We have $f$ defined from $\mathbb{R}→\mathbb{Z}$, where $f(x)=\lceil2x−1\rceil$.
- We need to find $f(A)$, where $A=\{x \in \mathbb{R}∣1\le x \le4\}$.
- And we also need to find $f^{-1}(B)$, where $B = \{−9,−8\}$.
I don't really understand how to solve this problem. I just need a little help and I'll do the calculations myself.
By definition we have $$f(x)=\left\lceil 2x-1\right\rceil=n$$ iff $n-1<2x-1\leq n$, which is the same as $$f(x)=n\qquad\Leftrightarrow\qquad{n\over2}<x\leq {n+1\over2}\ .\tag{1}$$ It follows that $f(x)=1$ for ${1\over2}<x\leq1$, that $f(x)=2$ for $1<x\leq{3\over2}$, and so on, until $f(x)=7$ for ${7\over2}<x\leq4$. This allows to conclude that $f(A)=\{1,2,3,4,5,6,7\}$.
On the other hand from $(1)$ it follows that $f(x)=-9$ iff $-{9\over2}<x\leq-4$, and $f(x)=-8$ iff $-4<x\leq-{7\over2}$. This allows to conclude that $f^{-1}(B)=\bigl]-{9\over2},-{7\over2}\bigr]$.