#include<stdio.h>
#include<conio.h>
#include<math.h>
float f (double x);

void main()
{
clrscr();
float xl,xu,xrold,xrnew;
int c=0;
xl=0.5;
xu=1.0;
xrold=(xl+xu)/2;
float EA;
do
{
c++;
if(f(xl)*f(xrold)<0)
xu=xrold;
if(f(xl)*f(xrold)>0)
xl=xrold;
xrnew=(xl+xu)/2;
EA=(xrnew-xrold)/xrnew;
if(EA<0)
EA=-EA;
xrold=xrnew;
}while(EA>0.1);
printf(“%f, %f,No of transitions %d”,xrold,EA,c);
getch();
}
float f(double x)
{
return -26+85*x-91*pow(x,2)+44*pow(x,3)-8*pow(x,4)+pow(x,5);
}

This question is solved by using the Bisection Method. In this question we have lower point and upper point. So xl is representing the lower limit and xu is representing upper limit. When we put xl in the function then result will be negative and when we put xu in function the result will be positive.

This can be checked by ensuring that f(xl)(xu)<0.

An estimate of the root xr is determined by xr=(xl+xu)/2.

In this program we use xrold and xrnew. xrold will be replaced by xrnew when a new point replaced the xl orxu. xr representing the mid point of xl and xu. If the xr<0 then the value of xl is replaced by xr and if the xr>0then the value of xu will be replaced by xr.

The f function is returning the result of -26+85x-91×2+44×3-8×4+x5.
The Rules for changing the Limits according to Numerical computing are:
Following evaluations to determine in which sub interval the root lie.
1) If f(xl)f(xr)<0, the root lies in the lower sub interval. Therefore ,set xu=xr.
2) If f(xl)f(xr)>0, the root lies in the upper sub interval. Therefore ,set xl=xr
3) If f(xl)f(xr)=0, the root equals xr; terminate the computation.