Inverse of a given function Let the function $f: R \to R$ be defined by $f(x)=x^2-2$. Find $f^{-1} \{- \infty&lt;x \leq2\}$

What is $f^{-1} \{- \infty&lt;x \leq2\}$ is supposed to mean here? If I take $ - \infty&lt;x \leq2\ $ as domain, the $f$ is not injective in that interval, hence $f^{-1}$ does not exist. Could someone help me with this notation?
Suppose $f$ is a function from $X$ to $Y$, and suppose $V$ is a subset of $Y$. Then $f^{-1}(V)$ is defined to be the preimage of $V$, i.e. it is the set of all $x$ whose image under $f$ is contained in $V$:
$$ f^{-1}(V) = \{ x \in X : f(x) \in V \}.$$

Note that this definition works for arbitrary $f$, bijective or not! There doesn't need to be an inverse function $f^{-1}$. We write it as $f^{-1}$(V)$ purely because it is a convention.

[If $f$ is bijective, then the inverse $f^{-1}$ does exist, and you can show that the $f^{-1}(V)$ that I defined is the same as the image of $V$ under the function $f^{-1}$, i.e $f^{-1}(V) = \{ f^{-1}(y) : y \in V \}$.]
It doesn't seem like standard notation.  I have two guesses:


It's a typo and they meant the domain of $f$ to be $-\infty &lt;x\leq \sqrt{2}.$
$f^{-1}$ may not be a function, but it is a relation.  And maybe they 
want the set of all $x$ such that $f(x) \in (-\infty,2]$.