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

1 Answer
May 18, 2018

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}.