Question 908773
I'm assuming the function is {{{f(x) = 1/(13-x)}}}


Since you cannot divide by zero, this means the denominator {{{13-x}}} cannot be zero.


If it were zero, then {{{13-x=0}}} ---> {{{x=13}}}


In reverse, this means that if {{{x=13}}} then the denominator {{{13-x}}} is zero.


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Because {{{x=13}}} causes the denominator {{{13-x}}} to be zero, we have to kick this number out of the domain. Any other number works.


So the domain is the set of real numbers x BUT x cannot equal 13.


In interval notation, the domain is *[Tex \Large \left(-\infty,13\right)\cup\left(13,\infty\right)]