Find the Domain of $$f(x)=x^{\frac{1}{x}}$$ I have Confusion on Negative Real numbers. if $x=-2$ then $f(x)$ is not Real, but if $x=-3$ ,$f(x)$ is Real. I am unable to figure out which set of negative real numbers come into the Domain.Books give the Domain as $\mathbb{R^+}$
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If you are thinking along the lines of $\sqrt[3]{-27}=-3$, that's not the same as $(-27)^{1/3}$.
$(-27\pm i\varepsilon)^{1/3}$ are both nowhere near a negative real (where $\varepsilon$ is a small positive real), so if $(-27)^{1/3}$ had meaning, a desire to be working with continuous functions would mean $(-27)^{1/3}$ is not a negative real.
These considerations should convince someone that it's not a good idea to say $\sqrt[3]{x}=x^{1/3}$ when $x$ is negative.
If you do insist that $\sqrt[3]{-27}=(-27)^{1/3}$, then why isn't it the same as $(-27)^{2/6}=\sqrt[6]{(-27)^2}$, which is a positive real? You run into trouble like this if you insist that $\sqrt[3]{x}=x^{1/3}$ when $x$ is negative.
2'5 9'2
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@alex.jordan I do not believe $e^{x\ln{1/x}}\neq{x}^{\frac{1}{x}}$ when $x<0$. Plus $x^{p/q}$ it is not equal to $\sqrt[q]{x^p}$. You must not do anything with $p/q$ when $x$ is negative. – Arbuja Dec 13 '15 at 01:08
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@Arbuja Yes, that's my opinion: you must have $x^{1/3}=x^{2/6}$, so you need that $x>0$. – egreg Dec 13 '15 at 10:56
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2@Arbuja When you apply fractional exponents, you need to assume the base is positive (or apply some contorted conditions). So $x\mapsto x^3$ is defined also for $x<0$; this does not mean that $x^3=x^{6/2}$ for all $x$. Unfortunately exponential notation is slightly ambiguous. – egreg Dec 13 '15 at 13:20
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@alex.jordan I apologize you did not directly answer the OP's post. The OP could misguided that there is no domain. – Arbuja Dec 13 '15 at 23:16
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1I don't see the appeal here, but perhaps I do not practice enough complex analysis. By the same logic, I apparently have to believe that $x\neq x^{1/1}$ because $x^{1/1}=x^{2/2}$ 'convinces me' that $x^{1/1}$ has no negative domain. I can understand wanting to keep the calculus of rational exponents crisp by restricting domains to be positive, though, which is apparently the intention here. – rschwieb Dec 14 '15 at 09:11
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I rolled back this answer to how I wrote it. Please, if you don't understand the answer, don't try to edit it. – 2'5 9'2 Dec 14 '15 at 22:39
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@rschwieb I'm not saying what you are saying about $x^{1/1}$ or $x^{2/2}$. Those are both just as you would expect them to be: $x^1$. What I am saying is the latter is only reliably equal to $\sqrt{x^2}$ in the case where $x$ is positive. I'm saying that no matter how much Arbuja religiously believes that $\sqrt[q]{x^p}$ and $x^{p/q}$ meant the same thing, they in fact do not. They have distinct definitions, which coincide when $x$ is positive. – 2'5 9'2 Dec 14 '15 at 23:41
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1@Arbuja I have failed to convince you that you have been trained incorrectly to believe that $x^{p/q}$ is the same thing as $\sqrt[p]{x^q}$ [even for reduced rational $p/q$.] So I will appeal to a CAS. Type "cube root of -8" into Wolfram Alpha. Then type "(-8)^(1/3)" into Wolfram Alpha. While I normally would never use WA to "prove" anything, hopefully this opens you to the possibility that you are using a nonstandard meaning for exponentiation. – 2'5 9'2 Dec 14 '15 at 23:47
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@alex.jordan Clicking on your links I got the cube root of $-8$ to be $-2$ but I got the result from $(-8)^{1/3}$ to be $2\sqrt[3]{-1}$ but the decimal form is a complex number. As I said before $e^{\ln{x}}\neq{x}$ for example taking $x=-1$ proves you should use identity for negative numbers. – Arbuja Dec 15 '15 at 00:07
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1@Arbuja Well, serves me right for trying to use WA to make a point. It's not the details of the output that matters; it's that the outputs are two different things. Anyway, you sincerely believe that $(-1)^{1/3}=-1$, and I don't. It's pointless to continue if you feel that $(-1)^{1/3}$ is $-1$, for example. We are using different definitions for $x^y$. I assert that mine is more standard, and is used by computers etc. Yours stems from extending the relation $x^{1/q}=\sqrt[q]{x}$ to apply to negative real $x$. The two meanings are incompatible for nonpositive $x$ and noninteger $y$. – 2'5 9'2 Dec 15 '15 at 00:52
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General observation: given a real valued function $f$, we can write $$ f(x)=\exp(\log(f(x))) $$ only when $f>0$.
