# how to find the domain of a function like this and graph it?

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

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 $$$${(\frac{x+2}{4})^{x^2} – (\frac{x+2}{4})^{2|x|+3}}> 0$$$$ $$$${(\frac{x+2}{4})^{x^2} > (\frac{x+2}{4})^{2|x|+3}}$$$$ $$$${{x^2} > {2|x|+3}}$$$$ $$$${{x^2} -{2|x|-3}>0}$$$$ $$$${ where \ for \ the \ equation }{\ \ \ }{ { x^2} -{2x-3}=0} {\ the \ solutions \ are}$$$$ $$$${(x-3)(x+1)}$$$$ $$$${(x=3)(x=-1)}$$$$

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

$$$${(\frac{x+2}{4})^{x^2} – (\frac{x+2}{4})^{2|x|+3}}= 0$$$$ when ? when there is 1 to the power of a number - 1 to the power of a number or when we have $${0 - 0 }$$ so $$$${(\frac{x+2}{4})}= 0$$$$ x=-2 $$$${(\frac{x+2}{4})}= 1$$$$ 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

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.