Locate any extrema and points of inflection for the graph of `y=x^2ln(x/4) ` :

The domain for the function is x>0.

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

`y'=2xln(x/4)+x^2((1/4)/(x/4)) `

`y'=2xln(x/4)+x ` This is continuous and differentiable for all x in the domain so we set it equal to zero:

`2xln(x/4)+x=0 ==> ln(x/4)=-1/2 `

` x/4=e^(-1/2) ==> x=4e^(-1/2)~~2.43 `

For 0<x<4e^(-1/2) the first derivative is negative, greater it is positive so **there is a minimum at `x=4e^(-1/2) ` which is the only extrema.**

Any inflection points can only occur if the second derivative is zero:

`y''=2ln(x/4)+(2x)(1/4)/(x/4)+1 `

`y''=2ln(x/4)+3 `

`2ln(x/4)+3=0 ==> ln(x/4)=-3/2 ==> x=4e^(-3/2)~~.89 ` **so there is an inflection point at** `x=4e^(-3/2) ` as the concavity changes from concave down to concave up.

The graph:

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