Question

已知A为大于0的常数，求函数F(X)=(AX^2+X+1)/(X+1)(X>=3)的最小值

怎么做的?

Answer 1

F(X)=(X+1)/(X+1)+(AX^2+AX-AX)/(X+1)
=1+AX(X+1)/(X+1)+(-AX)/(X+1)
=1+AX+(-AX-A+A)/(X+1)
=1+AX-A(X+1)/(X+1)+A/(X+1)
=1+AX-A+A/(X+1)
=1+AX+A-2A+A/(X+1)
=1-2A+A(X+1)+A/(X+1)
>=1-2A+A√2

Answer 2

(9A+4)/4