= Solution
\b[Not near the minimum at $1$.] The <derivative> being set to zero factors as
$$
f'(x)=\frac{2(x-1)^3(x+1)}{x^3},
$$
so $1$ has <multiplicity of a root> equal to three. The ordinary <Newton iteration at a multiple root> is generally only linearly convergent, with error ratio $(3-1)/3=2/3$ here.
In fact, cancellation in $x-f'(x)/f''(x)$ gives the exact error identity, away from the original undefined quotient at the root,
$$
x_{n+1}-1=(x_n-1)\frac{x_n+3}{x_n^2+2x_n+3}.
$$
For iterates approaching $1$, the factor tends to $2/3$, proving <linear convergence>. The other zero of $f'$ at $-1$ is simple and could instead give quadratic local convergence; the failure at $1$ is due to its multiplicity, not to every <stationary point> of this <rational function>.
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