Fast and robust root of a cubic polynomial with constraints
If your function $f$ is convex and increasing, or concave and decreasing, on an interval $[a,b]$ that contains a solution of $f(x) = 0$, Newton's method starting at $b$ will converge to a solution. Similarly if it is concave and increasing, or convex and decreasing, and you start at $a$. For a cubic it's easy to identify the critical points and inflection and thus find such an interval.
I think you can avoid the Newton-Raphson altogether, since cubic is solvable in radicals.
Here is the complete algorithm (working under constraints outlined in the problem), in Mathematica:
Clear[PositiveCubicRoot];
PositiveCubicRoot[p_, q_, r_] :=
Module[{po3 = p/3, a, b, det, abs, arg},
b = ( po3^3 - po3 q/2 + r/2);
a = (-po3^2 + q/3);
det = a^3 + b^2;
If[det >= 0,
det = Power[Sqrt[det] - b, 1/3];
-po3 - a/det + det
,
(* evaluate real part, imaginary parts cancel anyway *)
abs = Sqrt[-a^3];
arg = ArcCos[-b/abs];
abs = Power[abs, 1/3];
abs = (abs - a/abs);
arg = -po3 + abs*Cos[arg/3]
]
]
Then
In[222]:= PositiveCubicRoot[-2.52111798, -71.424692, -129.51520]
Out[222]= 10.499
However, if the Newton-Raphson method must be used, then a good initial guess is imperative. Binary division is a good method to isolate the root in the case at hand. We start with an arbitrary point $x$, I chose $x=1$, and double it while the polynomial at that point is negative. Then do binary division a certain number of times (the code below does it twice). Ultimately, polish it off with Newton-Raphson iterations:
In[283]:=
NewtonRaphsonStartingPoint[{p_, q_, r_}] := Module[{x1=0, x2=1,f1,f2,xm,fm},
f1 = r + x1 (q + (p + x1) x1);
While[(f2 = r + x2 (q + (p + x2) x2)) <= 0,
x1 = x2; f1 = f2; x2 = 2 x2];
Do[xm = (x1 + x2)/2; fm = r + xm (q + (p + xm) xm);
If[fm <= 0, f1 = fm; x1 = xm, f2 = fm; x2 = xm], {i, 2}];
(f2 x2 - f1 x1)/(f2 - f1)
];
NewtonRaphsonIterate[{p_, q_, r_}, x0_Real] :=
FixedPoint[
Function[x, x - (r + x (q + (p + x) x))/(q + (2 p + 3 x) x)], x0]
In[285]:=
NewtonRaphson[p_, q_, r_] :=
NewtonRaphsonIterate[{p, q, r}, NewtonRaphsonStartingPoint[{p, q, r}]]
In[286]:= NewtonRaphson[-2.52111798, -71.424692, -129.51520]
Out[286]= 10.499