y = x^(1/x) for (x > 0) where the tangent is horizontal

[jwpaine]

Find the coordinates of the point on the curve \(\displaystyle \L y = x^{1/x} (x > 0)\) where the tangent line is horizontal

I need to find the value of x first... here is my work thus far:

\(\displaystyle \L y = x^{1/x}\)

\(\displaystyle \L lny = \frac{1}{x}lnx\)

\(\displaystyle \L y' = [\frac{d}{dx}\frac{1}{x}lnx\,+\, \frac{1}{x}\frac{d}{dx}lnx]x^{1/x}\)

\(\displaystyle \L y' = [\frac{-1}{x^2}\frac{lnx}{1}\,+\, \frac{1}{x^2}]x^{1/x}\)

\(\displaystyle \L y' = [\frac{1\,\,-\,\,lnx}{x^2}]x^{1/x}\)

Now I have told myself that I need to find the zeroes of the numerator for (x > 0) in order to find the x value for the horizontal tangent (0 slope)

So I am only concerned with the numerator:


\(\displaystyle \L 0 = [1\,-\,lnx]x^{1/x}\)

Am I on the right path here? I'm not quite sure how to solve for x and y from this point on......

John

 

[pka]

\(\displaystyle \L\ln (e) = ?\)

 

[jwpaine]

\(\displaystyle \L\ln (e) = ?\)

1 - but I'm not sure how that helps me

John

 

[pka]

\(\displaystyle \L \left[ {1 - \ln (e)} \right] = 0\)

 

[jwpaine]

Ok, so for x = e, y'(e) = 0; do I now have to plug x = e back into the original equation (x)^(1/x) to get the value for y?

NO - I would plug e into (1/x)ln(x) to get y = 1/e

Correcto?

 

[skeeter]

y = e^1/e

 

[pka]

Try the point \(\displaystyle \L \left( {e,\sqrt[e]{e}} \right).\)

 

[jwpaine]

Of course, substituting into original function - thanks!

John