用二分法求方程在(-10,10)之间的根:2x^3-4x^2+3x-6=0.
解:x1<=x0=(x1+x2)/2<=x2
程序:
#include<stdio.h>
#include<math.h>
int main()
{
float x0,x1,x2,fx0,fx1,fx2;
do
{
printf("输入x1,x2的值:");
scanf("%f,%f", &x1, &x2);
fx1 = 2*x1*x1*x1 - 4 * x1*x1 + 3 * x1 - 6;
fx2 = 2 *x2*x2*x2 - 4 *x2*x2 + 3 * x2 - 6;
} while (fx1*fx2>0);