Let $f:\mathbb{R} \to \mathbb{R}$ be a function, so that $f(x)=\lfloor{2\lceil{\frac x2}\rceil}+\frac 12 \rfloor$. How do I find $f([-4,-2])$?

Question

By using logic I can see that the answer is $\{-4,-2\}$, but what is the correct way to format the solution?

Answer 1

Hint. Consider what happens when $x\in (2n,2n+2]$ for $n\in\mathbb{Z}$. Then $\frac{x}{2}\in (n,n+1]$ and $$f(x)=\lfloor 2\lceil\frac{x}{2}\rceil+\frac{1}{2} \rfloor= \lfloor 2(n+1)+\frac{1}{2} \rfloor=2n+2\implies f((2n,2n+2])=\{2n+2\}. $$ What is $f([-4,-2])=f(-4)\cup f((-4,-2])$?