当x＞1时,求f（x）＝x²-3x+1／x+1的值域
问题描述：
当x＞1时,求f（x）＝x²-3x+1／x+1的值域
答
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)^2x＞1x∈（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,+∞）