 Solution

Since $f^{\prime }(x)=2x-2$, Newton's method becomesFor $x_n>2$, subtraction from $2$ givesAlso $x_n-x_{n+1}=x_n(x_n-2)/(2(x_n-1))>0$. Hence $2<x_{n+1}<x_n$. 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 $\ell \ge 2$. Passing to the limit in the continuous update gives $2\ell (\ell -1)={\ell }^2$, or $\ell (\ell -2)=0$. Therefore $x_n\to 2$. The exact error formula also shows quadratic convergence near this root.