# `y = x/ln(x)` Locate any relative extrema and points of inflection.

## Expert Answers Find the extrema and points of inflection for the graph of ` y=x/(lnx) ` :

Extrema can only occur at critical points, or where the first derivative is zero or fails to exist.

Note that the domain for the function is x>0, `x ne 1 ` .

`y'=(lnx-1)/(lnx)^2 ` This function is defined for all x in the domain so we set it equal to zero. Note that a fraction is zero if the numerator, but not the denominator, is equal to zero.

`lnx-1=0 ==> lnx=1 ==> x=e ` . For 1<x<e the first derivative is negative and for x>e it is positive, so the only extrema is a minimum at x=e.

Any inflection point can only occur if the second derivative is equal to zero.

`y''=((ln^2x)/x+(2lnx)/x)/(ln^4x) ` or

`y''=(2+lnx)/(xln^3x) ` Which is negative for x<1, and positive for x>1. The graph is concave down on 0<x<1 and concave up on x>1, but x=1 is undefined so there are no inflection points.

The graph:

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