对任意的x1,x2∈R,若函数f(x)=2^x,试比较[f(x1)+f(x2)]÷2与f[(x1+x2)]÷2的大小关系
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![对任意的x1,x2∈R,若函数f(x)=2^x,试比较[f(x1)+f(x2)]÷2与f[(x1+x2)]÷2的大小关系](/uploads/image/z/8711696-56-6.jpg?t=%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x1%2Cx2%E2%88%88R%2C%E8%8B%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2%5Ex%2C%E8%AF%95%E6%AF%94%E8%BE%83%5Bf%EF%BC%88x1%EF%BC%89%2Bf%EF%BC%88x2%EF%BC%89%5D%C3%B72%E4%B8%8Ef%5B%EF%BC%88x1%2Bx2%EF%BC%89%5D%C3%B72%E7%9A%84%E5%A4%A7%E5%B0%8F%E5%85%B3%E7%B3%BB)
对任意的x1,x2∈R,若函数f(x)=2^x,试比较[f(x1)+f(x2)]÷2与f[(x1+x2)]÷2的大小关系
对任意的x1,x2∈R,若函数f(x)=2^x,试比较[f(x1)+f(x2)]÷2与f[(x1+x2)]÷2的大小关系
对任意的x1,x2∈R,若函数f(x)=2^x,试比较[f(x1)+f(x2)]÷2与f[(x1+x2)]÷2的大小关系
[f(x1)+f(x2)]÷2
= (2^x1 + 2^x2) /2
f((x1+x2)/2)
= 2^[(x1+x2)/2]
= 2^(x1/2) .2 ^(x2/2)
consider
(2^(x1/2) - 2^(x2/2)) ^2 ≥0
2^x1 + 2^x2 - 2[ 2^(x1/2).2^(x2/2)]≥0
=> 2^(x1/2) .2 ^(x2/2) ≤ (2^x1 + 2^x2)/2
ie
f((x1+x2)/2)≤ [f(x1)+f(x2)]÷2
【f(x1)+f(x2)】/2=(2^x1+2^x2)/2
f((x1+x2)/2)=2^((x1+x2)/2)=2^(x1/2)2^(x2/2)
【f(x1)+f(x2)】/2-f((x1+x2)/2)
==(2^x1+2^x2)/2-2^(x1/2)2^(x2/2)
=1/2[2^(x1/2)-2^(x2/2)]^2>=0
故【f(x1)+f(x2)】/2≥f((x1+x2)/2)
[f(x1)+f(x2)]/2>=f[(x1+x2)/2]
[f(x1)+f(x2)]/2=(2^x1+2^x2)/2>=根号[2^x1*2^x2]=2^[(x1+x2)/2]
f[(x1+x2)/2]=2^[(x1+x2)/2]
证毕
如果没学过基本不等式,那么用作差比较.
[f(x1)+f(x2)]/2-f[(x1+x2)/2]=(2^x1+2^x2)/2-2^[(x1+x2)/2]
=1/2*[2^(x1/2)-2^(x2/2)]^2>=0