Question

已知函数f(x)=(ax^2-1)•e^x a属于R （1）若函数在x=1时取得极值，求a的值

已知函数f(x)=(ax^2-1)•e^x a属于R
（1）若函数在x=1时取得极值，求a的值
（2）当a小于等于0时，求函数的单调区间

Answer 1

(1)f'(x) = 2axe^x + (ax² -1)e^x = (ax² + 2ax - 1)e^x
e^x总为正, f'(1) = (3a -1)e = 0
a = 1/3

(2)
f'(x) = (ax² + 2ax - 1)e^x
e^x总为正, 只需考虑ax² + 2ax - 1
(i) a = 0
f'(x) = -e^x < 0
f(x)为减函数

(ii) a < 0
y = ax² + 2ax - 1的图象为开口向下的抛物线, ax² + 2ax - 1 = 0
判别式∆ = 4a² + 4a = 4a(a+1)
(甲)-1 < a < 0时
∆ < 0, y = ax² + 2ax - 1的图象与轴无交点, f'(x) < 0, f(x)为减函数
(乙)a < -1时
∆ = 4a(a+1) > 0
ax² + 2ax - 1 = 0的解为:
x1 = -1 + √(a²+a)/a
x2 = -1 - √(a²+a)/a
a < 0, x1 < x2
当x1 < x < x2时, f'(x) > 0, f(x)为增函数
当x < x1 或 x > x2时, f'(x) < 0, f(x)为减函数