当X＞-1时,求f（x）=x＾2-3x+1/x+1的值域

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

令y=f（x）
y=(x²-3x+1)/(x+1)
令t=x+1
x>-1
x+1>0
t>0
y=[(t-1)²-3(t-1)+1]/t=(t²-2t+1-3t+3+1)/t=(t²-5t+5)/t=(t+5/t)-5
=[√t-√(5/t)]²+2√5-5
t>0,[√t-√(5/t)]²≥0,y的最小值当√t=√(5/t)即t=√5时,y有最小值=2√5-5
x》-1,当x=√5-1时,f（x）的值域是f（x）≥2√5-5

f(x)=((x+1)^2-5(x+1)+5)/(x+1)
=(x+1)-5+5/(x+1)
x>1
f(x)≥2√5-5
x

f(x)=[(x+1)(x-4)+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(x+1)+5/(x+1)-5>=2√5-5
值域[2√5-5,+∞)

f(x)=(x^2-3x+1)/(x+1) = (x^2+x-4x-4+5)/(x+1) = x - 4 + 5/(x+1)
f'(x) = 1 - 5/(x+1)^2 = { (x+1)^2 - 5 ] / (x+1)^2
x＞1
x∈（1,-1+√5）时,f'(x)＜0,f(x)单调减
x∈（-1+√5,+∞）时,f'(x)＞0,f(x)单调增
x=-1+√5时,最小值f(x)min= -1+√5 - 4 + 5/(-1+√5+1) = -5+2√5
值域【-5+2√5,+∞）
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