Answer
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$
Work Step by Step
We are given the function:
$f(x)=\dfrac{x^5+6x+17}{x^2-9}$
$f(x)$ is a rational function, therefore according to Theorem 2.10 b), the function is continuous for all $x$ for which the denominator is not zero.
We check the zeros of the denominator:
$x^2-9=0$
$(x-3)(x+3)=0$
$x-3=0\Rightarrow x=3$
$x+3=0\Rightarrow x=-3$
The interval on which the function is continuous is:
$(-\infty,-3)\cup(-3,3)\cup(3,\infty)$
