Given $f: \mathrm{R} \to \mathrm {R}$ , define $ f^2(x)= f(f(x)).$ Which of the following are true?
- If $f$ is strictly monotonic, then $f^2$ is strictly increasing.
- If $f^2(x) = -x $ for all $x \in \mathrm{R},$ then $ f $ is one- one.
- There does not exist a continuous function $f: \mathrm{R} \to \mathrm {R}$ such that $f^2(x)= -x$ for all $x \in \mathrm{R}.$
I think $1$ is true since function composition will always give increasing function for strictly monotone functions (either increasing like $x$, $\log x$ or decreasing like $e^{-x}$ or $1/x$) but unable to find function for option $2$ and $3$
I leave the first to you, the intuition is correct but you should formalize the math.
We answer the second and third. In points :
Let $f^2(x) = -x$ for some continuous function $f$.
Prove that $f$ is one-one as follows : start with $f(x) = f(y)$. Apply $f$ to both sides. Simplify using our assumption and conclude $x=y$, so $f$ is one-one.
Show that any one-one continuous function on $\mathbb R$ is strictly monotonic (use the Intermediate Values Theorem). Hence $f$ is strictly monotonic.
Conclude from part $1$ that $f^2$ is strictly increasing . But $-x$ is not, so that gives a contradiction.