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$f(x)=x^\frac{5}{3}-5x^\frac{2}{3}$

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?

enter image description here

or this?

enter image description here

or this?

enter image description here

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.

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2 Answers 2

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.

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$f(x)$ can be rewritten into form :

$f(x)=\sqrt[3] {x^2}\cdot(x-5)$

Graph :

enter image description here

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