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Given that $x^{\frac{2}{3}}=({x^2})^{\frac{1}{3}}$, I thought that $f(x)$ is continuous in $\mathbb{R}$ but when I plot this function, I get something like this in wolfram.

enter image description here

What happens to $f(-5)$? is not $f(-5)=25^{\frac{1}{3}}$. I find this plot odd. There is nothing for negative $x$ values. I thought the graph should be symmetrical about the $y$ axis.

So does it mean $f$ is not continuous on all of $\mathbb{R}$?

Abby Shearim
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  • You seem to be asking whether $x^{2/3}$ is defined on all of $\Bbb R$, rather than whether it's continuous. You're certainly right that it is defined (and continuous) for all real numbers. Moral: computers aren't perfect. – Greg Martin Mar 18 '20 at 06:26
  • $(x^{2})^{1/3}$ and $(x^{1/3})^{2}$ are both continuous functions on $\mathbb R$. – Kavi Rama Murthy Mar 18 '20 at 06:26
  • For me, it seems to be a bug of the Wolfram software. With the plotting tool from desmos.com I get a plot, like you would expect. – thinkingeye Mar 18 '20 at 06:27
  • Please click on the button on the right hand top corner that says "Real-valued plots" and select complex plot. You should see it continuous – Srini Mar 18 '20 at 06:34
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    Wolfram alpha has the principal root and real valued root installed. Enter ''plot x^(2/3) from -10 to 10'' to wolfram alpha and change to real valued root, and you see the plot defined on $\mathbb{R}\cap [-10,10]$ – Fakemistake Mar 18 '20 at 06:36

2 Answers2

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Not sure if this is the issue but, allowing for complex answers, there are actually three cube roots of every real number (except $0$). For negative real numbers, the principal cube root (one with the smallest argument) is a complex number with non-zero imaginary part. Some computer and calculator implementations are programmed to return this principal root when a cube root function is called. Since it's not real, it cannot be plotted on the real axis.

Also, when you take the exponent of $\frac 23$, there may be differing implementations of how you execute that. There is no reason to expect the value to be squared first (giving non-negative values) before taking the cube root, which is why the above "issue" manifests.

Deepak
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I think it's maybe the fault of the plotting program (maybe the software understood something else), I've used Desmos Graphic Calculator and find that this function is perfectly continuous in $\Bbb R$.

Here is the image

P.S.: If you are trying to find the continuity of some function try plotting the function manually if you are unable to solve it the go for software.

Sujit Bhattacharyya
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