Does the equality $f(a,b)=(f(a),f(b)) $ hold? I doubt it. Taking $f(x)= | x|$ and $a=-5, b=6$ gives us $f(-5,6)= (5,6)$, but $0 \notin (5,6)$.
This is indeed a very stupid question, but I am not confident enough.
Does the equality $f(a,b)=(f(a),f(b)) $ hold? I doubt it. Taking $f(x)= | x|$ and $a=-5, b=6$ gives us $f(-5,6)= (5,6)$, but $0 \notin (5,6)$.
This is indeed a very stupid question, but I am not confident enough.
Copyright © 2021 JogjaFile Inc.

It does, because $f$ is monotonic (and $|x|$ isn't).
As $f$ is strictly monotonic, for $c \in (a, b)$ we have $f(a) < f(c) < f(b)$, so $f(a, b) \subseteq (f(a), f(b))$.
For $x \in (f(a), f(b))$, as $f$ is one-one, we have $f(c) = x$ for some $c$. But would $c \leqslant a$ we would have $f(c) < f(a)$, so $x$ would be not in $(f(a), f(b))$. Similarly it's impossible $c \geqslant b$. So $c \in (a, b)$ and $(f(a), f(b)) \subseteq f(a, b)$.