Consider the problem of finding the roots of $f(x)$. We assume that there is a single root $x_*$ between $a$ and $b$, $a < x_* < b$.
Assume also that the sign of $f''(x)$ does not change for $x \in [a,b]$.
It is well known that if $f(a) f''(a) > 0$ then the Newton method converges to the solution without overshoot. See, for instance, https://en.wikipedia.org/wiki/Newton%27s_method#Analysis and Newton iteration converges monotonically
Now, what if $f(a) f''(a) < 0$?
Under which conditions can we establish that the Newton approximation method will converge after a single overshoot, which will occur precisely at the first iteration?
Let \begin{equation} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \end{equation} and $x_0=a$, \begin{equation} x_{1} = a - \frac{f(a)}{f'(a)} \end{equation}
Under which conditions does $x_1 > x_*$ and $f(x_1) f''(x_1) > 0$?
This question is a follow up to Darboux' theorem on the convergence of Newton's method
A reference with a theorem or insights would be very helpful. In particular, I was not able to extract the answer to the question above from the following paper
"Sur la méthode d'approximation de Newton", Nouvelles annales de mathématiques: journal des candidats aux écoles polytechnique et normale, serie 2, vol 8 (1869), pp.17-27
Concrete example
Let $0 < p < 1$ and $0.5 < q < 1$. Assume \begin{align} &f(x)=(8(q-0.5)^2{p}^3+(-34(q-0.5)^2-1.5){p}^2 + \nonumber \\ &\quad\quad\quad +(40(q-0.5)^2+6)p-16(q-0.5)^2-4)/(p-2)^2. \label{eq:cubic2} \end{align} We search for $x_*$ such that $f(x_*)=0$.
Then, $x_* \approx x_1$, \begin{align} x_* & \approx x_1 \\ & = x_0 - \frac{f(x_0)}{f'(x_0)} \\ &= 0.845 + \frac{1.23688 q^2-1.23688 q+0.31}{-2.38422 q^2+2.38422 q+2} \label{eq:pstarf} \end{align} where $x_0=0.845$.
It is easy to show that $x_0 < x_* \leq 1$.
How can I show that $x_1 > x_*$ using general properties of the global convergence of Newton approximation method?
Additional remark
Clearly, the root of $f(x)$ is the same as the root of $g(x)$, \begin{align} &g(x)=(8(q-0.5)^2{p}^3+(-34(q-0.5)^2-1.5){p}^2 + \nonumber \\ &\quad\quad\quad +(40(q-0.5)^2+6)p-16(q-0.5)^2-4). \end{align} Then, \begin{align} x_* & \approx x_0 - \frac{g(x_0)}{g'(x_0)} \\ &= 0.845-\frac{1.650041 (q - 0.5)^2 + 0.0010375}{0.3234 (q - 0.5)^2 - 3.465} \end{align} where $x_0=0.845$. However, Newton approximation convergence is much slower for $g(x)$ than $f(x)$. Still, for $g(x)$ we known that NAM will never overshoot, as $g(x_0) g''(x_0) > 0$. Is there a way to check in advance why/if $f(x)$ is a best input for Newton approximation than $g(x)$ with respect to convergence time, but that $g(x)$ is best with respect to number of overshoots?
