Use Newton's iteration method to find the roots of the following equation near 1.5: 2x3-4x2 + 3x-6 = 0 #include "math.h" main() {float x,x0,f,f1; x=1.5; do{x0=x; f=2*x0*x0*x0-4*x0*x0+3*x0-6; f1=6*x0*x0-8*x0+3; x=x0-f/f1; }while(fabs(x-x0)>=1e-5); printf ("%f\n",x); } I'd like to ask you how F1 = 6 * x0 * x0-8 * x0 + 3 comes from

Use Newton's iteration method to find the roots of the following equation near 1.5: 2x3-4x2 + 3x-6 = 0 #include "math.h" main() {float x,x0,f,f1; x=1.5; do{x0=x; f=2*x0*x0*x0-4*x0*x0+3*x0-6; f1=6*x0*x0-8*x0+3; x=x0-f/f1; }while(fabs(x-x0)>=1e-5); printf ("%f\n",x); } I'd like to ask you how F1 = 6 * x0 * x0-8 * x0 + 3 comes from

The steps of Newton's iterative method are as follows: first, give an initial value x0, and use it to iterate. The iterative method is to make the tangent of the curve at the point (x0, f (x0)) and get an intersection (x1,0) with the horizontal axis. X1 is the result of the first iteration, which is an approximation of the solution of the equation

How to solve the cubic equation 2x3-3x2 + 3x-1 = 0

The 2x ^ 3-3-3-3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x ^ 3-3 x-3 X-1 = 0 (2x ^ 3-3-2x ^ 3-2x ^ 3-2x ^ 3-3 x + 2x-1) + (2x-1) + (2x-1) x (2x (2x ^ 2-2-2-2-2-2x-3x-2-3x-3 + 3 / 4 > (x-1 / 2) + (x-1 / 2) + (x-1 / 2) + 3 / 4 / 4 > 0) the original equation is: the original equation = 2x-2x-1 = 2x-1-1 = 2x-1-1 = 1 = 0 (2x-1-1-1 = 1 = 0 (2x-1 = 0} = = (1 ± √ 3I) / 2