f(x)=4x立方-108x,求单调递减区间,极大值和极小值

df(x)/dx=12x²-108
令df/dx=0,得
df/dx=12x²-108=x²-9=0
(x+3)(x-3)=0
当x3时,df/dx>0,此两区间内部,f(x)单调递增
当-3≤x≤3时,df/dx≤0,在此区间内,f(x)单调递减
∴当x=-3时,f(x)取极大值,f(x)=216
当x=3时,f(x)取极小值,f(x)=-216