Question 783955
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The domain of a function is the set of real numbers for which the function is defined.  For polynomial functions, that is always the entire set of real numbers.  For rational functions, such as your example, it is the set of real numbers with any values that would cause the denominator polynomial to equal zero excluded.  For functions including radicals with an even index (square root, 4th root, 6th root, and so on), the domain is restricted to those values that make the radicand non-negative (that is to say, greater than or equal to zero).


The procedure for a radical function such as your example is to take the denominator expression, set it equal to zero, then solve for all real roots.  Then the domain is the set of all real numbers except for those real zeros you found for the denominator.


Specifically, in this case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{x^2\ -\ 14x\ +\ 49}{100x}]


So set the denominator equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  100x\ =\ 0]


And solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x\ =\ 0]


Hence *[tex \Large x\ =\ 0] is excluded.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{dom}\left(f\right)\ =\ \left\{x\ \in\ \mathbb{R}\ |\ x\ \not =\ 0\right\}]


Read:  "The domain of f is the set of all numbers *[tex \Large x] contained in the real numbers such that *[tex \Large x] is not equal to zero."


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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