= Solution
Apply the <mean value theorem>, the first-order case of <Taylor's theorem>, to $f'$ between $0$ and $x>1$. It yields
$$
f'(x)=3+xf''(\xi)\geq3+2x>5.
$$
Also $f(3/2)\geq-5+9/2+9/4=7/4>0$, so the initial point is to the right of the root $x^*$. For an increasing <strictly convex function>, its tangent at a point to the right of a root meets the axis between that root and the point. Hence the exact <Newton root-finding iteration> remains in $[x^*,3/2]$ and decreases towards $x^*$.
On every relevant <interval>, $|f''|<3$ and $f'>5$, so the stated <quadratic convergence> bound gives, with $e_n=|x_n-x^*|$,
$$
e_{n+1}\leq\tfrac3{10}e_n^2,\qquad e_0<\tfrac12.
$$
Iterating the bound gives
$$
e_1<0.075,\quad e_2<0.0016875,\quad
e_3<8.543\times10^{-7},\quad e_4<2.190\times10^{-13}<2^{-40}.
$$
Thus \b[four Newton corrections suffice to reduce the exact-iteration error bound below the floating-point spacing scale]. The <quadratic convergence> doubles the approximate number of correct digits at each late step. Beyond this point further corrections cannot reliably improve a generic floating-point answer. This is a bound on useful accuracy under the usual stable-evaluation model, not a claim that the data specify the exact iteration at which every implementation must stagnate.
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