# Question #4fbab

Mar 16, 2017

The domain of $f \left(x\right)$ is $x < - 4 \mathmr{and} x > 3$

#### Explanation:

Because the argument for the $\frac{1}{2}$ power cannot be negative we must find the part of the domain where ${x}^{2} + x - 12 \ge 0$ and exclude it.

Let's find the points where the argument is 0:

${x}^{2} + x - 12 = 0$

Factor:

$\left(x + 4\right) \left(x - 3\right) = 0$

This implies that:

$x + 4 = 0 \mathmr{and} x - 3 = 0$

$x = - 4 \mathmr{and} x = 3$

The function is negative between these values of x:

graph{x^2+x-12 [-8, 8, -14, 14]}

Therefore, the domain of $f \left(x\right)$ is $x < - 4 \mathmr{and} x > 3$