h'(x)=a-(2a-1)/x^2=[ax^2-(2a-1)]/x^2 在區(qū)間【1,2】h'(x)>0
1.a>0 h'(1)>=0 a
設(shè)h(x)=ax+(2a-1)/x,若函數(shù)h(x)在區(qū)間【1,2】上是增函數(shù),求實數(shù)a的取值范圍.
設(shè)h(x)=ax+(2a-1)/x,若函數(shù)h(x)在區(qū)間【1,2】上是增函數(shù),求實數(shù)a的取值范圍.
數(shù)學(xué)人氣:777 ℃時間:2019-10-23 07:00:24
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