已知函数f(x)=|x^3-3/2(a+1)x^2+3ax|,其中a>0,若f(x)有三个极值点,则实数a的取值范围是

设g(x)=x^3-3/2(a+1)x^2+3ax,那么f(x)=|g(x)|g'(x)=3x^2-3(a+1)x+3a=3(x-1)(x-a)当a=1时,g'(x)=3(x-1)^2≥0恒成立,g(x)无极值点,将g(x)翻折后得到f(x)=|g(x)|图像,g(x)的零点为f(x)的个极值点,此时,f(x)有1个极值点。当00,g(x)递增,10,g(x)递增,∴g(x)极小值=g(a),需g(x)极小值=g(a)=a^3-3/2(a+1)a^2+3a^2>0即2a^2-3(a+1)a+6a>0a^2-3a