Question 559661
The domain of f(x) = 5x/(x^3+36x) = {{{5x/(x^3+36x)}}}
includes all real numbers except those that make {{{x^3+36x=0}}}
Factoring, we get:
{{{x^3+36x=0}}} ==> {{{x(x^2+36)=0}}}
In the real numbers there are no solutions for
{{{x^2+36=0}}}, so the only solution would be {{{x=0}}}.
For {{{x=0}}}, the function does not exist; it's undefined.
For any other value of {{{x}}}, we can simplify to
f(x) = {{{5x/(x(x^2+36))}}} = {{{5cross(x)/(cross(x)(x^2+36))}}} = {{{5/(x^2+36)}}}
The graph of the function is the same as the graph of
f(x) = {{{5/(x^2+36)}}} , except that it does not exist for {{{x=0}}}.
There is a hole where (0,5/36) would be.
The number {{{x=0}}} is the only real number excluded from the domain of the function.