已知a属于R,讨论函数fx＝e^x（x²+ax+a+1）的极值点个数

f'(x) = (e^x)(x²+ax+a+1) + (e^x)(2x+a) = [x²+ (a + 2)x+ 2a+1]e^x
e^x > 0
x²+ (a + 2)x+ 2a+1的判别式为(a + 2)² - 4(2a + 1) = a(a - 4)
0 ≤ a ≤ 4时,f'(x) ≥ 0,无极值点
a < 0或a > 4时,x²+ (a + 2)x+ 2a+1与x轴有两个交点,外侧x²+ (a + 2)x+ 2a+1 > 0,f'(x) > 0; 内侧x²+ (a + 2)x+ 2a+1< 0,f'(x) < 0,有两个极值点 (左极大,右极小)