2t/1+t^2一次比二次形函数求值域在线等谢.

Question

2t/1+t^2 一次比二次形函数求值域 在线等 谢.

Answer 1

2t/(1+t^2)=(t^2+2t+1)/(1+t^2)-1=(t+1)^2/(1+t^2)-1
(t+1)^2/(1+t^2)≥0
2t/(1+t^2)≥-1
2t/(1+t^2)=-(t^2-2t+1)/(1+t^2)+1=-(t-1)^2/(1+t^2)+1
-(t-1)^2/(1+t^2)≤0
2t/(1+t^2)≤1
值域[-1,1]