Remedy to $1-f(x) = f(-x)$
Clearly, we just have relationships in between f (x) and also f (- x). The relationship suggests that 0 needs to have value 1/2. We can separate all non - absolutely no actual numbers right into disjoint sets of x and also - x and also specify the function f on each set independently. For each and every set, f (x) can be offered any kind of value and afterwards f (- x) has a solitary legitimate value. As stated by Grigory, the legitimate features can be qualified as any kind of weird function plus 1/2.
WolframAlpha gives a remedy to this (and also several various other) reappearance formulas :
Consider rather the features $g$ that please the identification $-g(x)=g(-x)$ for all $x$. If $(x,g(x))$ is a factor of the function $g$, after that $(-x,-g(x))$ is additionally a factor (given that $g(-x)=-g(x)$). Consequently, every function $g$ is symmetrical when revolved by $180$ levels concerning the factor $(0,0)$.
Just how do points transform for the identification $1-f(x)=f(-x)$? We just change the factor of proportion to $(0,1/2)$. Below the factor $(x,f(x))$ indicates the factor $(-x,1-f(x))$.
The function $f$ pleases the identification $1-f(x)=f(-x)$ for all actual numbers $x$ if and also just if it is symmetrical when revolved concerning the factor $(0,1/2)$ by $180$ levels.
There's mosting likely to be most of these features ; several of which will certainly be polynomials, several of which will certainly not.
Usually straightforward troubles similar to this ask you to locate a function that values the problem, not all of them. And Also (once more) generally you start by examining if a straightforward polynomial function of the first level can be a remedy.
$$f(x) = ax + b$$
$$f(x) + f(-x) = 1 \implies ax + b + a \cdot (-x) + b = 1 \implies 2b = 1 \implies b = 1/2$$
So the problem is pleased by any kind of function of the type :
$$f(x) = ax + 1/2$$