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

[anemone]

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!

 

[anemone]

No one answered last week's problem, here is the solution proposed by lfdahl:

Spoiler

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.:)