作业帮设函数f(x)的定义域为R,对于任意实数m,n,恒有f(m+n)=f(m)*f(n),且当x>0时,0
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设函数f(x)的定义域为R,对于任意实数m,n,恒有f(m+n)=f(m)*f(n),且当x>0时,0
设函数f(x)的定义域为R,对于任意实数m,n,恒有f(m+n)=f(m)*f(n),且当x>0时,0
(2)求证f(x)在R上单调递减
(3)若F(1)=1/2,试解不等式F(a^x)/F(4-a^2x) >1/4
设函数f(x)的定义域为R,对于任意实数m,n,恒有f(m+n)=f(m)*f(n),且当x>0时,0
（1）、令m=0,n=1,代入等式得f(0+1)=f(0)*f(1),即f(1)=f(0)*f(1),
由已知可得f(1)>0,所以上式两边同除以f(1)可得：f(0)=1
当y0,所以00,
对任意x>y,有x-y>0,f(x)>0,f(y)>0,
又f(x)/ f(y)= f(x)* f(-y)= f(x-y)
所以0< f(x)/ f(y) 0,
所以f(x) < f(y),
所以f(x)在R上单调递减.
（3）F(a^x)/F(4-a^2x) = F(a^x)*F(a^2x -4)= F(a^x+a^2x -4)
又f(2)=f(1+1)=f(1)*f(1)= 1/2*1/2=1/4,
所以F(a^x)/F(4-a^2x) >1/4可变形为F(a^x+a^2x -4) > f(2)
因为f(x)在R上单调递减,所以a^x+a^2x -4
这一看就是Y=a的x次方函数，其中a是大于0小于1的嘛。
d
in fact, f(x)=f(1)^x is exponential function.
first to show f(n)=f(1)^n for any integer n.
next f(1/m)=f(1)^{1/m} where m is integer.
then for any rational number n/m>0,
f(n/m)=f(1)^...
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in fact, f(x)=f(1)^x is exponential function.
first to show f(n)=f(1)^n for any integer n.
next f(1/m)=f(1)^{1/m} where m is integer.
then for any rational number n/m>0,
f(n/m)=f(1)^{n/m}.
Since the set of rationall number is dense in real line, for any real number x>0, we have
f(x)=f(1)^x.
for x<0, f(x-x)=f(0)=1=f(x)*f(-x) this implies that f(x)=f(-x)^{-1}=f(1)^{-x*-1}=f(1)^x;
So for any real number x, f(x)=f(1)^x
According the condition x>0, f(x)<1, we have 0
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