问题
解答题
以知函数f(x)=logax(a>0且a≠1,x∈R+),若x1,x2∈R+,判断
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答案
f(x1)+f(x2)=logax1+logax2=loga(x1x2)
∵x1,x2∈R+,
∴x1x2≤(
)2(当且仅当x1=x2时取“=”号).当a>1时,有loga(x1x2)≤loga(x1+x2 2
)2x1+x2 2
∴
loga(x1x2)≤loga(1 2
) ,x1+x2 2
(logax1+logax2)≤loga(1 2
),x1+x2 2
即
[f(x1)+f(x2)]≤f(1 2
)(当且仅当x1=x2时取“=”号)当0<a<1时,有loga(x1x2)≥loga(x1+x2 2
)2,x1+x2 2
∴
(logax1+logax2)≥loga(1 2
)2,x1+x2 2
即
[f(x1)+f(x2)]≥f(1 2
)x1+x2 2
(当且仅当x1=x2时取“=”号).