I need help with a question. It says:

Graph and state all important points and intervals: y=(lnx)/(x)

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- Apr 6th 2008, 04:33 PMCenturionMonkeyNeed help with ln and e stuff
I need help with a question. It says:

Graph and state all important points and intervals: y=(lnx)/(x) - Apr 6th 2008, 05:05 PMTheEmptySet
Well lets start by taking the derivative

$\displaystyle \frac{dy}{dx}=\frac{x \cdot \frac{1}{x}-(1) \cdot \ln(x)}{x^2}=\frac{1-\ln(x)}{x^2}$

So our critical numbers are the zero's of the derivative

$\displaystyle 1 -\ln(x)=0 \iff 1 =\ln(x) \iff e=x$

The inflection points are given by the 2nd derivative

$\displaystyle \frac{d^2y}{dx^2}=\frac{x^2 \cdot\frac{-1}{x}-(2x)(1-\ln(x)}{x^4}=\frac{-3x+\ln(x)}{x^4}$

setting equal to zero we get

$\displaystyle -3x+\ln(x)=0$

This has no Real solutions see graphs below

So the graph will look like the next one

Good luck