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
来源:学生作业帮 编辑:百度作业网作业帮 分类:英语作业 时间:2024/07/15 03:24:05
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.
![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](/uploads/image/z/7699638-30-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%28x%E2%88%88R%2Cx%E2%89%A00%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0x1%2Cx2%2C%E6%81%92%E6%9C%89f%28x1x2%29%3Df%28x1%29%2Bf%28x2%29%2C%E4%B8%94%E5%BD%93X%3E1%2Cf)
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!
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y=f (x)(x∈R,x≠0)对任意的非零实数x1,x2,恒有f(x1x2)=f(x1)+f(x2),试判断f
已知函数y=f (x)(x∈R,x≠0)对任意的非零实数x1,x2,恒有f(x1x2)=f(x1)+f(x2)
设函数y=f(x)(x∈R,且x≠0)对任意非零实数x1,x2,满足f(x1)+f(x2)=f(x1x2)
设函数f(x)定义域(0,+∞),且f(4)=1,对任意正实数x1x2,有f(x1x2)=f(x1)+f(x2),且当x
已知函数y=f(x)(x属于R,且x不等于零) 对任意非零实数x1,x2,恒有f(x1乘以x2) =f(x1)+f(x2
已知奇函数f(x)对任意正实数x1x2 (x1≠x2)恒有(x1-x2)[f(x1)-f(x2)]
已知函数f(x)对任意实数x1,x2都有f(x1x2)=f(x1)+f(x2)成立,则f(0)=?f(1)=?
设函数y=f(x)(x属于R,且x不等于0),对任意非零实数x1,x2.满足f(x1)+f(x2)=f(x1x2)
已知函数Fx的定义域是x≠0的一切实数,对定义域内的任意x1,x2都有f(x1x2)=f(x1)+f(x2),且当x>1
对任意的非零x1,x2,有f(x1x2)=f(x1)+f(x2),且f'(1)=1,证明:当x不等于0时,f'(x)=1
已知函数f(x)的定义域为R,对任意实数x1,x2,都有f(x1+x2)=f(x1)+f(x2)成立,且当x>0时,有f