= Solution
For $1<x_0<2$, the same <Newton's method> formula gives
$$
x_1-2=\frac{(x_0-2)^2}{2(x_0-1)}>0.
$$
Thus the first step \b[overshoots the root to $x_1>2$]. Part (e) then applies from index $1$: the remaining <sequence> decreases to $2$. The first step can be very large when $x_0$ is close to $1$, where the derivative vanishes, but every starting value strictly between $1$ and $2$ still gives \b[convergence to $2$]. The entire sequence is not decreasing, since $x_1>2>x_0$.
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