由题,f'(x)=3x^2-3,设切点为(x1,x1^3-3x1)
则k=3x1^2-3,
又k=(x1^3-3x1+2)/(x1-1)
∴3x1^2-3=(x1^3-3x1+2)/(x1-1)
∴x1=1或x1=-1/2
又x1不=1,
所以x1=-1/2
k=-9/4
l方程:y=-9*(x-1)/4-4
由题,f'(x)=3x^2-3,设切点为(x1,x1^3-3x1)
则k=3x1^2-3,
又k=(x1^3-3x1+2)/(x1-1)
∴3x1^2-3=(x1^3-3x1+2)/(x1-1)
∴x1=1或x1=-1/2
又x1不=1,
所以x1=-1/2
k=-9/4
l方程:y=-9*(x-1)/4-4