f(x)=(2x-1)/(x+1)
=2-[3/(x+1)]=2-g(x)
g(x)=3/(x+1)
g(x)在[-1,2]上是递减函数
=> -g(x)在[-1,2]上是递增函数
=> f(x)在[-1,2]上是递增函数
所以 f(x)在x=-1处取得最小值 (-无穷大)
在x=2处取得最大值 1
f(x)=(2x-1)/(x+1)
=2-[3/(x+1)]=2-g(x)
g(x)=3/(x+1)
g(x)在[-1,2]上是递减函数
=> -g(x)在[-1,2]上是递增函数
=> f(x)在[-1,2]上是递增函数
所以 f(x)在x=-1处取得最小值 (-无穷大)
在x=2处取得最大值 1