问题标题：
已知函数f(x)(x∈R,x≠0)对任意的非零实数x1,x2,恒有f(x1x2)=f(x1)+f(x2),且当X>1,f(x)>0.求证.f(x)在0到正无穷上为增函数.
问题描述：
已知函数f(x)(x∈R,x≠0)对任意的非零实数x1,x2,恒有f(x1x2)=f(x1)+f(x2),且当X>1,f(x)>0.
求证.f(x)在0到正无穷上为增函数.
杜淑光回答：
it'seasy!
letx1>x2>0,
f(x1)-f(x2)=f((x1/x2)*x2)-f(x2)
=f(x1/x2)+f(x2)-f(x2)
=f(x1/x2)
causex1>x2>0,sox1/x2>1,sof(x1/x2)>0,
sof(x1)-f(x2)>0,thatisf(x1)>f(x2)
f(x)istheincreasingfunctionwhenx>0
done
whenyoudealwithaproblemlikethisyoushouldrefertothequestionandusetheconditionaspossibleasyoucan!
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