X>-1,x+1>0
f(x)=x^2-3x+1/x+1=x-[4x-1/x+1]
=x-4+5/(x+1)=(x+1)+5/(x+1)-5>=2√[5(x+1)/(x+1)]-5
=2√5-5
f(x)=x^2-3x+1/x+1的值域
f(x)>=2√5-5
网络百科教团为你解答,如果懂了可以采纳,
第2步能详细点吗f(x)=x^2-3x+1/x+1=x-[4x-1/x+1]
f(x)=(x^2-3x+1)/(x+1)=[(x+1)^2-5(x+1)+5]/(x+1)=(x+1)+5/(x+1)-5因为x>-1所以x+1>0所以(x+1)+5/(x+1)>=2√[(x+1)*5/(x+1)]=2√5值域为[2√5-5,正无穷)