函数f(x)=x2-2x+22x-2(x>1)的最小值是()A.1B.-1C.-2D.2-数学

题目简介

函数f(x)=x2-2x+22x-2(x>1)的最小值是()A.1B.-1C.-2D.2-数学

题目详情

函数f(x)=
x2-2x+2
2x-2
(x>1)的最小值是(  )
A.1B.-1C.-2D.2
题型:单选题难度:偏易来源:不详

答案

(x)=
x2-2x+2
2x-2
可变形为f(x)=
(x-1)2+1
2(x-1)

即f(x)=class="stub"x-1
2
+class="stub"1
2(x-1)

∵x>1,∴class="stub"x-1
2
>0,class="stub"1
2(x-1)
>0,
class="stub"x-1
2
+class="stub"1
2(x-1)
≥2
(x-1)
2
class="stub"1
2(x-1)
=1,
当且仅当class="stub"x-1
2
=class="stub"1
2(x-1)
,即(x-1)2=1,x=2时,等号成立.
∴函数f(x)=
x2-2x+2
2x-2
(x>1)的最小值是1
故选A

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