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looking on wolphram alpha for the function $$f(x) = x^{1/x}$$ it says that the domain is for each $x>0$ but for example if I make $x=-3$ it becomes $$ (-3)^{1/(-3)} = \left(-\frac{1}{3}\right)^{1/3} = -0.69336.... $$

So where am I wrong? enter image description here

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    $\begingroup$ Part of it has to do with how you define a fractional exponent of a negative number. What do you suppose would be $(-0.5)^{-0.5}$? Would it be real? Would it be imaginary? Recall that there are $n$ different $n$'th roots of a number. If you want to talk about this as a function, which one of those $n$ roots would you take? The real one if it exists? The one with smallest argument? You're running into complex territory here. $\endgroup$ – JMoravitz Oct 23 at 22:42
  • $\begingroup$ yes assuming all available in R. If is not in R then its not in the domain $\endgroup$ – Andri Nic Oct 23 at 22:43
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    $\begingroup$ You completely missed the point of my questions. Read it again. $\endgroup$ – JMoravitz Oct 23 at 22:44
  • $\begingroup$ @JMoravitz yes, i agree that is not part of the domain when x is an even number, but when it is an odd number it's part of the domain right? $\endgroup$ – Andri Nic Oct 23 at 22:48
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    $\begingroup$ Not necessarily. There are those who prefer to say that $\sqrt[3]{-1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ for instance. It depends on how you define things... and you haven't yet defined things... $\endgroup$ – JMoravitz Oct 23 at 22:49
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As a real function, $x^y$ is defined as $$x^y\overset{\text{def}}{=}\mathrm e^{y\ln x}, \enspace\text{ so }\quad x^{\tfrac 1x}= \mathrm e^{\tfrac{\ln x}x},$$ which requires $x>0$ to be defined.

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$f(x) =x^{\frac{1}{x}} =e^{\frac{1}{x} \ln(x)} $

So :

The Domain is $ ] 0,+\infty [$

Because the domain of $ \ln(x) $ is $ ] 0,+\infty [$

And the domain of $\frac{1}{x}$ is $\mathbb{R^*} $

And the domain of $e $ is $\mathbb{R} $

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Its

$$(-3)^{1/(-3)}$$

not

$$-(3^{1/(-3)})$$

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    $\begingroup$ The argument can be made that $(-a)^{1/3} = (-1)^{1/3}(a)^{1/3}=-(a)^{1/3}$ if people ignore the ambiguity of principle roots and what not. It is conceivable that the manipulation done here was not only intentional, but possibly correct under certain sets of definitions. This is not the issue here, but rather, the issue is the difficulty of taking arbitrary noninteger powers of negative numbers which requires more care to define which this answer does nothing to address. $\endgroup$ – JMoravitz Oct 23 at 22:59
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Your mestake is : $(-3)^{-\frac{1}{3}} = (-\frac{1}{3})^{\frac{1}{3}}$ not $ - (\frac{1}{3})^{\frac{1}{3}} $

And

$(-\frac{1}{3})^{\frac{1}{3}}=0.34668063717532 + 0.600468477588 i$

So for $x<0 f(x) $no defined at $\mathbb{R} $

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    $\begingroup$ See my comment on QC_QOA's answer $\endgroup$ – JMoravitz Oct 23 at 23:03
  • $\begingroup$ that was a sintax error but the concept is same wolframalpha.com/input/?i=cuberoot%5B-1%2F3%5D, i corrected it $\endgroup$ – Andri Nic Oct 23 at 23:07
  • $\begingroup$ But $(-\frac{1}{3})^{\frac{1}{3}}=0.34668063717532 + 0.600468477588 i$ $\endgroup$ – Anas chaabi Oct 23 at 23:14
  • $\begingroup$ @JMoravitz I didn't understand your comment $\endgroup$ – Anas chaabi Oct 23 at 23:17

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