已知a∈R,函数2x^3-3（a+1）x^2+6ax,若|a|>1,求f(x)在闭区间[0,|2a|]上的最小值

由f'(x)=6x^2-6(a+1)x+6a=6[x^2-(a+1)x+a]=6(x-a)(x-1)=0,得极值点x=1,af(1)=2-3(a+1)+6a=3a-1f(a)=2a^3-3(a+1)a^2+6a^2=-a^3+3a^2端点值f(0)=0,f(|2a|)=2|a|^3-3(a+1)a^2+6a|a|因|a|>1若a>1,则函数在(1,a)单调减；...