Question 310403
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The domain of a rational function is all real numbers EXCEPT any number that would cause the value of the denominator to be zero.  Take the denominator and set it equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(x^2\ -\ 9\right)\ =\ 0]


and then solve for *[tex \Large x]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x\ +\ 3)(x\ -\ 3)\ =\ 0]


Therefore, *[tex \Large 0],  *[tex \Large -3], and *[tex \Large 3] must be excluded from the domain.  Using interval notation, the domain is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\infty,-3\right)\ \small\cup\LARGE\ \left(-3,0\right)\ \small\cup\LARGE \left(0,3\right)\ \small\cup\LARGE\ \left(3,\infty\right)\] 


Or in set-builder notation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \{x\,|\,x\,\in\,\mathbb{R},\,x\ \neq\ -3,\,x\ \neq\ 0,\,x \neq\ 3\}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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