Question

若函数f（x）=ax^3-3ax+2的单调递减区间为［-1,1］，则实数a的取值范围是

Answer 1

f'(x)=3ax^2-3a=3a(x^2-1)=3a(x+1)(x-1), 所以当x=1或-1是为0.因为单减区间是［-1,1］，所以a的取值范围是a>0.

Answer 2

先求导，得到导数=3aX^2-3a=3a(X-1)(X+1),因为函数f（x）=ax^3-3ax+2的单调递减区间为［-1,1 ],即导数在［-1,1］上小于0，因为(X-1)(X+1)在区间为［-1,1］上小于0，所以3a>0,即a>0

Answer 3

f（x）=ax^3-3ax+2,依题意有；f'（x）=3ax^2-3a=3a*（x^2-1),令f'（x）<0,要使单调递减区间为［-1,1］，则3a>0，a>0