求函數(shù)y=f(x)=x³-3x-1的單調(diào)區(qū)間和極值.
記y=f(x)
∵y'=f'(x)=3x²-3=3(x+1)(x-1)
∴當(dāng)x∈(-∞,-1]時f'(x)≥0；當(dāng)x∈[-1,1]時f'(x)≤0；當(dāng)x∈[1,-∞)時f'(x)≥0
∴f(x)在x∈(-∞,-1]上單調(diào)遞增,在x∈[-1,1]上單調(diào)遞減,在x∈[1,-∞)上單調(diào)遞增；f(x)在x=-1處取得極大值,即f(x)極大=f(-1)=1；f(x)在x=1處取得極小值,即f(x)極小=f(1)=-3.