It is known that [piecewise function] f (x) = - x + 3-3a, X ＜ 0 or a ^ x, X ≥ 0 (a ＞ 0 and a ≠ 1) is defined as R, which is a decreasing function

It is known that [piecewise function] f (x) = - x + 3-3a, X ＜ 0 or a ^ x, X ≥ 0 (a ＞ 0 and a ≠ 1) is defined as R, which is a decreasing function

f(0-)=3-3a
f(0)=a^0=1
Because it is a decreasing function on R, we have: F (0 -) > = f (0), that is, 3-3a > = 1, we get: a

It is known that the odd function y = f (x) with the domain (- 1,1) is a decreasing function, and f (A-3) + F (8-3a)

First, define the domain requirements: - 1

Given the function f (x) = LG1 + X / 1-x, find the domain of definition of (1). F (x); (2) the value range of X that makes f (x) > 0

Is it f (x) = LG1 + X / (1-x)?
(1)
∵ zero cannot be divisor
∴1-x!=0→x!=1
So f (x) is defined as {x | X! = 1}
(2)
∵ make f (x) > 0
∴lg1+x/(1-x)>0
And LG1 = 0
∴x/(1-x)>0 (x!=1)
→x(1-x)>0
Ψ x > 0 and X