采用二分法求方程的根

//此方法用来求方程x*(e^x)-1=0在一个[x1,x2]反外内的根，x1,x2的值在运行时输入//
#include<math.h>
#include<iostream>
using namespace std;
double f(double x)
{
    double y;
    y=exp(x)*x-1;
    return y;
}

double point(double x1,double x2)
{
    double y;
    y=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));
    return y;
}

double root(double x1,double x2)
{
    double x,y,y1;
    y1=f(x1);

    do{
        x=point(x1,x2);
        y=f(x);
        if(y*y1>0)
            {x1=x; y1=y;}
        else
            x2=x;
    }while(fabs(y)>=0.0001);
    return x;
}

void main()
{
    double x1,x2,f1,f2,x;
    do{
        cin>>x1>>x2;
        f1=f(x1);
        f2=f(x2);
    }while((f1*f2)>=0);
    x=root(x1,x2);
    cout<<x<<endl;
}