已知函数f(x)=ax^3-3x^2+1-3/a(a不得0).若a=1,过点M(2,m)(m不得-6)可做曲线y=f(x)三条切线,求m取值范围

f(x)=x^3-3x^2-2f'(x)=3x^2-6x设切点坐标为(n,f(n))f(n)=n^3-3n^2-2f'(n)=3n^2-6n切线方程为:y=(3n^2-6n)(x-n)+(n^3-3n^2-2)将点(2,m)代入方程:(即x=2,y=m)m=(3n^2-6n)(2-n)+(n^3-3n^2-2)即:m=-2n^3+9n^2-12n-2此方...