已知函数f(x)=ax^3+bx^2-3x在x=1或-1处取得极值. (1)求函数f(x)的解析式.

f'(x)=3ax^2+2bx-3
f'(-1)=3a-2b-3=0、f'(1)=3a+2b-3=0.a=1、b=0.
(1)f(x)=x^3-3x.
(2)设切点为(t,t^3-3t).
切线斜率为(t^3-3t-m)/(t-1).
f'(x)=3x^2-3,所以切线斜率为3t^2-3.
(t^3-3t-m)/(t-1)=3t^2-3、t^3-m=(t-1)(3t^2-3).
2t^3-3t^2+m+3=0在三个根.
设h(t)=2t^3-3t^2+m+3、h'(t)=6t^2-6t=6t(t-1).
极大值为h(0)=m+3、极小值为h(1)=m+2.
若2t^3-3t^2+m+3=0在三个根,则h'(0)=m+3>0且h'(1)=m+2