> 数学 >
对于任意的x1,x2∈R.若函数f(x)=2^x,试比较[f(x1)+f(x2)]/2与f[(x1+x 2)/2]的大小
人气:322 ℃ 时间:2020-06-12 04:55:34
解答
底数2>1,2^x恒>0
[f(x1)+f(x2)]/2/f[(x1+x2)/2]
=[(2^x1+2^x2)/2]/2^[(x1+x2)/2]
=[2^(x1-1)+2^(x2-1)]/2^[(x1+x2)/2]
=2^[x1-1 -(x1+x2)/2] +2^[x2-1-(x1+x2)/2]
=2^[(x1-x2)/2 -1]+2^[(x2-x1)/2 -1]
=(1/2)[2^[(x1-x2)/2] +1/2^[(x1-x2)]/2]
由均值不等式得2^[(x1-x2)/2]+1/2^[(x1-x2)/2]≥2,当且仅当x1=x2时取等号.
(1/2)[2^[(x1-x2)/2] +1/2^[(x1-x2)]/2]≥1
[f(x1)+f(x2)]/2≥f[(x1+x2)/2],当且仅当x1=x2时取等号
推荐
猜你喜欢
© 2025 79432.Com All Rights Reserved.
电脑版|手机版