Question 235932
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The domain of a function is the set of values of the independent variable for which the function is defined.  What this means is that if you have a number that you would like to substitute for *[tex \Large x] in your function *[tex \Large p(x)\ =\ x^2\ -\ 2x\ +\ 10] and a real value for *[tex \Large p(x)] results, then the value you selected for *[tex \Large x] IS in the domain set.  On the other hand, if *[tex \Large p(x)] is not defined for that value of *[tex \Large x], then that value of *[tex \Large x] IS NOT in the domain.


Examining the given function, we can see that there are no instances of the independent variable in a denominator.  If that were the case, then we would have to exclude any value of the independent variable that would make any denominator equal to zero -- and hence make the entire function undefined.  We also have no radicals with even-numbered indices, so we do not have the problem of trying to take the square or higher even numbered root of a negative number.  There are also no logarithmic, trigonometric, or inverse trigonometric functions that might cause us to have domain restrictions.


In short, the domain of this function, namely *[tex \Large p(x)\ =\ x^2\ -\ 2x\ +\ 10] is the set of all real numbers.  This can be expressed in interval notation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\infty,\,\infty\right)]


Or set builder notation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \{x\,|\,x\,\in\,\R\}]


In general, any polynomial equation of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p(x)\ =\ a_nx^n\ +\ a_{n-1}x^{n-1}\ +\ a_{n-2}x^{n-2}\ +\ \cdots\ +\ a_0x^0]


(*[tex \Large a_0] through *[tex \Large a_n] are constant coefficients) has a domain of all real numbers, *[tex \Large \left(-\infty,\,\infty\right)]


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