求导y'=a(3x^2-1)<=0
(1)当a>0时,3x^2<=1,解得:-根号3/3<=x<=根号3/3
(2)当a<0时,3x^2>=1,解得:x>=根号3/3,X<=-根号3/3
又:递减区间为(-根号3/3,根号3/3),则a取值范围为:a>0.
函数y=a[(x^3)-x]在区间(-根号3/3,根号3/3)单调递减,则有
a[(-√3/3)^3-(-√3/3)]>a[(√3/3)^3-(√3/3)]
2a[(-√3/3)^3+(√3/3)]>0
a>0
因为一但a的正负性确定了,函数y=a[(x^3)-x]的单调区间也随着确定!