How to find the pre-image of a relation given the interval?

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$$ \begin{aligned} &\begin{array}{l} \text { 2) Given the following relations: } \\ \qquad f=\left\{(x, y) \text { , } x, y \in Z, y=x^{4}+4\right\}, \text { a relation from } Z \text { to } Z \text { . } \end{array}\\ &\begin{array}{l|l} \mathrm{g}=\{(x, y) & \left.x, y \in \mathbf{R}, x^{2}+y^{2}=4\right\}, \text { a relation from } \mathbf{R} \text { to } \mathbf{R} . \\ \mathrm{h}=\{(x, y) & \left.x, y \in \mathbf{R}, x^{2}=y-1\right\}, \text { a relation from } \mathbf{R} \text { to } \mathbf{R} . \end{array} \end{aligned} $$ $$ \text { c) For each of } f, g, \text { and } h, \text { find the preimage of the interval } B=[-2,2] $$

Can someone please explain how to find the pre-image of the interval $B[-2,2]$?

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For example, in the first case, you are looking for all integers $x\in\mathbb{Z}$ such that $-6\leq x^4\leq -2$, which is clearly the empty set. In the second case, $g^{-1}([-2,2])$ is the set of all real numbers $x\in\mathbb{R}$ such that $(x, \pm\sqrt{4-x^2})$ is a point of the circle of radius $2$ around the origin, hence $x\in[-2,2]$. As for $h$, the closed interval $[-1,1]$ will work.