Would this be an acceptable answer for the inverse of floor function This problem is from Discrete Mathematics and its Applications

And the  book's definition on inverse

Would an acceptable answer to 43b just be the set itself again? What I like to think of the inverse is what input did you pass to get this output. One can pass {-1, 0, 1} into the floor function to get {-1, 0, 1}. Is this the right way to think about this problem? Or is it better to introduce 3 separate variables, say x, y, and z and show the interval of values that can be passed into the floor function to get this set. 

There is an overloading here, of the symbol $f^{-1}$.
If $R$ is a binary relation, then $R^{-1}=\{(a,b)\mid (b,a)\in R\}$, and when $f$ is an injective function we can show that $f^{-1}$ is an injective function as well (here a function is just a set of ordered pairs with a particular property). In the case we call $f^{-1}$ the inverse function.
But if $f$ is not an injective function then $f^{-1}$ is not a function, it is a binary relation. In that case we use $f^{-1}$ to denote the preimage function, which is a function mapping subsets of the range (or codomain) to subsets of the domain. Namely, $f^{-1}(A)=\{x\in X\mid f(x)\in A\}$.
Note, that if $f$ is indeed injective, then for every singleton, $\{b\}$ its preimage is at most a $\{a\}$ for some $a$ in the domain.
You are asked to find the preimage, which is the set of all values being mapped into the given set. So writing $g^{-1}(\{-1,0,1\})=\{-1,0,1\}$ is blatantly wrong, since there are many numbers being mapped to those three values, not just the three.
