C: Using Newton's Iterative method to find the root of the equation near 1.5:2x^3-4x^2+3x-6=0.

Newton's Iterative method was used to find the root of the equation near 1.5:2x^3-4x^2+3x-6=0.

Solution: Newton's Iterative method is also called Newton tangent method. Setf =2x^3-4x^2+3x-6,F1 is the derivative of the equation, thef1 = 6x^2-8x+3, and f1= (f (x0) -0)/(X0-X1), deduced:x1 = x0-f/F1

Program:

#include <stdio.h>

#include <math.h>

int main ()

{

Double x0,x1,f,f1;

x1 = 1.5;

Do

{

x0 = x1;

f = 2*x0*x0*x0-4 * x0*x0 + 3 * x0-6;

F1 = 6 * x0*x0-8 * x0 + 3;

X1 = x0-f/F1;

} while (Fabs (x0-x1) >= 1e-5);

printf ("The root of equation is%5.2f\n", x1);//the root of equation is the root of the equation

return 0;

}

Results:

The root of equation is 2.00

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C: Using Newton's Iterative method to find the root of the equation near 1.5:2x^3-4x^2+3x-6=0.