【y=(2x^2-x+1)/(2x-1) (0.5<x≤1.5) 求值域】

来源：百度知道编辑：UC知道时间：2024/06/02 14:47:41

y=(2x^2-x+1)/(2x-1) (0.5<x≤1.5) 求值域

过程

∵y=(2x^2-x+1)/(2x-1) (0.5<x≤1.5)
即y
=[x(2x-1)+1]/(2x-1)
=x+1/(2x-1)
=x-1/2+1/[2(x-1/2)]+1/2{x>1/2,x-1/2>0}
≥2√(1/2)+1/2{仅当x=（1+√2)/2时取等}
=√2+1/2

又x→1/2时y→∞
所以值域[√2+1/2,+∞)

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