求函数y=(x^2-3x+1)/(x+1)(x>-1)的最小值
问题描述：
求函数y=(x^2-3x+1)/(x+1)(x>-1)的最小值
答
y=(x²-3x-4+5)/(x+1)
=[(x-4)(x+1)+5]/(x+1)\
=(x-4)(x+1)/(x+1)+5/(x+1)
=x-4+5/(x+1)
=(x+1)+5/(x+1)-5
x>-1
x+1>0
所以(x+1)+5/(x+1)>=2√[(x+1)*5/(x+1)]=2√5
所以y>=2√5-5
所以最小值=2√5-5