How do you find the domain of ${(x^{2} - x - 12)}^{- 4}$ $(x^2-x-12)^-4$ ?

Question

1759 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-18T23:11:59

${(x^{2} - x - 12)}^{- 4}$ $(x^2-x-12)^-4$ can be written as $\frac{1}{{(x^{2} - x - 12)}^{4}}$ $1/(x^2-x-12)^4$

This will be undefined if $(x^{2} - x - 12) = 0$ $(x^2-x-12) = 0$ , but defined for all other values of $x$ $x$ in $RR$ .

$x^2-x-12 = (x-4)(x+3)$ is zero when $x=4$ or $x=-3$ , so these are the only prohibited values of $x$ and the domain of the function is:

$RR \ {-3,4}$

that is ${x in RR: x != -3 ^^ x != 4}$

How do you find the domain of (x^2-x-12)^-4(x2−x−12)−4(x^2-x-12)^-4?