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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
人气:329 ℃ 时间:2019-08-21 12:56:41
解答
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
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