Question

一道数学题,求大神来函数y=x2+x分之3(x>0)的最小值为

Answer 1

y=(ax^2+x+1)/(x+1)=ax^2/(x+1)+1因为a>0且x>=3,所以y>1y=ax^2/(x+1)+1=a/(1/x+1/x^2)+1可见x越大,y就越大,函数在范围内单调递增,x取最少值3时,y取得最少值y=9a/4+1