Code two required condition in finding root with Fixed point method - c++

Question

I'm trying to find a root with simple fixed-point method by C++, but the point is that Xr is a root of f(x) and a inflection point as well. In addition, A equation is a little bit more complex than the normal Fixed-Point method.
The equation is added constant c for check how quickly converge to the root xr.
I was going to find a root and then check if the root is a inflection point or not, but it is not working and I can't find the problem in my code.
I need your help.
The real Problem is
Consider the root finding problem f(x)=0 with root xr, with f'(x)=0.
Convert it to the simple fixed-point problem.
x=x+c*f(x)=g(x)
with c a nonzero constant. How should c be chosen to ensure rapid convergence of
x(n+1)=x(n)+c*f(x(n)) ( x(n+1) means the value of the n+1th of X )
to c (provided that x0 is chosen sufficiently close to xr?). Apply your way of choosing c to the root-finding problem x*x*x-5=0. Start your program with x0=1.0 and run with several values of c and discuss about the observed trend in your results (in other words, the effect of c value on convergence behavior)
#include <stdio.h>
#include <conio.h>
#include <math.h>
#include <stdlib.h>
double gx(double x, double c)
{
return(x + c*(x*x*x - 5));
}
double gxpr(double x, double c)
{
return(x + c*(3 * x*x));
}
void Simple_Fixed_Point(double x, double c)
{
int i = 1;
long double x2=0.0;
long double x3=0.0;
long double ea=0.0;
long double ea2 = 0.0;
long double es = pow(10, -6);
printf("Simple Fixed Point Method\n");
Lbl:
x2 = gx(x,c);
printf("iteration=%d Root=%.5f Approximate error=%.15f\n", i++,
x2, ea);
if (ea=fabs((x2 - x)/x2*100) <es)
{
goto Lbm;
}
else
{
x = x2;
goto Lbl;
}
Lbm:
x3 = gxpr(x2, c);
if (ea2 = fabs((x3 - x2) / x3 * 100) < es)
{
goto End;
}
else
{
x2 = x3;
goto Lbm;
}
End:
getch();
}
int main(void)
{
Simple_Fixed_Point(1.0, 1.0);
return(0);
}

Answer 1

Hope this helps you:
//f(x+dx) = f(x) + (dfdx) * dx;
eps = 1.0;
dx = 1e-7; //something small
x = x0;
while (eps > mineps) {
f1 = f(x);
f2 = f(x + dx);
f3 = f(x + dx + dx);
d2fdx2 = (f3 - f2 - f2 + f1) / dx / dx;
dfdx = (f2 - f1) / dx;
x -= (relax1 * f1 / dfdx + relax2 * dfdx / d2fdx2); //relax - something less 1
eps = max(abs(dfdx), abs(f1));
}