函数y=x2+8x-1(x>1)的最小值为()A.4B.6C.8D.12-数学

题目简介

函数y=x2+8x-1(x>1)的最小值为()A.4B.6C.8D.12-数学

题目详情

函数y=
x2+8
x-1
(x>1)
的最小值为(  )
A.4B.6C.8D.12
题型:单选题难度:中档来源:不详

答案

∵x>1,
∴y=
x2+8
x-1
=
x2-1
x-1
+class="stub"9
x-1

=(x+1)+class="stub"9
x-1

=(x-1)+class="stub"9
x-1
+2

≥2
(x-1)×class="stub"9
x-1
+2
=8;
故选C.

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