用二分法求方程（x+1)(x-2)(x-3)=1在区间（-1,0）的近似解(精确度0.1）为什么在球

用二分法求方程（x+1)(x-2)(x-3)=1在区间（-1,0）的近似解(精确度0.1）为什么在球

#include#includefloat f(float x){float y;y=(x+1)*(x-2)*(x-3)-1;return(y);}float xpoint(float x1,float x2){float y;y=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));//求 x轴点 交点坐标return(y);}float root(float x1,float x2){float x,y,y1;y1=f(x1);do{x=xpoint(x1,x2);y=f(x);if(y*y1>0){y1=y;x1=x;}else x2=x;}while(fabs(y)>=1e-1);return(x);}void main(){float x1,x2,f1,f2,x;do{printf(请输入方程解得范围x1,x2(注意中间用逗号隔开):\n);scanf(%f,%f,&x1,&x2);f1=f(x1);f2=f(x2);}while(f1*f2>=0);x=root(x1,x2);printf(A root of equation is %.1f\n,x);}判断单调是为了保证在某一个区间只有唯一解.

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