函数f(x)=loga(x-3),p(x,y)在f(x)上,q(x-2,-y)是函数g(x)上点

1 设(x`,y`)在g(x)上
则y`=-y
x`=x-2
既y=-y`
x=x`+2
又(x,y)在f(x)上
故y=loga(x-3)
既-y`=loga(x`+2-3)
y`=-loga(x`-1)
既g(x)=-loga(x-1)
（2） 首先由f(x) g(x)定义域知x>3
f(x)>g(x)
loga(x-3)>-loga(x-1)
loga(x-3)+loga(x-1)>0
loga(x-3)(x-1)>0
如果0<a<1
则(x-3)(x-1)<1
x^2-4x+2<0
2-√2<x<2+√2 又x>3
故3<x<2+√2
如果a>1
则(x-3)(x-1)>1
x>2+√2 或x<2-√2
又x>3
故x>2+√2

p(x,y)在f(x)上, y=loga(x-3)
q(x-2,-y)是函数g(x)上点
则：-y=g(x-2)
g(x-2)=-loga(x-3)
g(x)=-loga(x-1)

2) f(x)>g(x)
loga(x-3)>-loga(x-1)
loga(x-3)+loga(x-1)>0

若a>1, 则(x-3)(x-1)>1, x-3>0, x-1>0
解得：x >2+√2
若a<1, 则 (x-3)(x-1)<1, x-3>0,x-1>0
解得：3<x<2+√2