Calculate the $F^{−1}$ of a function

Question

Let $V:=\{x \in \mathbb{R}:-5 \leq x \leq 5\} .$ Consider the function $F: V \rightarrow V$ defined by $$ F(x)=\left\{\begin{array}{ll} -1, & \text { if }-5 \leq x \leq 1 \\ 2, & \text { if } 1<x \leq 3 \\ 0, & \text { if } 3<x<4 \\ x, & \text { if } 4 \leq x \leq 5 \end{array}\right. $$

$I:=\{x\in\mathbb{R}\ :\ -5\leq x\leq1\}$. Choose the set that $F^{−1}(I)$ is equal to.

Answer 1

Of all the possible values of $F$ the only value in $(0,4)$ is $2$. Hence, $F^{-1}(I)=\{x: F(x) \in (0,4)\}=(1,3]$. The correct answer is b).

For the edited version of the question $F^{-1}(I)=[-5,1] \cup (3,4)$.