Why is f (x) = LG ((square of X + 1) - x) an odd function

Why is f (x) = LG ((square of X + 1) - x) an odd function

f(x)+f(-x)
=lg[√(x²+1)-x]+lg[√(x²+1)+x]
=lg{[√(x²+1)-x][√(x²+1)+x]}
=lg(x²+1-x²)
=lg1
=0
f(-x)=-f(x)
Domain √ (X & sup2; + 1) - x > 0
The domain of definition is r, symmetric about the origin
So it's an odd function

If f (x) = LG [2x / (1 + x) + a] is an odd function, find the value of A

∵f(-x)=-f(x)
∴f(x)+f(-x)=0
∴lg[2x/(1+x)+a]+lg[-2x/(1-x)+a]=0
That [(2 + a) ^ 2-1] x ^ 2 + (1-A ^ 2) = 0 holds for any X in the domain
(2 + a) ^ 2-1 = 0 and (1-A ^ 2) = 0
The solution is a = - 1

It is known that f (x) is an odd function defined on R with period 2, when x ∈ [- 1,1], f (x) = x & # 178;
When x ∈ [1,3], find the expression of F (x),
Find f (3.5), f (- 3)

If it is an odd function, X ∈ [- 1,1], f (x) = x & # 178; itself is wrong, it should be x ^ 3