Question

怎样建立函数的零点与方程根的联系?

Answer 1

函数f(x)=0,

解 f(x)=0的方程根x

e.g.,

f(x)=ax+b=0 ==>x=-b/x

f(x)=ax*x+bx+c=0 ==>

x=[-b±(b^2-4ac)^(1/2)]/2a

f(x)=asin(x)+b=0 ==> x=arcsin(-b/a)