# Why is the cube-root of $x$ 'odd'?

I am trying to understand why $\sqrt[3]{x}$ is an odd function; can anyone explain how I could come to this conclusion?

• $f$ is odd iff $f(-x)=-f(x)$ for all $x$. $f(-x)=\sqrt[3]{-x}=\sqrt[3]{(-1)(x)}=\sqrt[3]{-1}\sqrt[3]{x}=-\sqrt[3]{x}=-f(x)$.
– mjh
Aug 27, 2014 at 18:02
• In particular $g(x)=x^3$ is odd, and, when an odd function has an inverse, it's inverse must be odd. Aug 27, 2014 at 18:14
• @thomas-andrews Ah, that makes sense. Thanks.
– tsz
Aug 27, 2014 at 18:15

A function $f$ is odd if for all $x$ in the domain of the function, we have $f(-x)=-f(x)$. In the context of your problem, note that $\sqrt[3]{-x} = \sqrt[3]{(-1)x} = \sqrt[3]{-1}\sqrt[3]{x} = -\sqrt[3]{x}$
An inverse of an odd function is odd: $\ f(-x)\, =\, f(-f^{-1}\!fx) \overset{f^{-1}\rm\ odd} = f f^{-1}(-fx)\, =\, -fx$
Remark $\$ Group-theoretically this may be viewed as a special case of the following observation: $\$ if $\ g^{-1} = g\$ then $\,g\,$ commutes with $f^{-1}\!$ $\iff$ $\,g$ commutes with $\,f.\,$ Indeed, with $\,g(x) = -x,\,$ and the group operation of function composition (denoted by multiplication) we have
$$\ (\!\underbrace{f^{-1} g = g f^{-1}}_{\underbrace{f^{-1}(-x)\ =\ -f^{-1}(x)}}_{\large f^{-1}\ odd}\!\!)^{-1} \Rightarrow\! \underbrace{gf = f g}_{\underbrace{-f(x)\ =\, f(-x)}_{\Large f\ odd}}\qquad$$