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

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

2 Answers 2


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 $\sqrt[3]{-x} = \sqrt[3]{(-1)x} = \sqrt[3]{-1}\sqrt[3]{x} = -\sqrt[3]{x}$

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    $\begingroup$ That makes all the right sense to me. Thank you. $\endgroup$
    – tsz
    Aug 27, 2014 at 18:04
  • $\begingroup$ Glad to help! Even though mjh provided nearly an identical answer 6 seconds before I did $\endgroup$
    – graydad
    Aug 27, 2014 at 19:04

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

$$\ (\!\underbrace{f^{-1} g = g f^{-1}}_{\underbrace{f^{-1}(-x)\ =\ -f^{-1}(x)}}_{\large f^{-1}\ odd}\!\!)^{-1} \Rightarrow\! \underbrace{gf = f g}_{\underbrace{-f(x)\ =\, f(-x)}_{\Large f\ odd}}\qquad $$


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