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若∏/4<x<∏/2,则函数y=tan2x(tanx)^3的最大值是多少
人气:165 ℃ 时间:2020-02-05 17:30:17
解答
y=2(tanx)^4/[1-(tanx)^2]
令t=(tanx)^2
因为tanx>1
t>1
y=2t^2/(1-t)
=2(t^2-1+1)/(1-t)
=2[(t^2-1)/(1-t)+1/(1-t)]
=2[-t-1-1/(t-1)]
=-2[t+1+1/(t-1)]
=-2[2+(t-1)+1/(t-1)]
因为t-1>0
所以(t-1)+1/(t-1)》2
y《-2(2+2)
y《-8
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