# How do you find the domain and range of (x^2-x-12)^-(1/4)?

Aug 23, 2017

See explanation.

#### Explanation:

The function can be written as:

## $f \left(x\right) = \frac{1}{\sqrt{{x}^{2} - x - 12}}$

To find the domain of rhis function we have to think of the set of arguments ($x$), for which the function's value is defined.

This function is defined for those values of $x$, fior which

## ${x}^{2} - x - 12 > 0$

$\Delta = {\left(- 1\right)}^{2} - 4 - 1 \cdot \left(- 12\right) = 1 + 48 = 49$

$\sqrt{\Delta} = 7$

${x}_{1} = \frac{1 - 7}{2} = - 3$

${x}_{2} = \frac{1 + 7}{2} = 4$

graph{x^2-x-12 [-32.48, 32.47, -16.24, 16.24]}

From the graph we can see that the domain is:

D=(-oo;-3) uu (4;+oo)

To find the range we have to analyze the end behaviour of the function.

If $x$ goes to $- 3$ from the left side or to $4$ from the right side, then the denominator goes to zero, so the function's value goes to $+ \infty$

If $x$ goes to $+ \infty$ and $- \infty$ then $f \left(x\right)$ goes to zero, so the function's range is: