设函数f(x)=ax^3-6ax^2+3bx+b,其图像在x=2处的切线方程为3x+y-11=0.（1）求函数f(x)的

（1）f(x)'=3ax^2-12ax+3b,f(2)'=-3,f(2)=5
f(2)'=12a-24a+3b=-12a+3b=-3 ```````(1)
f(2)=8a-24a+7b=-16a+7b=5```````````(2)
由(1)(2)得a=1,b=3
f(x)=x^3-6x^2+9x+3
(2)由(1)知f(x)'=3x^2-12x+9
y=1/3f(x)'+5x+m=x^2+x+3+m