Question

求函数y=X+1分之ax^2+x+1(a>0且x>=3)的最小值

Answer 1

y=(ax^2+x+1)/(x+1)

=ax^2/(x+1)+1

因为a>0且x>=3,所以y>1

y=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