Write the program for finding aprroxiamate root of a function by Bisection method

The method is applicable for numerically solving the equation f(x) = 0 for the real variable x, where f is a continuous function defined on an interval [a, b] and where f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root since, by the intermediate value theorem, the continuous function f must have at least one root in the interval (a, b).

At each step the method divides the interval in two by computing the midpoint c = (a+b) / 2 of the interval and the value of the function f(c) at that point. Unless c is itself a root (which is very unlikely, but possible) there are now only two possibilities: either f(a) and f(c) have opposite signs and bracket a root, or f(c) and f(b) have opposite signs and bracket a root.^[5]
 The method selects the subinterval that is guaranteed to be a bracket as the new interval to be used in the next step. In this way an interval that contains a zero of f is reduced in width by 50% at each step. The process is continued until the interval is sufficiently small.

Explicitly, if f(a) and f(c) have opposite signs, then the method sets c as the new value for b, and if f(b) and f(c) have opposite signs then the method sets c as the new a. (If f(c)=0 then c may be taken as the solution and the process stops.) In both cases, the new f(a) and f(b) have opposite signs, so the method is applicable to this smaller interval.

#include<iostream>
#include<math.h>
using namespace std;

  float func(float x)  //User Define Funtion
{  float m= x*x*x-x-11;
      return m;    }

int main()
{ float a, b,c;
     
       a = 2;
       b = 3;
  
  cout<<"Funtion is X^3-X-11"<<endl;
 
  cout <<"initialization by two numbers "<<endl;
 
  cout<<"First initialization number = 2"<<endl;
  
  cout<<"secound initialization number = 3"<<endl;
  

   if( func(a)*func(b)>=0 )

    { cout<<"Roots are not present between initialization"<<endl;
           return 0;
     }
                      
    for(int i=1;i<10;i++)     
   
   {    c= (a+b)/2.0;
        if (func(c)*func(b)<0)
              {  a=c; }
         else {  b=c; }
       
        if(func(c)==0)
         {break;}
    } 
      
   cout<<"\nApprox root of this function between 2 and 3 by using bisected method is "<<c;
  return 0;

}

• click on below link and run this program
• https://onlinegdb.com/SyYBY9gr_
  

https://onlinegdb.com/Bkdv4ceHd