设函数f(x)满足f(x^2-3)=loga ^[(x^2)/(6-x^2)] (a>o,a不等于1)
2个回答
1 设x^2-3=t
x^2=t+3
f(t)=loga[(t+3)/(6-t-3)
=loga[(t+3)/(3-t)]
f(x)=loga[(x+3)/(3-x)]
=loga(x+3)-loga(3-x)
f(-x)=log(-x+3)-loga(3+x)=-f(x)
f(0)=0
f(x)是奇函数
2 f(x)=loga[(x+3)/(3-x)]>Loga(2x)
loga[(x+3)/((3-x)2x))>0
(x+3)/((3-x)2x)>1
借此不等式可得x取值范围
3 （x+3)/(3-x)=a^f(x)
x+3=a^f(x)*3-a^f(x)*x
x*(a^f(x)+1)=3*(a^f(x)-1)
x=3*(a^f(x)-1)/(a^f(x)+1)
f-(x)=3*(a^x-1)/(a^x+1)
(a^x-1)/(a^x+1)
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