What is the domain and range of #p(x)= root3(x-6)/sqrt(x^2 - x - 30)#?

1 Answer
May 18, 2018

Answer:

The domain of #p# can be defined as #{x in RR : x>6}#
and the range as #{y in RR: y>0}#.

Explanation:

First, we can simplify #p# as given thusly:

#(root(3)(x-6))/(root()(x^2-x-30))= (root(3)(x-6))/(root()((x-6)(x+5)))#.

Then, further simplifying, we discern that
#(root(3)(x-6))/(root()((x-6)(x+5)))=((x-6)^(1/3))/((x-6)^(1/2)(x+5)^(1/2))#,

which, by means of dividing exponents, we deduce

#p(x)=1/(root(6)(x-6)root()(x+5))# .

By seeing #p# like this, we know that no #x# can make #p(x)=0#, and indeed #p(x)# cannot be negative because the numerator is a positive constant and no even root (i.e. #2# or #6#) can yield a negative number. Therefore the range of #p# is #{y in RR: y>0}#.

Finding the domain is no more difficult. We know that the denominator cannot equal #0#, and by observing which values for #x# would lead to thus, we find that #x# must be greater than #6#. Thereby the domain of #p# is #{x in RR : x>6}#.