(a) If \(f(x) = \frac{x - 3}{2x - 1} , x \neq \frac{1}{2}\) and \(g(x) = \frac{x - 1}{x + 1}, x \neq -1\), fing \(g \circ f\). (b)(i) Sketch the curve \(y =...

(a)

\[\begin{aligned} (g\circ f)(x)&=g(f(x))\\ &=\frac{f(x)-1}{f(x)+1}\\ &=\frac{\dfrac{x-3}{2x-1}-1}{\dfrac{x-3}{2x-1}+1}\\ &=\frac{\dfrac{x-3-2x+1}{2x-1}}{\dfrac{x-3+2x-1}{2x-1}}\\ &=\frac{-x-2}{3x-4}. \end{aligned}\]

Hence,

\[\boxed{(g\circ f)(x)=-\frac{x+2}{3x-4}}\]

The restrictions are \(x\ne\frac12\) and \(x\ne\frac43\).

(b)(i) For \(y=9x-x^3\),

\[y=x(9-x^2)=x(3-x)(3+x).\]

Therefore, the curve cuts the \(x\)-axis at \((-3,0)\), \((0,0)\), and \((3,0)\).

\[\frac{dy}{dx}=9-3x^2.\]

At turning points,

\[9-3x^2=0\implies x=\pm\sqrt3.\]

\[\begin{aligned}y\bigl(\sqrt3\bigr)&=6\sqrt3\approx10.39,\\y\bigl(-\sqrt3\bigr)&=-6\sqrt3\approx-10.39.\end{aligned}\]

Also, \(\dfrac{d^2y}{dx^2}=-6x\); hence \((\sqrt3,6\sqrt3)\) is a maximum point and \((-\sqrt3,-6\sqrt3)\) is a minimum point. The curve is symmetric about the origin.

(b)(ii) The part of the curve from \(0\) to \(3\) lies above the \(x\)-axis. By symmetry, the two bounded areas, from \(-3\) to \(0\) and from \(0\) to \(3\), are equal.

\[\begin{aligned}\text{Total area}&=2\int_0^3(9x-x^3)\,dx\\&=2\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3\\&=2\left(\frac{81}{2}-\frac{81}{4}\right)\\&=\frac{81}{2}.\end{aligned}\]

\[\boxed{\text{Total area}=\frac{81}{2}\text{ square units}=40.5\text{ square units}.}\]