(1)g(x)= -f(-x),x+1>0且-x+1>0 => -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,∵-1
0,∴(x+1)^2≥-x+1 =>x≤-3或≥0,又-1(2)f(x)+g(x)=loga(x+1)-loga(-x+1)=loga[(x+1)/(-x+1)]≥m恒成立
当x∈[0,1)时,h(x)=(x+1)/(-x+1)=(x-1+2)/(-x+1)=-1-2/(x-1)∈[1,+∞),又a>1,∴f(x)+g(x)∈[0,+∞),m≤[f(x)+g(x)]的最小值=0,即m≤0