Question 901601
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 19]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{x\ -\ 19}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{9}{sqrt{x\ -\  19 }}]


The argument for the square root cannot be negative, nor can the denominator expression be zero.  Hence the square root argument must be strictly positive, i.e. *[tex \Large x\ >\ 19].


As an interval, the domain is *[tex \Large (19,\infty)]


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