（1）f(x)=x³-3x,求导得f'(x)=3x²-3
令f'(x)≥0以求f(x)的单调增区间,得3x²-3≥0,解得x≤-1或x≥1
令f'(x)≤0以求f(x)的单调减区间,得3x²-3≤0,解得-1≤x≤1
从而可知,当x∈（-3,3/2）时,
f(x)在x∈(-3,-1]∪[1,3/2)上单调递增,在x∈[-1,1]上单调递减.
经计算：f(-3)= -18,f(1)= -2,f(1)=2
f(-3)
已知函数f(x)=x三次方-3X
（1）f(x)=x³-3x,求导得f'(x)=3x²-3
令f'(x)≥0以求f(x)的单调增区间,得3x²-3≥0,解得x≤-1或x≥1
令f'(x)≤0以求f(x)的单调减区间,得3x²-3≤0,解得-1≤x≤1
从而可知,当x∈（-3,3/2）时,
f(x)在x∈(-3,-1]∪[1,3/2)上单调递增,在x∈[-1,1]上单调递减.
经计算：f(-3)= -18,f(1)= -2,f(1)=2
f(-3)