Solution to $1-f(x) = f(-x)$ Can we find $f(x)$ given that $1-f(x) = f(-x)$ for all real $x$?
I start by rearranging to: $f(-x) + f(x) = 1$. I can find an example such as $f(x) = |x|$ that works for some values of $x$, but not all. Is there a method here? Is this possible?
 A: Clearly, we only have relations between f(x) and f(-x). The relation means that 0 has to have value 1/2. We can divide all non-zero real numbers into disjoint pairs of x and -x and define the function f on each pair separately. For each pair, f(x) can be given any value and then f(-x) has a single valid value. As mentioned by Grigory, the valid functions can be characterised as any odd function plus 1/2.
A: Usually simple problems like this ask you to find a function that respects the condition, not all of them. And (again) usually you start by checking if a simple polynomial function of the first degree could be a solution.
So, if
$$f(x) = ax + b$$
Then
$$f(x) + f(-x) = 1 \implies ax + b + a \cdot (-x) + b = 1 \implies 2b = 1 \implies b = 1/2$$
So the condition is satisfied by any function of the type:
$$f(x) = ax + 1/2$$
A: Consider instead the functions $g$ that satisfy the identity $-g(x)=g(-x)$ for all $x$.  If $(x,g(x))$ is a point of the function $g$, then $(-x,-g(x))$ is also a point (since $g(-x)=-g(x)$).  Therefore, every function $g$ is symmetric when rotated by $180$ degrees about the point $(0,0)$.
How do things change for the identity $1-f(x)=f(-x)$?  We merely shift the point of symmetry to $(0,1/2)$.  Here the point $(x,f(x))$ implies the point $(-x,1-f(x))$.

The function $f$ satisfies the identity $1-f(x)=f(-x)$ for all real numbers $x$ if and only if it is symmetric when rotated about the point $(0,1/2)$ by $180$ degrees.

There's going to be many of these functions; some of which will be polynomials, some of which will not.
A: WolframAlpha provides a solution to this (and many other) recurrence equations:
http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x)
A: $$f(x)=\frac{1}{2}+\text{(any odd function)}.$$
For example, $f(x)=\frac{1}{2}+x$ or, say, $f(x)=\frac{1}{2}+99x^3+7x^5$.
A: Let $$f(x) = \begin{cases} 1 \quad x>0, \\ 1/2 \quad x=0, \\ 0 \quad x<0\end{cases}.$$
If $x > 0$, then $-x < 0$ so that $f(x)+f(-x)=1+0=1$
Likewise with $x<0$.
If $x=0$, then $f(x)+f(-x)=(1/2)+(1/2)=1$.
A: We have $f(x)-1/2=-(1-f(x))+1/2=-f(-x)+1/2=-(f(-x)-1/2)$;
hence $f(x)-1/2$ is an odd function.
So, $f(x)=1/2+\phi(x)$ for some odd function $\phi$.
Clearly, any such function satisfies the original equation.
