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

Question

Abigail Allen · Accepted Answer

Domain and range: #[0,infty)# 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.

Isaiah Anthony · Answer

undefined Domain: $ x \geq 0 $  
Range: $ y \geq 0 $

Avery Alexander · Answer

undefined The domain of the function $ y = \sqrt{x^3} $ is all real numbers $ x $ such that $ x^3 $ is non-negative, because the square root of a negative number is not defined in the real number system. Since the square root function only accepts non-negative inputs, the domain is restricted to $ x \geq 0 $.

The range of the function $ y = \sqrt{x^3} $ is all real numbers $ y $ such that $ y $ is non-negative, because the square root of a non-negative number is always non-negative. Therefore, the range is $ y \geq 0 $.

Eli Assouline · Answer

undefined The domain of $ y = \sqrt{x^3} $ is all real numbers $ x $ such that $ x^3 \geq 0 $, which means $ x $ can be any real number.

The range of $ y = \sqrt{x^3} $ is all real numbers $ y $ such that $ y \geq 0 $, since the square root of any real number is non-negative.