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Re: y = x^(1/x)
Posted:
May 27, 2001 9:50 AM
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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:
yÃÂô = 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
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