What is the domain and range of #y= sqrt(x^3)#?

1 Answer
Oct 13, 2015

Answer:

Domain and range: #[0,infty)#

Explanation:

Domain: we have a square root. A square root only accepts as input a non-negative number. So we have to ask ourselves: when is #x^3 \ge 0#? It's easy to observe that, if #x# is positive, then #x^3# is positive too; if #x=0# then of course #x^3=0#, and if #x# is negative, then #x^3# is negative, too. So, the domain (which, again, is the set of numbers such that #x^3# is positive or zero) is #[0,\infty)#.

Range: now we have to ask which values the function can assume. The square root of a number is, by definition, not negative. So, the range can't go below #0#? Is #0# included? This question is equivalent to: is there a value #x# such that #sqrt(x^3)=0#? This happens if and only if there is an #x# value such that #x^3=0#, and we've already seen that the value exists and is #x=0#. So, the range starts from #0#. How further does it go?

We can observe that, as #x# gets large, #x^3# get even larger, growing to infinity. Same goes for the square root: if a number gets larger and larger, so does its square root. So, #sqrt(x^3)# is a combination of quantities which grow boundless to infinity, and thus the range has no bounds.