Question

y=x^1/3 on what interval is this function continuous? the answer is (-) infinity to (+) infinity. however how can you take a cubed root of 0?

Answer 1

no cube for 0

Answer 2

You can, since:
\[
0^3=0\cdot0\cdot0=0
\]So, it is continuous for:
\[
x\in\mathbb{R}
\]

Answer 3

thats what i was assuming but with a power of 1/3 that insinuates cubed root. and you cant take root of 0. does that rule not apply if it is written as 1/3

Answer 4

Nope, sorry, here's the definition of a cube root:
\[
\sqrt[3]{a}=b\\
a\in\mathbb{R}
\]Is the solution to the equation:
\[
b^3=a\\
b\in\mathbb{R}
\](That's the typical definition, you could also show using roots of unity that the other two solutions are the case, but these are complex)

Answer 5

So, 0 is the solution to the equation:
\[
\sqrt[3]{0}=a\implies
a^3=0
\]But we know:
\[
0^3=0
\]And
\[
0\in\mathbb{R}
\]So:
\[
a=0=\sqrt[3]{0}
\]