f(x)=ax²-丨x丨+2a-1=
ax²-x+2a-1,x≥0;
ax²+x+2a-1,x0;
ax+1+(2a-1)/x,x0時h(x)=x+(2-1/a)/x-1
所以要使h(x)在[1,2]上遞增
只需√(2-1/a)≤1
解得:a≤1
已知f(x)=ax^2-丨x丨+2a-1,設h(x)=f(x)/x,若函數(shù)h(x)在區(qū)間1<=x<=2上是增函數(shù),求實數(shù)a的取值范圍.
已知f(x)=ax^2-丨x丨+2a-1,設h(x)=f(x)/x,若函數(shù)h(x)在區(qū)間1<=x<=2上是增函數(shù),求實數(shù)a的取值范圍.
一些符號不會打,請見諒,有問題可以問我我在線
一些符號不會打,請見諒,有問題可以問我我在線
數(shù)學人氣:794 ℃時間:2020-04-14 01:11:48
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