)

\begin{align} \text{Let } y & = {3 \over (3x - 5)^3} \\ & = 3(3x - 5)^{-3} \\ \\ {dy \over dx} & = 3(-3)(3x - 5)^{-4} . (3) \phantom{00000} [\text{Chain rule}] \\ & = -27 (3x - 5)^{-4} \\ & = -{27 \over (3x - 5)^4} \\ \\ {d^2 y \over dx^2} & = -27 (-4) (3x - 5)^{-5} . (3) \phantom{00000} [\text{Chain rule}] \\ & = 324 (3x - 5)^{-5} \\ & = {324 \over (3x - 5)^5} \end{align}

(f) Use quotient rule for the differentiation (product rule is possible as well)

$$ \text{Let } y = {3x - 7 \over \sqrt{2x + 9}} $$ \begin{align} u & = 3x - 7 &&& v & = \sqrt{2x + 9} \\ & &&& & = (2x + 9)^{1 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = {1 \over 2} (2x + 9)^{-{1 \over 2}} . (2) \\ & &&& & = (2x + 9)^{-{1 \over 2}} \\ & &&& & = {1 \over \sqrt{2x + 9}} \end{align} \begin{align} {dy \over dx} & = {(\sqrt{2x + 9})(3) - (3x - 7)\left({1 \over \sqrt{2x + 9}}\right) \over (\sqrt{2x + 9})^2} \phantom{00000} [\text{Quotient rule}] \\ & = {{3\sqrt{2x + 9} \over 1} - {3x - 7 \over \sqrt{2x + 9}} \over 2x + 9 } \\ & = {{3\sqrt{2x + 9}(\sqrt{2x + 9}) \over \sqrt{2x + 9}} - {3x - 7 \over \sqrt{2x + 9}} \over 2x + 9 } \\ & = {{3(2x + 9) - (3x - 7) \over \sqrt{2x + 9}} \over 2x + 9} \\ & = {3(2x + 9) - (3x - 7) \over (\sqrt{2x + 9})(2x + 9)} \\ & = {6x + 27 - 3x + 7 \over (2x + 9)^{1 \over 2} (2x + 9)} \\ & = {3x + 34 \over (2x + 9)^{3 \over 2} } \\ & = {3x + 34 \over \sqrt{(2x + 9)^3}} \end{align}
\begin{align} u & = 3x + 34 &&& v & = \sqrt{(2x + 9)^3} \\ & &&& & = (2x + 9)^{3 \over 2} \\ {du \over dx} & = 3 &&& {dv \over dx} & = {3 \over 2} (2x + 9)^{1 \over 2} . (2) \\ & &&& & = 3(2x + 9)^{1 \over 2} \end{align} \begin{align} {d^2 y \over dx^2} & = {(2x + 9)^{3 \over 2}(3) - (3x + 34)[3(2x + 9)^{1 \over 2}] \over [\sqrt{(2x + 9)^3}]^2 } \phantom{00000} [\text{Quotient rule}] \\ & = {3(2x + 9)^{3 \over 2} - 3(3x + 34)(2x + 9)^{1 \over 2} \over (2x + 9)^3 } \\ & = {3(2x + 9)^{1 \over 2} [(2x + 9) - (3x + 34)] \over (2x + 9)^3 } \\ & = {3(2x + 9)^{1 \over 2} (2x + 9 - 3x - 34) \over (2x + 9)^3 } \\ & = {3(2x + 9)^{1 \over 2} (- x - 25) \over (2x + 9)^3 } \\ & = {3(-x - 25) \over (2x + 9)^{-{1 \over 2}} (2x + 9)^3} \\ & = {-3x - 75 \over (2x + 9)^{5 \over 2}} \\ & = {-(3x + 75) \over \sqrt{(2x + 9)^5}} \\ & = -{3x + 75 \over \sqrt{(2x + 9)^5}} \end{align}

\begin{align} y & = {x^2 - 1 \over x} \\ & = {x^2 \over x} - {1 \over x} \\ & = x - x^{-1} \\ \\ {dy \over dx} & = 1 - (-1)x^{-2} \\ & = 1 + x^{-2} \\ & = 1 + {1 \over x^2} \\ \\ \text{Let } & {dy \over dx} = 5, \\ 5 & = 1 + {1 \over x^2} \\ 4 & = {1 \over x^2} \\ 4x^2 & = 1 \\ x^2 & = {1 \over 4} \\ x & = \pm \sqrt{1 \over 4} \\ & = \pm {1 \over 2} \end{align} \begin{align} \text{Substitute } & x = {1 \over 2} \text{ into eqn of curve,} &&& \text{Substitute } & x = -{1 \over 2} \text{ into eqn of curve,} \\ y & = \left(1 \over 2\right) - \left(1 \over 2\right)^{-1} &&& y & = \left(-{1 \over 2}\right) - \left(-{1 \over 2}\right)^{-1} \\ & = -{3 \over 2} &&& & = {3 \over 2} \\ \\ \therefore & \phantom{.} \left({1 \over 2}, -{3 \over 2}\right) &&& \therefore & \phantom{.} \left(-{1 \over 2}, {3 \over 2} \right) \end{align}

\begin{align} y & = ax^3 + bx \\ \\ \text{Using } & (3 , 1), \\ 1 & = a(3)^3 + b(3) \\ 1 & = 27a + 3b \phantom{000} \text{--- (1)} \\ \\ {dy \over dx} & = a(3)x^2 + b \\ & = 3ax^2 + b \\ \\ \text{Let } & x = 3 \text{ and } {dy \over dx} = 3, \\ 3 & = 3a(3)^2 + b \\ 3 & = 3a(9) + b \\ 3 & = 27a + b \\ \\ b & = 3 - 27a \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 1 & = 27a + 3(3 - 27a) \\ 1 & = 27a + 9 - 81a \\ 81a - 27a & = 9 - 1 \\ 54a & = 8 \\ a & = {8 \over 54} \\ & = {4 \over 27} \\ \\ \text{Substitute } & a = {4 \over 27} \text{ into (2),} \\ b & = 3 - 27 \left(4 \over 27\right) \\ & = -1 \\ \\ \therefore a & = {4 \over 27}, b = -1 \end{align}

The coordinates of P and Q can be found by solving simultaneous equations (using the equations of line & curve).

