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
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.
求证.f(x)在0到正无穷上为增函数.
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it's easy!
let x1>x2>0,
f(x1)-f(x2)=f((x1/x2)*x2)-f(x2)
=f(x1/x2)+f(x2)-f(x2)
=f(x1/x2)
cause x1>x2>0,so x1/x2>1,so f(x1/x2)>0,
so f(x1)-f(x2)>0,that is f(x1)>f(x2)
f(x) is the increasing function when x>0
done
when you deal with a problem like this you should refer to the question and use the condition as possible as you can!
let x1>x2>0,
f(x1)-f(x2)=f((x1/x2)*x2)-f(x2)
=f(x1/x2)+f(x2)-f(x2)
=f(x1/x2)
cause x1>x2>0,so x1/x2>1,so f(x1/x2)>0,
so f(x1)-f(x2)>0,that is f(x1)>f(x2)
f(x) is the increasing function when x>0
done
when you deal with a problem like this you should refer to the question and use the condition as possible as you can!
1年前
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