When a≠2/3, find the monotone interval and extreme value of function f(x).
Solution: (1) When a = 0, f (x) = x2ex, f'
(x)=(x2+2x)
Ex, therefore f'
( 1)=e。
Therefore, the slope of the tangent of curve y = f (x) at (1, f( 1)) is e. 。
(2)f '
(x)=[x2+(a+2)x-2a2+4a]
ex，
Ling f'
(x) = 0, x =-2a, or x = a-2. From a≠23, -2a ≠ a-2.
The following discussion is divided into two situations:
① if a > 23, -2a < a-2. When x changes, f'
The changes of(x) and f (x) are as follows:
x
(-∞，-2a)
-2a
(-2a，a-2)
a-2
(a-2，+∞)
f '
(10)
+
—
+
f(x)
↗
maximum
↘
minimum value
↗
The function f(x) takes the maximum value f at x =-2a.
(-2a)= 3ae-2a；
The minimum value f is obtained at x = a-2.
(a-2)=(4-3a)e
a-2；
② if a < 23, -2a > a-2. When x changes, f'
The changes of(x) and f (x) are as follows:
x
(-∞，a-2)
a-2
(a-2，-2a)
-2a
(-2a，+∞)
f '
(10)
+
—
+
f(x)
↗
maximum
↘
minimum value
↗
The function f(x) takes the minimum value f (-2a) = 3ae-2a at x =-2a;
The maximum value f (a-2) = (4-3a) e is obtained when x = a-2.
a-2。