Substitute the x-coordinates of each point into the gradient function, ${dy \over dx}$, in order to find the gradient at the respective point.

\begin{align} \text{Eqn of curve: } \phantom{000} xy + 10 & = 0 \\ xy & = -10 \\ y & = -{10 \over x} \\ y & = -10 x^{-1} \\ \\ {dy \over dx} & = -10(-1)x^{-2} \\ & = 10x^{-2} \\ & = {10 \over x^2} \\ \\ \\ \text{Eqn of curve: } \phantom{000} xy + 10 & = 0 \phantom{000} \text{--- (1)} \\ \\ \text{Eqn of line: } \phantom{000} 2x + 3y & = 7 \\ 2x & = 7 - 3y \\ x & = {7 \over 2} - {3 \over 2}y \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ \left({7 \over 2} - {3 \over 2}y \right) y + 10 & = 0 \\ {7 \over 2}y - {3 \over 2}y^2 + 10 & = 0 \\ -{3 \over 2}y^2 + {7 \over 2}y + 10 & = 0 \\ -3y^2 + 7y + 20 & = 0 \\ 3y^2 - 7y - 20 & = 0 \\ (3y + 5)(y - 4) & = 0 \end{align} \begin{align} 3y + 5 & = 0 && \text{ or } & y - 4 & =0 \\ 3y & = -5 &&& y & = 4 \\ y & = -{5 \over 3} \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ x & = {7 \over 2} - {3 \over 2}\left(-{5 \over 3}\right) &&& x & = {7 \over 2} - {3 \over 2}(4) \\ & = 6 &&& & = -{5 \over 2} \\ \\ \therefore & \phantom{.} P\left(6, -{5 \over 3}\right) &&& \therefore & \phantom{.} Q\left(-{5 \over 2}, 4\right) \\ \\ \text{Substitute } & x = 6 \text{ into } {dy \over dx}, &&& \text{Substitute } & x = -{5 \over 2} \text{ into } {dy \over dx}, \\ {dy \over dx} & = {10 \over (6)^2} &&& {dy \over dx} & = {10 \over \left(-{5 \over 2}\right)^2} \\ & = {5 \over 18} &&& & = {8 \over 5} \end{align}

(i)

\begin{align} y & = 10 - (2 - x)^4 \\ \\ {dy \over dx} & = 0 - 4(2 - x)^3 . (-1) \phantom{00000} [\text{Chain rule}] \\ & = 4(2 - x)^3 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4(2 - x)^3 \\ 0 & = (2 - x)^3 \\ 0 & = 2 - x \\ x & = 2 \\ \\ \therefore a & = 2 \\ \\ \text{For } & x < 2, \\ (2 - x)^3 & > 0 \\ \implies {dy \over dx} & > 0 \\ \\ \therefore y \text{ is in} & \text{creasing} \end{align}

(ii)

\begin{align} {dy \over dx} & = 0 - 4(2 - x)^3 . (-1) \phantom{00000} [\text{Chain rule}] \\ & = 4(2 - x)^3 \\ \\ \text{For } & x > 2, \\ (2 - x)^3 & < 0 \\ \implies {dy \over dx} & < 0 \\ \\ \therefore y \text{ is de} & \text{creasing} \end{align}

Note that time, $x$ minutes, is greater or equals to 0

\begin{align} T & = \sqrt[3]{x} - {100 \over x + 2} + 75 \\ & = x^{1 \over 3} - 100(x + 2)^{-1} + 75 \\ \\ {dT \over dx} & = {1 \over 3}x^{-{2 \over 3}} - 100(-1)(x + 2)^{-2} \\ & = {1 \over 3} \left(1 \over \sqrt[3]{x^2}\right) + 100(x + 2)^{-2} \\ & = {1 \over 3\sqrt[3]{x^2}} + {100 \over (x + 2)^2} \\ \\ \text{For } x & > 0, \phantom{.} \sqrt[3]{x^2} > 0 \text{ and } (x + 2)^2 > 0 \\ \\ \implies {1 \over 3\sqrt[3]{x^2}} & + {100 \over (x + 2)^2} > 0 \\ & \phantom{000000.} {dT \over dx} > 0 \\ \\ \therefore \text{The func} & \text{tion is an increasing function} \end{align}

Note that time, $t$ months, is greater or equals to 0

\begin{align} g(t) & = {20 \over 3(2t + 1)} \\ & = {20 \over 3} (2t + 1)^{-1} \\ \\ g'(t) & = {20 \over 3} (-1) (2t + 1)^{-2} . (2) \phantom{00000} [\text{Chain rule}] \\ & = -{40 \over 3} (2t + 1)^{-2} \\ & = -{40 \over 3} \left[ 1 \over (2t + 1)^2 \right] \\ & = -{40 \over 3(2t + 1)^2} \\ \\ \text{For } t & \ge 0, (2t + 1)^2 > 0 \\ \\ \implies &-{40 \over 3(2t + 1)^2} < 0 \\ & \phantom{0000000.} g'(t) < 0 \phantom{000000} [g(t) \text{ is a decreasing function}] \\ \\ \text{As time} & \text{ goes on, the population of the town will decrease} \end{align}