$f(x)=x^\frac{5}{3}-5x^\frac{2}{3}$
is the same as :
$f(x)=(\sqrt[3]x)^5-(\sqrt[3]{5x})^2$
Except, with the first equation, my calculator returns an error for negative values of $x$ (We are assuming $x\in\mathbb{R}$ here)
The second equation seems to have no problems with negative values of $x$.
Various graphing technologies have failed to provide a conclusive result.
Does it look like this?
or this?
or this?
It's driving me mad. I'm sure there's a simple explanation, or something obvious that I'm missing here.
EDIT: I think I made a mistake. $(\sqrt[3]{5x})^2$ should be $5(\sqrt[3]{x})^2$. That seems to clear up some of the issues.
Best Answer
The problem is that $a^{\frac 1b}$ is well defined for $a \lt 0$ in $\mathbb R$ if $b$ is a positive integer but not if $b$ is something else. So if you do the divide in $a^{\frac 53}$ first (because of the implied parentheses) and do it as a float instead of a rational, you have a real exponent and can't handle negative bases.