y'=3x^2-2x-1=(3x+1)(x-1)=0--> x=-1/3, 1
极大值f(-1/3)=-1/27-1/9+1/3+3=5/27+3>0
极小值f(1)=2>0
因此只在x
函数y=x^3-x^2-x+3的零点个数为
x=-1/3, 1极大值f(-1/3)=-1/27-1/9+1/3+3=5/27+3>0极"}}}'>
y'=3x^2-2x-1=(3x+1)(x-1)=0--> x=-1/3, 1
极大值f(-1/3)=-1/27-1/9+1/3+3=5/27+3>0
极小值f(1)=2>0
因此只在x