Locate extrema and inflection points for `y=(ln(x))/x ` :

The domain is x>0.

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

`y=1/x*ln(x) `

`y'=1/x*(1/x)+(-1/x^2)lnx `

`y'=1/x^2(1-lnx) `

Setting equal to zero we get:

`1/x^2=(lnx)/x^2 ==> lnx=1 ==> x=e `

For 0<x<e the first derivative is positive and for x>e it is negative, so there is **a maximum at x=e and this is the only extrema.**

Inflection points can only occur if the second derivative is zero:

`y''=-2/(x^3)(1-lnx)+1/x^2(-1/x) `

`y''=-1/x^3(2-2lnx+1)=-1/x^3(3-2lnx) `

So `3-2lnx=0==> lnx=3/2 ==> x=e^(3/2)~~4.482 `

**There is an inflection point at** ` x=e^(1.5)` where the graph changes from concave down to concave up.

The graph: