已知0<a<1,f(a^x)=x+1/x.
(1)求f(x)的解析式,并求出f(x)的定义域.
(2)判断并证明f(x)在[1/a,+∞)上的单调性.
各位,帮帮忙~最好有过程
參考答案:a^x=t
x=loga(t)
f(t)=loga(t)+1/loga(t)
f(x)=loga(x)+1/loga(x)
loga(x)≠0
f(x)的定义域{x|x>0且x≠1}
1/a≤x1<x2
loga(x1)>loga(x2)
loga(x1) ≥loga(1/a)=-1
loga(x2) ≥loga(1/a)=-1
loga(x2)*loga(x1)≥1
f(x1)-f(x2)=loga(x1)-loga(x2)+1/loga(x1)-1/(loga(x2)
=[loga(x1)-loga(x2)][loga(x1)*loga(x2)-1]/loga(x1)loga(x2)
>0
f(x1)>f(x2)
函数在[1/a,+∞)上的单调递增
