What are the maximum and minimum values of the function y = x^(1/x)?

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SUMMARY

The discussion focuses on finding the absolute maximum and minimum values of the function \( y = x^{\frac{1}{x}} \). The problem was proposed by user lfdahl as part of the Secondary School/High School Problem of the Week (POTW). The community is encouraged to engage with this mathematical challenge, although no solutions were provided in the thread. The problem emphasizes the importance of understanding the behavior of the function across its domain.

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anemone
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This week's problem was submitted by lfdahl and we truly appreciate his taking the time to propose a quality problem for us to use as our Secondary School/High School POTW.:)

Find the absolute maximum and the absolute minimum value(s) of the function of $y=x^{\dfrac{1}{x}}_{\phantom{i}}$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered last week's problem, here is the solution proposed by lfdahl:

Let $y = x^{1/x}$. Then $lny = \frac{lnx}{x}$. Differentiating this with respect to $x$ we obtain

\[\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}lny=\frac{\mathrm{d} }{\mathrm{d} x}\frac{lnx}{x}=\frac{\frac{1}{x}\cdot x-lnx\cdot 1}{x^2}=\frac{1-lnx}{x^2} \\\\ \Rightarrow \frac{\mathrm{dy} }{\mathrm{dx} } = y\frac{1-lnx}{x^2}=x^{1/x}\frac{1-lnx}{x^2}\]

Note that $\dfrac{dy}{dx}=0$ iff $1-\ln x=0\,\,\rightarrow x=e$.

Since $1-lnx > 0$ for $0<x<e$ and $1-lnx < 0$ for $x>e$, the absolute maximum value of $x^{1/x}$ occurs at
$x = e$ and is $e^{1/e}$. $x^{1/x}$ has no absolute minimum value.

Again, we'd like to express our thanks to lfdahl for his suggested problem.:)
 

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