Mathematics 113 Online
OpenStudy (anonymous):

f(x)=(x+3)/(4-√(x^2-9)) what is the domain

myininaya (myininaya):

you don't want the bottom to be zero you also don't want the thingy under the square root to be negative so first when is the bottom 0 $4-\sqrt{x^2-9}=0 => 4=\sqrt{x^2-9} =>16=x^2-9$ $=>x^2-25=0 => x=\pm 5$ so fraction is undefined when x=5 or x=-5 now what is the domain for sqrt{x^2-9} so we have that we need x^2-9 to be positive or equal to zero x^2-9 =0 when x=-3 or x=3 ----|---|--- -3 3 test intervals around these numbers + - + so the domain of sqrt{x^2-9} is (-inf,-3] U [3,inf) but we do not include -5 or 5 for f since that would make f undefined so we say (-inf,-5)U (-5,-3] U[3,5) U (5,inf)

OpenStudy (anonymous):

thanks!! i appreciate your help <3