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, so 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，

Let 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, the changes of f' (x) and f(x) are as follows:

x(-∞，-2a)-2a(-2a，a-2)a-2(a-2，+∞)

f '(x)+0—0+

F (x) ↗ Maximum ↘ Minimum ↗

The function f(x) obtains the maximum value f (-2a) = 3ae-2a at x =-2a;

The minimum value f (a-2) = (4-3a) ea-2 is obtained at x = a-2;

② if a < 23, -2a > a-2. When x changes, the changes of f' (x) and f(x) are as follows:

x(-∞，a-2)a-2(a-2，-2a)-2a(-2a，+∞)

f '(x)+0—0+

F (x) ↗ Maximum ↘ Minimum ↗

The function f(x) takes the minimum value f (-2a) = 3ae-2a at x =-2a;

The maximum value f (a-2) = (4-3a) ea-2 is obtained when x = a-2.