Thread: problem with the bisection method

1. problem with the bisection method

HI!
I wrote a program that finds a root with the bisection method.
The problem is that i cant find the bug :P
here is the code...
can you give me your ideas for solving the problem?

Code:
```#include<stdio.h>
double my_function(double x);
double a,b,e,meso,fm,fa,fb;
int main()
{
printf("GIVE {a} AND {b} OF THE INTERVAL [a,b] \n,AND   ALSO THE TOLERANCE e:");
printf("EXAMBLE:IF a=1 AND b=2 GIVE 1 2\n");
scanf("%lf%lf%lf ",&a,&b,&e);
fa=my_function(a);
fb=my_function(b);
while((fa>0)&&(fb>0) || (fa<0)&&(fb<0))
{
printf("THE VALUES A AND B DO NOT DIFFER IN SIGN\n");
printf(" (+a)*(+b) OR (-a)*(-b) WORNG\n");
printf("GIVE {a} AND {b} OF THE INTERVAL [a,b] \n,AND ALSO THE TOLERANCE e:");
scanf("%lf%lf%lf",&a,&b,&e);
fa=my_function(a);
fb=my_function(b);
}
meso=(a+b)/2;

while(my_function(meso)!=0 || ((b-a)>e) )
{
fm=my_function(meso);
fa=my_function(a);			               fb=my_function(b);

if((fa>0)&&(fm>0) || (fa<0)&&(fm<0))
{
b=meso;
}
else
{
a=meso;
}
meso=(a+b)/2;
}

if(my_function(meso)==0)
printf("%lf IS THE ROOT OF THE FUNCTION F(X)",meso);
else if(((b-a)<e))
printf("THE ROOT APPROXIMATIVE IS %lf",meso);
return 0;
}

double my_function(double x)
{   double fx;
fx=x*x*x+4*x*x-10;
return fx;
}```

2. I just answered this on another board!!!!