Question

【在线等解答】

X1 , X2 是 f(x)=(a/3) x^3+(b/2)x^2-a^2 x (a>0)两个极值点。且 |X1|+|X2|=2
①求a。②证：|b|≤ 4√3/9

Answer 1

解：

（1） f ' (x) = ax^2 + bx - a^2

X1 + X2 = - b/a
X1 * X2 = - a → | X1 | + | X2 | = | X1 - X2 | = √ ( X1 + X2 ) ^2 - 4*X1*X2 = 2

∴ ( - b/a )^2 - 4 (-a) = b^2 / a^2 + 4a = 4

∴ b^2 = 4 a^2 ( 1 - a ) ≥ 0

∴ 0 < a ≤ 1

(2) 设 b^2 = g (a) = 4a^2 - 4a^3
g ' (a) = 8a - 12a^2

令 g ' (a) > 0
则 x ∈ ( 0,2/3)

∴ 增区间( 0 , 2/3 ) 减区间( 2/3 , 1 )
g(a) max = g(2/3) = 16/27

∴ b^2 ≤ 16/27

∴ b ≤ 4/3√ 3 = 4√ 3 / 9

望采纳~~~~