The function FX = log (1 / 2) (x + 1) / (x-1) is known to judge the parity. It is proved that FX is an increasing function at (1, + ∞)

The function FX = log (1 / 2) (x + 1) / (x-1) is known to judge the parity. It is proved that FX is an increasing function at (1, + ∞)

Ah, the basic question
(1) F (x) = log (x + 1) / (x-1) base 1 / 2
Then f (- x) = log [(- x) + 1] / [(- x) - 1]
=log(1-x)/(-1-x)
=log(x-1)/(1+x)
=-log(x+1)/(x-1)
=-f(x)
So the function is odd
(2) Since the base is 1 / 2, we only need to prove that G (x) = (x + 1) / (x-1) decreases in this interval
g(x0=（x+1)/(x-1)=(x-1+2)/(x-1)=1 + 2/(x-1)
Obviously, in (1, + ∞), when x ↑, 2 / (x-1) ↓, G (x) ↓, f (x) ↑