函数f(x)的定义域为R,并满足以下条件：1,对任意x属于R,有f(x)大于零2,对任意x,y属于R,有f(xy)=[f(x)]^y;1问;1,求f(0)的值2,求证：f(x)在R上是单调增函数2f(b)
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1.令x=0 f(0)=[f(0)]^0=1 f(0)=1
2.取y>0 x=1/3 f(y*1/3)=f(1/3)^y
任取x1>x2 f(x1)/f(x2)=f(3x1/3)/f(3x2/3)=
[f(1/3)]^3x1/[f(1/3)]^3x2=[f(1/3)]^3(x1-x2) 3(x1-x2)>0 f(1/3)>1[f(1/3)]^3(x1-x2)>1 f(x1)>f(x2) 所以是单调增的
3.f(a)/f(b) +f(c)/f(b)=[f(1/3)]^3(a-b) +[f(1/3)]^3(c-b)基本不等式
≥2根号下[f(1/3)]^3(a+c-2b) b=根号下ac a+c-2根号下ac=(根a-根c)^2>0[f(1/3)]^3(a+c-2b)>1 2根号下[f(1/3)]^3(a+c-2b)>2所以f(a)+f(c)>2f(b)