= Solution
On $I=[1,7/5]$, $f'<0$ and $f''<0$, since $x^2<2$. Hence $|f'|$ increases and has minimum $|f'(1)|=14$. Also
$$
|f''(x)|=30x^2(2-x^2)
=30\bigl(1-(x^2-1)^2\bigr)\le30,
$$
with equality at $x=1$. Therefore
$$
\boxed{\frac{\max_I|f''|}{\min_I|f'|}=\frac{30}{14}=\frac{15}{7}.}
$$
There is a unique root $r$ in $I$: $f(1)=2>0$, $f(7/5)=-5.678464<0$, and the <function> is strictly decreasing. Moreover $f(6/5)=-1.382016<0$, so $r<6/5=x_0$.
It remains to justify that the iteration stays where the error estimate applies. Because $f$ is a <concave function>, its graph lies below each <tangent line>. At any $x\in[r,6/5]$, the tangent has nonpositive value $f(x)$ at $x$, negative slope, and nonnegative value at $r$. Its zero $N(x)=x-f(x)/f'(x)$ therefore satisfies
$$
r\le N(x)\le x.
$$
By <mathematical induction>, the <Newton root-finding iteration> stays in $[r,6/5]$ and decreases to some limit $L\ge r$. Continuity in $x_{k+1}-x_k=-f(x_k)/f'(x_k)$ implies $f(L)=0$, so uniqueness gives $L=r$.
Now the supplied error bound applies at every step:
$$
\boxed{|x_{k+1}-r|\le\frac{15}{14}|x_k-r|^2.}
$$
This proves at least <quadratic convergence>, with an invariant <interval> and a nonvanishing <derivative> explicitly established.
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