f(x)=(2(x+1)-3)/(x+1)=2-3/(x+1)设5≥x2>x1≥3 x2-x1>0 x1+1>0 x2+1>0f(x2)-f(x1)=3/(x1+1)-3/(x2+1)=3(x2-x1)/(x1+1)(x2+1)>0是增函数所以f(x)的最大值f(5)=2-1/2=3/2f(x)的最小值f(3)=2-3/4=5/4
