解:∵y=f(x)=(2x+5)/(x-3)=[(2x-6)+11]/(x-3)=2+11/(x-3), ∴森御塌得x=11/(y-2)+3,即f^-1(x)=11/(x-2)+3 则f^-1(x+1)=11/(x-1)+3 又得x=11/(y-3)+1, 即y=g(x)=11/拆键(x-3)+1∴g(4)=11/此圆(4-3)+1=11+1=12