Answer:
[tex]x\in [-1 \ ; \ 1) \ \cup \ (1 \ ;\ 3] \ \ or \ \ \left [ \begin{array}{ccc} x\geq -1 \\ \!\!\!\!x\neq 1 \\ \!\!\!\!x\leq 3\end{array}[/tex]
Step-by-step explanation:
[tex]\displaystyle y=2x^2+2 \ \ ; \ \ -2\leqslant x\leqslant 4 \\\\ Find \ intwerval \ if : \\\\ -1<y \leq 3 \Leftrightarrow -1<2x^2-2x\leqslant 3 \\\\ Then : \\\\ \left [{{x^2-2x>-1} \atop {x^2-2x\leq 3 } \right. \Leftrightarrow \left [ {{(x-1)^2>0} \atop {(x-3)(x+1) \leq 0}} \right. \Leftrightarrow \\\\\\ \left [ {{x\neq 1} \atop {(x-3)(x+1)\leq 0}} \right. \ \ ; \ \ and \ \-2\leqslant x\leqslant 4 \\\\ Then: \\\\ signs : +++[-1]---(1)---[3] +++ \\\\[/tex]
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[tex]x\in [-1 \ ; \ 1) \ \cup \ (1 \ ;\ 3][/tex]
Let's check whether [tex]-2\leqslant x\leqslant 4[/tex] or [-2 ; 4 ] is included in the interval
----------[-1]--------(1)--------[3]------
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---[-2]---[-1]---------(1)-----------[3]--------[4]
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The interval enters the interval x completely then the answer is :
[tex]x\in [-1 \ ; \ 1) \ \cup \ (1 \ ;\ 3] \ \ or \ \ \left [ \begin{array}{ccc} x\geq -1 \\ \!\!\!\!x\neq 1 \\ \!\!\!\!x\leq 3\end{array}[/tex]
