Question 171810
I'll do the first two to get you going in the right direction.



a)




{{{(5)/(2n+3)}}} Start with the given expression



{{{2n+3=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of n that make the denominator zero, then we must exclude them from the domain.




{{{2n=0-3}}}Subtract 3 from both sides



{{{2n=-3}}} Combine like terms on the right side



{{{n=(-3)/(2)}}} Divide both sides by 2 to isolate n




{{{n=-3/2}}} Reduce






Since {{{n=-3/2}}} makes the denominator equal to zero, this means we must exclude {{{n=-3/2}}} from our domain



So our domain is:  *[Tex \LARGE \textrm{\left{n|n\in\mathbb{R} n\neq\frac{-3}{2}\right}}]



which in plain English reads: n is the set of all real numbers except n CANNOT equal {{{-3/2}}}



So our domain looks like this in interval notation



*[Tex \Large \left(-\infty, \frac{-3}{2}\right)\cup\left(\frac{-3}{2},\infty \right)]



note: remember, the parenthesis <font size=4><b>excludes</b></font> -3/2 from the domain



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b)





Looking at {{{(5n+1)/7}}}, we can see that there are no square roots, logs, and other functions where there are restrictions on the domain.


Also, we can see that the function does not have a division by n (or any combination of variables and constants).
So we don't have to worry about division by zero.



Since we don't have any restrictions on the domain, this shows us that the domain is all real numbers. In other words, we can plug in <b>any</b> number in for n





So the domain of the function in set-builder notation is:



*[Tex \LARGE \textrm{\left{n|n\in\mathbb{R}\right}}]



In plain English, this reads: n is the set of all real numbers (In other words, n can be <b>any</b> number)



Also, in interval notation, the domain is:


*[Tex \Large \left(-\infty,\infty \right)]