已知函数f(x)=loga(x+1)(a>1)且f(x)与g(x)的图像关于原点对称。(1)解不等式2f(x)+g(x)>=o
已知函数f(x)=loga(x+1)(a>1)且f(x)与g(x)的图像关于原点对称。(1)解不等式2f(x)+g(x)>=o
日期:2011-07-20 15:08:58 人气:1
(1)g(x)= -f(-x),x+1>0且-x+1>0 => -1<x<1,
2f(x)+g(x)≥o即2loga(x+1)-loga(-x+1)≥0即loga[(x+1)^2/(-x+1)]≥0
∵a>1,∴(x+1)^2/(-x+1)≥1,∵-10,∴(x+1)^2≥-x+1 =>x≤-3或≥0,又-1<x<1,∴0≤x<1
(2)f(x)+g(x)=loga(x+1)-loga(-x+1)=loga[(x+1)