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....