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\begin{equation} f(x) = \frac{x^2 - 7}{\sqrt{(\frac{x+2}{4})^{x^2} – (\frac{x+2}{4})^{2|x|+3}}} \end{equation}

I tried to get the domain by finding where the denominator is bigger than zero since at zero it is undefinable where its positive {the denominator) since the square root sign \begin{equation} {(\frac{x+2}{4})^{x^2} – (\frac{x+2}{4})^{2|x|+3}}> 0 \end{equation} \begin{equation} {(\frac{x+2}{4})^{x^2} > (\frac{x+2}{4})^{2|x|+3}} \end{equation} \begin{equation} {{x^2} > {2|x|+3}} \end{equation} \begin{equation} {{x^2} -{2|x|-3}>0} \end{equation} \begin{equation} { where \ for \ the \ equation }{\ \ \ }{ { x^2} -{2x-3}=0} {\ the \ solutions \ are}\end{equation} \begin{equation} {(x-3)(x+1)} \end{equation} \begin{equation} {(x=3)(x=-1)} \end{equation}

\begin{equation} {{x^2} -{2|x|-3}=0} \end{equation} \begin{equation} {here \ I \ deduced \ that \ we \ only \ take \ when \ x >=0 \ for this equation \ because \ of the \ square \ and \ || \ sign }\end{equation} \begin{equation} {{x^2} -{2|x|-3}>0} \end{equation} so we get the following domain for the Inequality (3,∞)

\begin{equation} {(\frac{x+2}{4})^{x^2} – (\frac{x+2}{4})^{2|x|+3}}= 0 \end{equation} when ? when there is 1 to the power of a number - 1 to the power of a number or when we have ${0 - 0 }$ so \begin{equation} {(\frac{x+2}{4})}= 0 \end{equation} x=-2 \begin{equation} {(\frac{x+2}{4})}= 1 \end{equation} x=2 and we can see that the numbers between them are eligible when Substituting but still not sure how to take the whole period (-2,2) as part of the domain

any tips and explanations and corrections would help thanks

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1 Answer 1

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Between the points (-2,0) and (2,1) the base of the power is in (0,1).

The exponents have $x^2<1$ , while $2|x|+3 >1$.

It follows that the first term is greater than the second, because $$ \forall \ (x,p,q),\quad 0<(x,p)<1, q>1 \quad x^p>x,\quad x^q<x$$.

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