Please correct the answer for column Ea%.For 1st iteration it will display "---" For 2nd iteration, 1.0333-1/1.0333 times 100 = 3.22, For 3rd iteration, 1.0641-1.0333/1.0641 times 100 = 2.89. As observed for Ea the values used are from the previous row. The output table at inputs 1,5,4 is: The code: #include #include #include #include #include using namespace std::chrono; using namespace std;  static double function(double x); // function f(x) void menu();  int main() {     double a; // Lower Guess or beginning of interval     double b; // Upper Guess or end of interval     double c = 0; // variable for midpoint     double terms;     double newc;        // Taking Input     cout << "Enter Xn-1: ";     cin >> a;     cout << "\nEnter Xn: ";     cin >> b;     cout << "\nEnter number of iterations: ";     cin >> terms;      // Check for opposite sign (Intermediate Value Theorem)     if (function(a) * function(b) > 0.0f)     {         cout << "\nFunction has same signs at ends of interval";         return -1;     }      int iter = 1;     cout << setw(3) << "\n\n N (iterations)" << setw(9) << "Xn-1" << setw(17) << "Xn" << setw(18) << "Xn+1" << "        Ea%   " << endl;      auto start = high_resolution_clock::now();      while (terms >= iter) // terminating condition     {         double newC = a-(function(a)*(b-a))/(function(b)-function(a));          if (c == 0) {             cout << setprecision(4) << setw(3) << iter << setw(19) << a << setw(17) << b << setw(18) << newC << "        " << "----" << endl;         }         else {             double ea = (c - a) / c ;             cout << setprecision(4) << setw(3) << iter << setw(19) << a << setw(17) << b << setw(18) << newC << "        " <<  fixed << setprecision(4) <<100*ea << endl;         }          c = a-(function(a)*(b-a))/(function(b)-function(a));         // check for opposite sign         if (function(a) * function(c) < 0.0f)         {             b = c;         }         else         {             a = c;         }         iter++;     }      auto stop = high_resolution_clock::now();     auto duration = duration_cast(stop - start);        return 0; }  static double function(double x) {     return pow(x, 3) - x - 1; }

Please correct the answer for column Ea%.For 1st iteration it will display "---" For 2nd iteration, 1.0333-1/1.0333 times 100 = 3.22, For 3rd iteration, 1.0641-1.0333/1.0641 times 100 = 2.89. As observed for Ea the values used are from the previous row. The output table at inputs 1,5,4 is:

The code:

#include <iostream> #include <math.h> #include<iomanip> #include<chrono> #include <cstdlib> using namespace std::chrono; using namespace std;  static double function(double x); // function f(x) void menu();  int main() {     double a; // Lower Guess or beginning of interval     double b; // Upper Guess or end of interval     double c = 0; // variable for midpoint     double terms;     double newc;        // Taking Input     cout << "Enter Xn-1: ";     cin >> a;     cout << "\nEnter Xn: ";     cin >> b;     cout << "\nEnter number of iterations: ";     cin >> terms;      // Check for opposite sign (Intermediate Value Theorem)     if (function(a) * function(b) > 0.0f)     {         cout << "\nFunction has same signs at ends of interval";         return -1;     }      int iter = 1;     cout << setw(3) << "\n\n N (iterations)" << setw(9) << "Xn-1" << setw(17) << "Xn" << setw(18) << "Xn+1" << "        Ea%   " << endl;      auto start = high_resolution_clock::now();      while (terms >= iter) // terminating condition     {         double newC = a-(function(a)*(b-a))/(function(b)-function(a));          if (c == 0) {             cout << setprecision(4) << setw(3) << iter << setw(19) << a << setw(17) << b << setw(18) << newC << "        " << "----" << endl;         }         else {             double ea = (c - a) / c ;             cout << setprecision(4) << setw(3) << iter << setw(19) << a << setw(17) << b << setw(18) << newC << "        " <<  fixed << setprecision(4) <<100*ea << endl;         }          c = a-(function(a)*(b-a))/(function(b)-function(a));         // check for opposite sign         if (function(a) * function(c) < 0.0f)         {             b = c;         }         else         {             a = c;         }         iter++;     }      auto stop = high_resolution_clock::now();     auto duration = duration_cast<microseconds>(stop - start);        return 0; }  static double function(double x) {     return pow(x, 3) - x - 1; }