Hence in your case $\Bbb R^{+}$ is clearly contained in the domain of the function $f(x)=x^{\frac1 x}$. It remains to show that no other point is admitted in the domain.
$0$ is clearly out.
Then if $x<0$ what you have is a power of a real negative number. But in the system of real numbers, the power is defined only for positive base. Hence no negative number could be taken in account.
Thus the domain of your function is given by all and only positive real numbers, $\Bbb R^{+}$.
Joe
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@Arthur Do you mean $-\sqrt[3]{1/3}$? Anyway, the question is very confusing, why is f(x) not being real a problem? Given the confusing state of the question, I don't dislike this answer. – user2345215 Nov 16 '14 at 01:52
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the domain of a function f(x) is not the domain of e^log(f(x)), although in this case the result is the same – Mosk Nov 16 '14 at 01:56
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2@Joe, we cant change the Definition of the function, because if we do so, the Domain changes. For example Domain of Definition of $$f(x)=\log_{10}\left(x^2+2x+1\right)$$ is $\mathbb{R}-{-1}$, But when we write $f(x)$ as $$f(x)=2\log_{10}(x+1)$$ The Domain has Changed to $(-1 : \infty)$ – Ekaveera Gouribhatla Nov 16 '14 at 03:04
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1He hasn't changed the definition, as the definition of $x^{1/x}$ is $e^{\frac 1 x \ln x}$ if x is irrational and positive. Also, $\log_{10}(x + 1)^2 = 2 \log_{10} |x + 1|$, so the domain is the same. – GFauxPas Nov 16 '14 at 17:41
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@Alex.Jordan posted a nice answer but there is a domain that exists for ${x}^{1/x}$ if ${1/x}$ is reduced. Thus I don't believe his answer is proper to the OP's question.
When exponent of $x^{p/q}$ ,which is $p/q$, is reduced completely and $q$ is odd, $x^{p/q}$ satisfies the identities of $\sqrt[q]{x^p}$ and $\left(\sqrt[q]{x}\right)^{p}$ for the negative domain. Thus for ${x}^{1/x}$, we must find the values of $1/x$ where the output has an odd denominator and is completely reduced.
Also one should also know that for $f(x)^{g(x)}$ that $f(x)^{g(x)}\neq{e^{\ln(f(x))g(x)}}$ when $x<0$. This because $\ln(x)$ for the real logarithm is undefined at negative numbers. So they're is no reason to use this identity for figuring out the negative domain. Infact we must analyze $x^{1/x}$ strictly for the real domain.
One example is for $e^{\ln{x}}=x$ for $x<0$. When $x=-1$ the two terms are not the same.
Using all of my previous points, one can find the true domain of $x^{\frac{1}{x}}$ is...
$(0,\infty)\bigcup{\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}}\bigcup\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}$
For $f(x)^{g(x)}$, when one tries to search for the output of $g(x)$ that has a fraction reduced with an odd denominator; one can find the domain of $f(x)^{g(x)}$ when $f(x)$ is negative.
Be sure to check:
What is the domain of $x^x$ when $ x<0$
http://math.stackexchange.com/questions/398224/what-is-the-domain-of-xx-when-x0
http://math.stackexchange.com/questions/394110/can-the-graph-of-xx-have-a-real-valued-plot-below-zero
http://math.stackexchange.com/questions/695701/how-can-we-describe-the-graph-of-xx-for-negative-values
– Arbuja Dec 15 '15 at 13:02