Question 477204
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You have to avoid a zero denominator so we start with the restriction: *[tex \Large x\ \neq\ \pm\sqrt{5}]


Next, note that on the interval *[tex \Large -\sqrt{5}\ \leq\ x\ \leq\ \sqrt{5}], *[tex \Large x^2\ -\ 5\ \leq\ 0], but since *[tex \Large x\ +\ 4 > 0] on this same interval, *[tex \Large \frac{x\ +\ 4}{x^2\ -\ 5}\ < 0] on that interval.  Since *[tex \Large y\ =\ \sqrt{x}] is undefined for *[tex \Large x\ <\ 0], the interval *[tex \Large -\sqrt{5}\ \leq\ x\ \leq\ \sqrt{5}] must be excluded from the domain.


On the interval *[tex \Large -\infty\ <\ x\ <\ -4], *[tex \Large x^2\ -\ 5\ >\ 0] and *[tex \Large x\ +\ 4\ <\ 0], hence *[tex \Large \frac{x\ +\ 4}{x^2\ -\ 5}\ < 0] on that interval, and the function is again undefined.  Hence, exclude *[tex \Large -\infty\ <\ x\ <\ -4] from the domain.


So the domain is all real numbers except real numbers in the two excluded ranges:


Domain: *[tex \Large \left\{x\,|\,x\ \in\ \mathbb{R},\ x\ \in \left\[-4,-\sqrt{5}\right)\ \small\ \cup \Large\ \left(\sqrt{5},\infty\right)\right}]]


Since this is a positive square root and we have shown that there is a value in the domain, namely -4 that results in a zero value for the function, the range is *[tex \Large \{y\,|\,y\ \in\ \mathbb{R},\ y\ \geq\ 0\}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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