# Question #c73aa

Nov 25, 2014

Since the denominator cannot be zero,

${x}^{2} - x - 12 = \left(x + 3\right) \left(x - 4\right) \ne 0 \implies x \ne - 3 , 4$

So, the domain of $f$ is

${D}_{f} = \left(- \infty , - 3\right) \cup \left(- 3 , 4\right) \cup \left(4 , \infty\right)$

To find the range, you need to know what its graph looks like.

Since the graph extends from $- \infty$ to $\infty$, the range of $f$ is

${R}_{f} = \left(- \infty , \infty\right)$

I hope that this was helpful.