f'(x) = 3x^2 +2ax +1
f''(x) = 6x +2a
x∈ (-2/3, -1/3) , f''(x) <0
6x +2a <0
a< -3x
ie
1 <a<2追问
能否加以解释
说一下大概思路
追答不好意思，应该这样
找出 极大 x1， 极小 x2 : 使得 x1≤-2/3 or x2≥ -1/3
f(x) =x^3+ax^2+x+1
f'(x) = 3x^2 +2ax +1
f'(x)=0
3x^2+2ax +1 =0
x1=[-a-√(a^2-3 ) ]/3 , x2= [-a+√(a^2-3 ) ]/3
f''(x) = 6x +2a
f''[[-a-√(a^2-3 ) ]/3 ] <0 ( max)
f''[[-a+√(a^2-3 ) ]/3 ] >0 (min)
存在单调递减 (-2/3, -1/3)
=>
x1(max) ≤ -2/3 or x2 (min) ≥-1/3
[-a-√(a^2-3 ) ]/3 ≤ -2/3 (1) or
[-a+√(a^2-3 ) ]/3 ≥ -1/3 (2)
and
定义域
√(a^2-3 ) => -√3≤a≤√3 (3)
from (1)
[-a-√(a^2-3 ) ]/3 ≤ -2/3
-a-√(a^2-3 ) ≤ -2
√(a^2-3 ) ≥ 2+a
(a^2-3 ) ≥ (2+a)^2
4a≤ -7
a≤ -7/4 （舍去) 不在 -√3≤a≤√3
from (2)
[-a+√(a^2-3 ) ]/3 ≥ -1/3
-a+√(a^2-3 ) ≥ -1
√(a^2-3 ) ≥ a-1
a^2-3 ≥ (a-1)^2
2a ≥ 2
a≥1 (4)
(3) and (4)
-√3≤a≤√3 and a≥1
1≤a≤√3
ie
f(x) =x^3+ax^2+x+1 存在单调递减 (-2/3, -1/3)
1≤a≤√3
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