Question 220503
Here are some key things to remember:


1) You cannot take the square root of a negative number. So this means that {{{sqrt(x)}}} forces {{{x}}} to be positive or zero. This means that {{{x>=0}}}



2) You cannot divide by zero. So {{{x^2-9<>0}}} (ie the denominator can't be zero). This means that {{{x<>-3}}} or {{{x<>3}}} (these values would make the denominator equal to zero. Since {{{x>=0}}}, we don't need to worry about {{{x<>-3}}}


Putting 1) and 2) together gets us the domain (in interval notation): <font size=6>[</font>*[Tex \LARGE 0,3]<font size=6>)</font>*[Tex \LARGE \cup]<font size=6>(</font>*[Tex \LARGE 3,\infty]<font size=6>)</font>



This basically says that the domain is any non-negative number but {{{x<>3}}}