The functions [tex]$f(x)=x^2-3$[/tex] and [tex]$g(x)=-x^2+2$[/tex]are shown on the graph.
For the following system of inequalities, the solution set must be obtained.
[tex]$\begin{aligned}&y \leq x^2-3 \\&y > -x^2+2\end{aligned}$[/tex]
the function [tex]$f(x)$[/tex]Except the function, nothing has changed [tex]$g(x)$[/tex] Because the inequality strictly exceeds the border, we must eliminate it.[tex]$-x^2+2$[/tex]
On the other hand, we redefine the inequalities and search for a test point to establish the system of inequalities' solution set. In this situation, the[tex]$(0,0)$[/tex] and look for the orientation of the solution.
Let's look at this.
For [tex]$y \leq x^2-3$[/tex], we have
[tex]$y-x^2+3 \leq 0$[/tex]
Replacing the[tex]$(0,0)$[/tex], we have
[tex]$0-0^2+3 \leq 0 \Longrightarrow 3 \leq 0$[/tex] Because of the contradiction, the inequality is still true for values outside of the parabola.
Similarly, we have for [tex]$y > x^2+2 \Longrightarrow y-x^2-2 > 0$[/tex]
Replacing the[tex]$(0,0)$,[/tex]we have
[tex]$0-0^2-2 > 0 \Longrightarrow-2 > 0$[/tex].Since there is a contradiction here, zero is not part of the answer. Therefore, the collection of points outside the parabola corresponds to the solution.
The system's graphical solution is as follows based on the aforementioned.
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