Question

已知f(x)=2/3*x^3-2ax^2-3x，a属于全体实数，（1）若f(x)在区间(-1,1)上为减函数，求a的取值

Answer 1

f(x) = (2/3)x³ - 2ax² - 3x
f'(x) = 2x² - 4ax - 3
f'(- 1) < 0
2 + 4a - 3 < 0
f'(1) < 0
2 - 4a - 3 < 0
解得-1/4<a<1/4

Answer 2

-1/4<a<1/4

Answer 3

f(x) = (2/3)x³ - 2ax² - 3x
f'(x) = 2x² - 4ax - 3
f'(- 1) < 0
2 + 4a - 3 < 0
f'(1) < 0
2 - 4a - 3 < 0
解得- 7/10 < a < - 9/40