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## Homework Statement

Find all extrema and inflection points of the function ## y = \frac {x} {ln(x)} ##

## Homework Equations

I did the first and second derivative by hand, and they worked out in CAS as well...

## y = \frac {x} {ln(x)} ##

## y' = \frac {lnx - 1} {(ln(x))^2} ##

## y'' = \frac {2 - ln(x)} {x(ln(x))^3} ##

## The Attempt at a Solution

EXTREMA:

## y' = \frac {lnx -1} {(ln(x))^2} = 0 ##

## ln(x) - 1 = 0 ##

## ln(x) = 1 ##

## x = e ##

...-....+...

<-------- e -------->

## y = \frac {e} {ln(e)} ##

## y = e ##

Therefore, a minima exists at ##(e,e)##

INFLECTION POINT:

## y'' = \frac {2 - ln(x)} {x(ln(x))^3} = 0 ##

## y'' = 2 - ln(x) = 0 ##

## ln(x) = 2 ##

## x = e^2 ##

...+....-...

<------ e^2 ------>

## y = \frac {e^2} {ln(e^2)} ##

## y = \frac {e^2} {2} ##

Therefore, an inflection point exists at ## (e^2,\frac {e^2} {2}) ##

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Looking at the graph of ## y = \frac {x} {lnx} ##, the extrema is correct.

Also, I noticed by solving for the denominators = 0, I find that there is a vertical asymptote at x = 1, which seems like it should be the inflection point... not ## (e^2,\frac {e^2} {2}) ##, as the graphing is just continually concave up at ##e^2##, not changing concavity.

I also know the domain is ##(0,1)##U##(1,\infty)## due to the denominator, and property of ##ln(x)##.

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EDIT:

I think I may have gotten it... I forgot to use the critical point for ## y'' = \frac {2 - ln(x)} {x(ln(x))^3}##, when the bottom is equal to zero, which you solve analytically.

##x(ln(x))^3) = 0,## when ## x = 1##

So this tells us there is an inflection point at 1, but what does this mean of the ##e^2## result, I don't see how that is invalid, yet it doesn't satisfy the equation.

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