Найти минимум на одном отрезке

#include <iostream>
#include <math.h>
#include <clocale>
#define M_PI  3.14159265358979323846
using namespace std;
 
double function(double);
 
cout << "Задание повышенной сложности: " << endl;
    double c, d;
    cout << "Enter c,d: ";
    cin >> c >> d;
    double step = 0.001;
    double min = function(c);
    for (c; c <= d; c += step) {
        if (function(c) < min) 
            min = function(c);
        cout << "Minimum of function in range [" << c << ";" << d << "] is " << min << endl;
    }
    return 0;
}
 
double function(double u) {
    return sin(u) * u;
}
    ////////////////////////////////////////////////////////////////////////////////////////////////////////////

    for (c; c < d+step/2; c += step) {
        if (function(c) < min) 
            min = function(c);
    }    
    cout << "Minimum of function in range [" << c << ";" << d << "] is " << min << endl;

if(c>d)
{
double tmp = c;
c = d;
d = tmp;
}

#include <iostream>
#include <clocale>
#include <iomanip>
#include <cmath>    // for trigonometry functions
using namespace std; // std c++ libs implemented in std
 
 
double Minimum_value(double);
 
int main() {    
        cout << "Задание повышенной сложности: " << endl;
    double c, d;
    cout << "Enter c,d: ";
    cin >> c >> d;
    if (c < d)
    {
        double tmp = d;
        d = c;
        c = tmp;
    }
    double step = 0.001;
    double min = Minimum_value(c);
    for (c; c < d + step / 2; c += step) {
        if (Minimum_value(c) < min)
            min = Minimum_value(c);
    }
    cout << "Minimum of function in range [" << d << ";" << c << "] is " << min << endl;
 
    return 0;
}
 
double Minimum_value(double u) {
    return sin(u) * u;
}

f(c>d)
{
double tmp = c;
c = d;
d = tmp;
}

#include <iostream>
#include <cmath>    // for trigonometry functions
using namespace std; // std c++ libs implemented in std
 
 
double Minimum_value(double);
 
int main()
{
    double c, d;
    cout << "Enter c,d: ";
    cin >> c >> d;
    cin.get();
    if (c > d)
    {
        double tmp = d;
        d = c;
        c = tmp;
    }
    int n = 20;//можете в принципе вернуть и версию с шагом...
    double min = Minimum_value(c);
    for (int i = 0; i <= n; i++) {
        if (Minimum_value(c+i*(d-c)/n) < min)
            min = Minimum_value(c+i*(d-c)/n);
    }
    cout << "Minimum of function in range [" << c << ";" << d << "] is " << min << endl;
    cin.get();
 
    return 0;
}
 
double Minimum_value(double u) {
    return sin(u) * u;
}