解：∵y=f(x)=(2x+5)/(x-3)=[(2x-6)+11]/（x-3)=2+11/(x-3), ∴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/(x-3)+1∴g(4)=11/(4-3)+1=11+1=12
