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

We are asked to locate any relative extrema or inflection points for the graph of `y=xlnx ` :


The domain of the function is x>0.


Extrema can only occur at critical points; that is when the first derivative is zero or fails to exist.


`y'=lnx+x*1/x ==> y'=lnx + 1 `


This function is continuous and differentiable for all x in the domain, so setting y'=0 we get:


`lnx+1=0 ==> lnx=-1 ==> x=1/e~~0.368 `


For 0<x<1/e...

We are asked to locate any relative extrema or inflection points for the graph of `y=xlnx ` :


The domain of the function is x>0.


Extrema can only occur at critical points; that is when the first derivative is zero or fails to exist.


`y'=lnx+x*1/x ==> y'=lnx + 1 `


This function is continuous and differentiable for all x in the domain, so setting y'=0 we get:


`lnx+1=0 ==> lnx=-1 ==> x=1/e~~0.368 `


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


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


`y''=1/x>0 forall x ` so there are no inflection points.


The graph:


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