已知f(x)=2/3x^3-2ax^2-3x (a属于R)

f(x)=2/3x^3-2ax^2-3x
f(x)‘=2x^2-4ax-3 在区间(-1,1)上
<=0
f(x)'开口向上
f(1)‘<=0 4a+1>=0
f(-1)<=0 4a-1<=0
-1/4<=a<=1/4

(1)f'(x)=-2x^-4 - 4ax -3
=> 4ax^5+3x^4>-2 = x^4(4a+3)>-2
= 4a+3> -2/x^4 因为x (-1,1)
所以 4a+3≥-2 a>=-5/4

1. f`(x)=2x2-4ax-3
f`(-1)=2+4a-3<=0
a<=1/4
f`(1)=2-4a-3<=0
a>=-1/4
-1/4=< a<=1/4
2. f(x)在区间(-1,1)上为减函数,无极值点.