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Does the domain of the function $y=\sqrt[3]{x^3+1}$ include $x<-1$? If yes, why is Mathematica and Wolfram Alpha not plotting that part of the function?

Sekots Reivan
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  • Yes. And what do you mean Wolfram Alpha is not plotting the function for $x < -1$? Here: http://www.wolframalpha.com/input/?i=y+%3D+cube+root+of+%28x%5E3+%2B+1%29&lk=4&num=1 – 0XLR Jul 21 '15 at 04:15
  • Oh, I put (x^3+1)^(1/3). http://www.wolframalpha.com/input/?i=y+%3D++%28x%5E3+%2B+1%29%5E%281%2F3%29 – Sekots Reivan Jul 21 '15 at 04:17
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    Click "Use the real-valued root instead". By default, Mathematica chooses [the convention that is compatible with complex arguments](https://en.wikipedia.org/wiki/Cube_root#Complex_numbers). –  Jul 21 '15 at 04:20
  • possible duplicate of [cubic root of negative numbers](http://math.stackexchange.com/questions/25528/cubic-root-of-negative-numbers) See also http://math.stackexchange.com/questions/273149/plot-of-x1-3-has-range-of-0-inf-in-mathematica-and-r and on the Mathematica site, http://mathematica.stackexchange.com/questions/3886/finding-real-roots-of-negative-numbers-for-example-sqrt3-8 – Jonas Meyer Jul 21 '15 at 04:35

1 Answers1

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The function:

$y=(x^3+1)^{\frac{1}{3}}$

Should include a domain of all real numbers because negative numbers also can have a cube root. So, yes, it should include $x \lt -1$. I'm not sure why those websites are acting up.

iadvd
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user76267
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