#### Answer

g is continuous everywhere.

#### Work Step by Step

Polynomial functions are continuous everywhere, and by theorem 8, the quotient of two polynomials is also continuous everywhere where the quotient is defined.
The only possible problem is at x=3. $g(3)$ is defined to be 5. We apply the Continuity Test (see p.95), by testing whether $\displaystyle \lim_{x\Rightarrow 3}g(x)$ exists, and if it equals $g(3).$
$\displaystyle \lim_{x\Rightarrow 3}g(x)=\lim_{x\Rightarrow 3}\frac{x^{2}-x-6}{x-3}\qquad$... factor the trinomial,
$=\displaystyle \lim_{x\Rightarrow 3}\frac{(x-3)(x+2)}{(x-3)}\qquad$... cancel common factor
$=\displaystyle \lim_{x\Rightarrow 3}(x+2)\qquad$... evaluate directly
$=5$
Since $\displaystyle \lim_{x\Rightarrow 3}g(x)$ exists, and equals $g(3)=5,$
we find that g is continuous everywhere.