设函数f(x)=2x^3+3ax^2+3bx+8c在x=1及x=2取得极值（1）f(x)增区间（2)若对x属[0,3】都有f(x)＜c^成立,

f(x)=2x^3+3ax^2+3bx+8c,求导f(x)'=6x^2+6ax+3b,又,在x=1与x=2取到极值,故f(x)'=k(x-1)(x-2)=6x^2+6ax+3b,得到 kx^2-3kx+2k=6x^2+6ax+3b,比较系数,得：k=6,-3k=6a,2k=3b故,a=-3,b=4.所以f(x)=2x^3-9x^2+12x+8c——（...