= Solution
Since $f'(x)=2x-2$, <Newton's method> becomes
$$
\boxed{x_{n+1}=x_n-\frac{x_n(x_n-2)}{2(x_n-1)}=\frac{x_n^2}{2(x_n-1)}}\qquad(x_n\ne1).
$$
For $x_n>2$, subtraction from $2$ gives
$$
x_{n+1}-2=\frac{(x_n-2)^2}{2(x_n-1)}>0.
$$
Also $x_n-x_{n+1}=x_n(x_n-2)/(2(x_n-1))>0$. Hence \b[$2<x_{n+1}<x_n$]. Equivalently, the positive excess above $2$ is multiplied by the <function> from part (c): $x_{n+1}-2=g(x_n)(x_n-2)$, with $0<g(x_n)<1/2$.
Thus the iterates form a decreasing <sequence> bounded below by $2$. By the <monotone convergence theorem for sequences> they have a limit $\ell\geq2$. Passing to the limit in the continuous update gives $2\ell(\ell-1)=\ell^2$, or $\ell(\ell-2)=0$. Therefore \b[$x_n\to2$]. The exact error formula also shows <quadratic convergence> near this root.
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