Question 151704
{{{f(x)=(5x)/(x^2+25)}}} Start with the given function



{{{x^2+25=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.



{{{x^2=-25}}} Subtract 25 from both sides.



{{{x=sqrt(-25)}}} Take the square root of both sides.



{{{x=sqrt(-1)*sqrt(25)}}} Factor {{{sqrt(-25)}}} into {{{sqrt(-1)*sqrt(25)}}}



{{{x=i*sqrt(25)}}} Replace {{{sqrt(-1)}}} with "i"



{{{x=5*i}}} or {{{x=-5i}}} Take the square root of 25 to get 5 or -5



Since the values {{{x=5*i}}} or {{{x=-5i}}} make the denominator zero, this means that there are no real x values that make the denominator zero (since {{{5i}}} and {{{-5i}}} are both complex). 



So you can plug in any real number for x and you'll get a result for f(x). 




If that doesn't make any sense to you, then try to think of it like this:


{{{x^2}}} is <font size=4><b>always</b></font> positive. So there are no real values that make {{{x^2=-25}}} true. That's why the domain includes all real numbers.




-------------------------------------------------------------


Answer:


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



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



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



Also, in interval notation, the domain is:


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