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
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