设函数f(x)=x^3-3ax2+3bx的图像与直线12x+y-1=0相切于点(1,-11)

1)直线12x+y-1=0的斜率为-12
f'(x)=3x^2-6ax+3b,--> f'(1)=3-6a+3b=-12
f(1)=1-3a+3b=-11
两式相减得：2-3a=-1--> a=1,--> b=-3
2)f(x)=x^3-3x^2-9x
f'(x)=3x^2-6x-9=3(x-3)(x+1)=0--> x=3,-1
x>3 or x