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

1 Answer
Aug 23, 2017

Answer:

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)#