怎么求(t^2+(2根号2)t)/(t^2+1)最值

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怎么求(t^2+(2根号2)t)/(t^2+1)最值怎么求(t^2+(2根号2)t)/(t^2+1)最值怎么求(t^2+(2根号2)t)/(t^2+1)最值a=(t^2+2√2t)/(t^2+1)at

怎么求(t^2+(2根号2)t)/(t^2+1)最值
怎么求(t^2+(2根号2)t)/(t^2+1)最值

怎么求(t^2+(2根号2)t)/(t^2+1)最值
a=(t^2+2√2t)/(t^2+1)
at^2+a=t^2+2√2t
(a-1)t^2-2√2t+a=0
t是实数则方程有解
所以判别式大于等于0
所以8-4a(a-1)>=0
a^2-a-2