Question

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

Answer 1

See explanation.

Explanation:

The function can be written as:

`f(x)=1/(root(4)(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=(-1)^2-4-1*(-12)=1+48=49`

`sqrt(Delta)=7`

`x_1=(1-7)/2=-3`

`x_2=(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 `+oo`

If `x` goes to `+oo` and `-oo` then `f(x)` goes to zero, so the function's range is:

`R=(0;+oo)`