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

Oct 13, 2015

Domain and range: $\left[0 , \infty\right)$
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 ${x}^{3} \setminus \ge 0$? 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 $\left[0 , \setminus \infty\right)$.
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.