What is f(x)?

It is impossible to say what $f(x)$ is.

$f(x)$ could be equal to $x$ in certain intervals while it could be equal to $-x$ in other intervals.

Consider this example.

$f(x) = \begin{cases} x, \quad \mbox{if }\quad \pi\leq x \leq 2\pi \\ -x, \quad \mbox{if }\quad x<\pi \quad \text{or}\quad x > 2\pi. \end{cases}$ .

In fact there are infinite possibilities for $f(x)$ and it is impossible to tell what $f(x)$ is without any additional information.

Actually it is possible to tell what $f$ is, in the sense that we give a complete characterization of all possible $f$ .

Partition the reals into two subsets $A,B$ arbitrarily. Let $f(x) = x \, \forall x \in A$ and $f(x) = -x \, \forall x \in B$ . This is all the solutions.

To see this, observe that $f(x)^2 = x^2$ implies $|f(x)| = |x|$ , so $f(x) = x \vee -x$ . However, there is no way to relate $f$ taken on two different values from just the given equation, so the best we can do is that, $f(x) = x \vee -x$ for all $x$ . The above simply formalizes this, choosing which give positive values and which give negative values.

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Since for each real number besides $0$ we have two choices for the function, in total we have $2^\mathfrak{c}$ such functions where $\mathfrak{c}$ is the cardinality of the real numbers. Georg Cantor proved that $\omega < 2^\omega$ for any cardinal number $\omega$ , and in particular $\mathfrak{c} < 2^\mathfrak{c}$ ; thus we have more functions satisfying this condition than there are real numbers!

Consider the function $f(x)=x$ when $x$ is rational, and $f(x)=-x$ when $x$ is irrational. This function satisfies the equation but does not satisfy $[1]$ , $[2]$ , or $[3]$ .