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Topic: y = x^(1/x)
Replies: 14   Last Post: Jun 12, 2001 1:42 AM

 Messages: [ Previous | Next ]
 Marko Marin Posts: 48 Registered: 12/8/04
Re: y = x^(1/x)
Posted: May 27, 2001 9:50 AM

James wrote:

> Whilst messing about on my graphical calculator, I found that the graph of
> the xth root of x:
>
> y = x^(1/x)
>
> peaks when x = e.
> Just out of interest, why is this? Can anyone offer proof? I know that
> this is probably baby stuff for most of you... sorry if I waste your time :)
>
> Thanks,
> James

Write your function in exponential-logarithmic form:

y = exp(1/x*ln(x)), there ln(x) is natural logarithm and is defined for x > 0.

then differentiate using chain rule:

yÃÂÃÂ´ = exp(1/x*ln(x)) * (1/^x^2 - 1/x^2*ln(x))

the extremal points (maxima, minima or terrace points) occur where is the
derivate zero,

y' = 0, now because exponential function is always > 0 ==>

1/x^2 * (1 - ln(x)) = 0, for x = e

now is this the maxima?

for 0 < x < e the derivate is positive because 1 - ln(x) > 0 in that interval,
that means that your function is monotonic increasing in that interval.

for e < x < +oo the derivate is negative because 1 - ln(x) < 0 in that interval,
that means that your functions is monotonic decreasing in that interval.

now observe that lim x -> +oo y' = 0, so the points x = e is global maximum

so function x^(1/x) attains it's higest value at point x = e.

This has interesting consequences, namely we can prove that the most "economic"
number base is e.
But that another story...

Cheers, Marko

Date Subject Author
5/27/01 abc
5/27/01 gyan doshi
5/28/01 Dave L. Renfro
5/28/01 David W. Cantrell
5/29/01 Dave L. Renfro
5/30/01 David W. Cantrell
5/30/01 Dave L. Renfro
5/27/01 Marko Marin
5/27/01 Norm Dresner
5/27/01 John Prussing
5/28/01 Paul V. S. Townsend, M.Sc.
5/28/01 David W. Cantrell
5/28/01 Oscar Lanzi III
5/29/01 Martin Cohen
6/12/01 Bill Taylor