Does there exists a function $f$ such that $f \circ f (x)= -x$

Question

If $f: R \to R$ is a function such that $f \circ f (x) = -x$ then $f$ is injective. (True/false)

I couldn't find any function such that $f \circ f(x) = -x$

Does such function exist? Or any hint would be helpful to form such function.

Answer 1

True, $$(f\circ f)(x)=-x $$ is a bijection from the reals to reals. In general, if $$ g\circ f $$ is a bijection, $f$ must be an injection (while $g$ must be surjective).

Answer 2

I am going to prove that there exist no continuous functions and no Darboux functions(functions which have the Intermediate Value Property) which satisfy this condition.
Case $1$: $f$ is continuous. Assume there exists such $f$. As another user has already observed, $f$ is bijective.
A continous and bijective function is strictly monotonic $\implies f$ is strictly monotonic.
But the composition of two strictly monotonic functions of the same type is strictly increasing, so $f\circ f$ is strictly increasing . However, $g:\mathbb{R} \to \mathbb{R}$, $g(x)=-x,$ is strictly decreasing, contradiction.
Case 2: $f$ has the IVP. A bijective function with the IVP is continuous and we reach a contradiction in the same way as in Case 1.