令f′(x)=12x³-6x²-6x=12x(x+1/1)(x-1)=0得x=0,x=-1/2或x=1,当x＜-1/2或0＜x＜1时,f′(x)＜0,函数单调递减；当x＞1或-1/2＜x＜0时,f′(x)＞0,函数单调递增.所以函数有极大值f(0)=0,但无最大值；由于f(-1/2...