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Why is the graph of $y=x^{2.9}$ intangible on the negative side?

Why is the graph intangible on the negative side

Blue
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    How do you define $x^{2.9}$? – user170231 Jul 09 '23 at 17:07
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Jul 09 '23 at 17:08
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    If x<0, x^(2.9)=x^(29/10) which is the 10th root of x^29, But x<0 implies x^29<0 so the tenth root of that is not a real number. – coffeemath Jul 09 '23 at 17:11
  • Why do you think $-5^{-2.9}$ is defined (and a real number)? – aschepler Jul 09 '23 at 17:42
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    Note $(-5)^{-2.9}$ is undefined (in the reals), but you irrelevantly computed $-(5^{-2.9}) = -0.00939$ – GEdgar Jul 09 '23 at 17:53

1 Answers1

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In $\mathbb{R}$, one can define $x^r$ with $r\notin\mathbb{Z}$ in two ways.

  1. Define, for all real number $r$ that is not an integer, $$x^r=e^{r\ln{x}}$$ where $e$ is the exponential base and $\ln$ the natural logarithm.

As the natural logarithm is not defined when $x\leq0$ this means that the function $x\mapsto x^r$ with $r\notin\mathbb{Z}$ is only defined on $\mathbb{R}^+_\star$ if $r<0$, or $\mathbb{R}^+$ if $r>0$ (because of limit of $\exp(t)$ when $t\rightarrow-\infty$).

  1. However there is a way to accept the case $x<0$ : if $r\in\mathbb{Q}$ with $r=\dfrac{1}{q}$ and $q$ odd, one can define $x^{1/q}$ as the unique real number $y$ (and possibly negative) where $y^q=x$. This can be extended to $r=\dfrac{p}{q}$ if $p\wedge q=1$ and $q$ is odd (this is how Geogebra works, actually).

If $r=2.9=\dfrac{29}{10}$, the denominator is even so $x$ can't be negative

hdci
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  • The logarithmic definition for power is not the only one in common use for rational non-integer values. Fractional powers of form $x^(p/q)$ with $q$ odd are defined for negative $x.$ – coffeemath Jul 10 '23 at 08:39
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    Agreed (that's the second part of my answer) – hdci Jul 10 '23 at 15:43
  • [sorry, didn't read that. got thrown by your opening "By definition..." which leads one to believe there is not a different way to define it.] – coffeemath Jul 10 '23 at 16:31
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    True, I update my answer to remove this misunderstanding – hdci Jul 10 '23 at 17:22