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

1 Answer
Feb 16, 2018

Answer:

#Dom=RR\setminus {3,-4}#
#Ran=(0,infty)#

Explanation:

(I guess you are working on #RR#, if not please tell me...)

We can rewrite this as

#\frac{1}{(x^2-x-12)^4}#

Let #f(x)=(x^2-x-12)^4# and #g(y)=\frac{1}{y}#

The domain of the composition of functions #gf# is given by

#\{x\in Dom(f) : f(x)\in Dom(g)\}#

Since the domain of a polynomial is the whole #RR# and the domain of #g# is #RR\setminus \{0\}#, we have that the domain of #gf# is

#{x\in RR : x^2-x-12\ne 0}#

Since #x^2-x-12=0# if and only if #x=-4# or #3#, we have that

#Dom=RR\setminus {3,-4}#

Now for the range, again we have that it is #g(Ran(f)\cap Dom(g))# .
The range of a quadratic polynomial with two distinct roots is

#[a,infty)#

for some #a<0# (the vertex of the parabola defined by this quadratic equation).

Since now #4# is even, we have that

#[a,infty)^4=[0,infty)#

since an even power of a negative number is positive. So from that we see that

#Ran=(0,infty)#

since for all #x in (0,infty)# there is #y in (0, infty)# such that #x=frac{1}{y}#, namely #x=frac{1}{y}#, and for the rest there is none.