= Solution
{c}
The <polynomial> factors as $p(x)=(3x-2)^2(3x+1)$, so $\alpha=2/3$ is a <multiple root> of multiplicity two. The <Newton root-finding iteration> is
$$
x_{k+1}=x_k-\frac{p(x_k)}{p'(x_k)},\qquad p'(x)=81x^2-54x.
$$
For $x\ne0,2/3$, cancelling a factor gives
$$
\boxed{g(x)=\frac{2x}{3}+\frac19+\frac{2}{27x}.}
$$
This expression extends continuously to $g(2/3)=2/3$. Let $I=[1/3,1]$. On this <closed interval>,
$$
g'(x)=\frac23-\frac{2}{27x^2},\qquad 0\le g'(x)\le\frac{16}{27}<1.
$$
Thus $g$ is increasing, and
$$
g(I)=[g(1/3),g(1)]=[5/9,23/27]\subset I.
$$
By the <mean value theorem>, $|g(x)-g(y)|\le(16/27)|x-y|$ for all $x,y\in I$. The <contraction mapping theorem> states that a <contraction mapping> from a nonempty <complete metric space> into itself has a unique <fixed point>, and every <fixed-point iteration> converges to it. The <closed interval> $I$ with ordinary <Euclidean distance> is complete, and $g(2/3)=2/3$, so
$$
\boxed{x_k\longrightarrow\frac23\quad\text{for every }x_0\in[1/3,1].}
$$
In fact $|x_k-2/3|\le(16/27)^k|x_0-2/3|$. The <Newton iteration at a multiple root> here has <linear convergence>, since $g'(2/3)=1/2$. Thus \b[the convergence assertion is true for the simplified <fixed-point iteration>], and for the usual <Newton root-finding iteration> that stops upon reaching an exact <root of a polynomial>. If one insists on applying the unsimplified quotient at every step, it is undefined for the allowed starting value $x_0=2/3$, since both $p$ and $p'$ vanish there; that literal version of an infinite <sequence> requires the stated stopping convention or continuous extension.
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