Proving that a function is odd

Assume that there exists a function f:RR that is bijective and satisfies
f(x)+f1(x)=x
for all x. Here f1 is the inverse function. Show that f is odd.

This was a brain-teaser given to me by a friend. Two other related questions are:

  1. Show that f is discontinuous
  2. Give an example of such a function (if indeed one exists).

Edit: As an initial idea, maybe approaching the problem graphically would help? A function and its inverse are reflections of each other about y=x on the xy plane. Does this lead to anywhere?

Answer

Here is finally a constructive example of a solution, continuous except on a countable set.

Let ϕ=1+52, and let
f(x)={0if x=0ϕxif |x|[ϕ3k,ϕ3k+1),kZϕxif |x|[ϕ3k+1,ϕ3k+2),kZϕ2xif |x|[ϕ3k+2,ϕ3k+3),kZ

Then f is a bijection, whose inverse is
f1(x)={0if x=0ϕ2xif |x|[ϕ3k,ϕ3k+1),kZϕ1xif |x|[ϕ3k+1,ϕ3k+2),kZϕ1xif |x|[ϕ3k+2,ϕ3k+3),kZ
Checking each case shows that f(x)+f1(x)=x, as required.

Plot of f

Attribution
Source : Link , Question Author : Iconoclast , Answer Author : yoann

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