求下列函数的单调区间和极值①f(x)=2x^3-6x^2-18x+5 ②f(x)...

2022-11-07 10:08发布

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①f(x)=2x^3-6x^2-18x+5f(x)=6x^2-12x-18f(x)=12x-12令f(x)=6x^2-12x-18=0得x=-1或x=3f(-1

①f(x)=2x^3-6x^2-18x+5f(x)=6x^2-12x-18f(x)=12x-12令f(x)=6x^2-12x-18=0得x=-1或x=3f(-1
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2022-11-07 10:43 .采纳回答
①f(x)=2x^3-6x^2-18x+5f'(x)=6x^2-12x-18f''(x)=12x-12令f'(x)=6x^2-12x-18=0得x=-1或x=3f''(-1)=-12-12=-240所以函数f(x)=2x^3-6x^2-18x+5的极大值点为(-1,15),极小值点为(3,-49)当x∈(-∝,-1)∪(3,+∝)时,f'(x)=6x^2-12x-18>0,函数单调递增当x∈(-1,3)时,f'(x)=6x^2-12x-18

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