求函数y=(x-1)e^arctanx 的单调区间及极值

1个回答

  • ∵y=(x-1)e^arctanx

    ∴y′=(x-1)′e^arctanx+(x-1)[e^arctanx]′

    =e^arctanx+(x-1)e^arctanx• [arctanx]′

    =e^arctanx+(x-1)e^arctanx• [1/(1+x^2)]

    =e^arctanx(1+(x-1)/(1+x^2))

    令y′=0

    ∵e^arctanx>0即求1+(x-1)/(1+x^2)=(1+x^2+x-1)/(1+x^2)=(x^2+x)/(1+x^2)=0

    得驻点x1=-1.x2=0

    当x〈-1.x>0时,y′>0.函数y单调递增

    当-1<x<0时,y′<0.函数y单调递减

    ∴∶x=-1时,y取得极大值-2e^(arctan(-1))=-2e^(-∏/4)

    x=0时,y取得极小值-e^(0)=-1

    综上.函数y=(x-1)e^arctanx 的单调增区间为x∈[-∞,-1]∪[0,+∞]

    函数y=(x-1)e^arctanx 的单调减区间为x∈[-1,0]

    y的极大值为-2e^(-∏/4)

    y的极小值为-1