应用牛顿迭代法求x3+2x2+3x+4=0方程的根。c++代码实现

下面是使用牛顿迭代法求解x3+2x2+3x+4=0方程的根的C++代码：

#include <iostream>
#include <cmath>

using namespace std;

double f(double x)
{
    return pow(x, 3) + 2 * pow(x, 2) + 3 * x + 4;
}

double df(double x)
{
    return 3 * pow(x, 2) + 4 * x + 3;
}

int main()
{
    double x0 = 0, x1 = 0, eps = 1e-6;
    int iter = 0, max_iter = 100;

    cout << "Enter initial guess: ";
    cin >> x0;

    do {
        x1 = x0 - f(x0) / df(x0);
        iter++;

        if (fabs(x1 - x0) < eps) {
            cout << "Root = " << x1 << endl;
            cout << "Number of iterations = " << iter << endl;
            return 0;
        }

        x0 = x1;
    } while (iter < max_iter);

    cout << "Solution did not converge" << endl;

    return 0;
}

其中，f(x)和df(x)分别是方程和其导函数，x0是初始猜测值，eps是误差容限，iter是迭代次数，max_iter是最大迭代次数。程序中使用do-while循环计算迭代值，直到满足误差容限或达到最大迭代次数为止。