Newton rapshon method in numerical method in C++

Newton rapshon method
#include
#include
#include
#define f(x) (pow(x,3)-2*pow(x,2)-5*x+6)
#define df(x) (3*pow(x,2)-4*x-5)
/*#define f(x) (pow(x,2)-2*x+1)
#define df(x) (2*x-2)*/
void main()
{
float a,b,root,error=0.000001;
int maxit=20,count=1;
input:
printf("Enter one initial guess: ");
scanf("%f",&a);
if (f(a)==0)
{
root=a;
goto display;
}
if(df(a)==0)
goto input;
mid:
b=a-f(a)/df(a);
if(fabs((b-a)/b)
#include
#define max 10
void main()
{
int n,i,j;
float x[max],f[max],fp,lf,sum,xp;
clrscr();
printf("\n\n\n\n\n\nInput number of data points, n ");
scanf("%d",&n);
printf("\nInput data points x[i] and values f[i] \n");
printf("(one set in each line)\n");
for(i=0;i {
scanf("%f %f",&x[i], &f[i]);
}
printf("\nInput value at which interpolation is required, xp ");
scanf("%f",&xp);
sum=0.0;
for(i=0;i {
lf=1.0;
for(j=0;j {
if(i!=j)
{
lf=lf*(xp-x[j])/(x[i]-x[j]);
}
}
sum=sum+lf*f[i];
}
fp=sum;
printf("\nLagrangian Interpolation \n\n");
printf("Interpolated function value \n");
printf("at x = %f is %f \n",xp,fp);
getch();
}