1.1. A curve has equation $y=f(x)$, where

$f(x)=\frac{7x{e}^{x}}{\sqrt{{e}^{3x}-2}}x>ln\sqrt[3]{2}$

Show that

${f}^{'}(x)=\frac{7{e}^{x}({e}^{3x}(2-x)+Ax+B)}{2{({e}^{3x}-2)}^{\frac{3}{2}}}$

where $A$ and $B$ are constants to be found.

1.2. Hence show that the $x$ coordinates of the turning points of the curve are solutions of the equation

$x=\frac{2{\begin{matrix} & & e\end{matrix}}^{3x}-4}{{\begin{matrix} & & e\end{matrix}}^{3x}+4}$

1.3. The equation $x=\frac{2{\begin{matrix} & & e\end{matrix}}^{3x}-4}{{\begin{matrix} & & e\end{matrix}}^{3x}+4}$ has two positive roots $\alpha$and $\beta$ where $\beta > \alpha$

A student uses the iteration formula

${x}_{n+1}=\frac{2{\begin{matrix} & & e\end{matrix}}^{3{x}_{n}}-4}{{\begin{matrix} & & e\end{matrix}}^{3{x}_{n}}+4}$

in an attempt to find approximations for $\alpha$ and $\beta$

Diagram 1 shows a plot of part of the curve with equation $y=\frac{2{\begin{matrix} & & e\end{matrix}}^{3x}-4}{{\begin{matrix} & & e\end{matrix}}^{3x}+4}$ and part of the line with equation $y=x$

Using Diagram 1 draw a staircase diagram to show that the iteration formula starting with ${x}_{1}=1$ can be used to find an approximation for $\beta$

1.4. Use the iteration formula with ${x}_{1}=1$, to find, to $3$ decimal places,

(i) the value of ${x}_{2}$

(ii) the value of $\beta$

1.5. Using a suitable interval and a suitable function that should be stated show that $\alpha =0.432$ to $3$ decimal places.