Why does this graph only the positive side

Question

The following equation graphs a love heart,

$x^2+(y- \sqrt[3]{x^2})^2=1$

And since (please correct me if I'm wrong)

$\sqrt[3]{x^2}=x^{2/3}$

I should be able to write it like this:

$x^2+(y- x^{2/3})^2=1$

However when I graph the latter I only get the positive side of the heart, not the negative side. Can someone tell me why this is?

Hint: Do you think that $\sqrt[4]{x^2}=x^{2/4}=x^{1/2}=\sqrt x$ for all real numbers $x$? — Hagen von Eitzen, Jul 23 '14 at 08:41

goblin GONE · Accepted Answer · 2014-07-27T13:05:57.840

0

I think the graphing software is just wrong in this case.

Consider $x=-1.$

Then $x^2 = 1$, so $(x^{2})^{1/3} = 1$.
Also $x^{1/3} = -1$, so $(x^{1/3})^2 = 1.$

Hence $(x^{2})^{1/3} = (x^{1/3})^2$ for $x$ equal to $-1$.

This suggests to me that $\sqrt[3]{x^2}=x^{2/3}$ holds for all real $x$, not just the positive ones.

Edit. One possibility is that the moment you raise $x$ to the power of a non-integer, the graphing software automatically interprets $x^r$ as $\exp(r \log x),$ which only makes sense if $x$ is positive. According to this convention, $x^{2/3}$ is defined only for positive $x$, while $\sqrt[3]{x^2}$ is defined for all $x$.

edited Jul 27 '14 at 13:05

answered Jul 23 '14 at 08:53

goblin GONE

67,744

If this is the case I'm surprised things like this have not been patched, especially given I tried this on several graphers. – John Smith Jul 23 '14 at 08:59
@JohnSmith, it is possible that I'm wrong, of course. – goblin GONE Jul 23 '14 at 09:57
Yeah but it's the only explanation I have at the moment. – John Smith Jul 27 '14 at 12:45

Why does this graph only the positive side

1 Answers1