y = x^2 ln(x/4) Locate any relative extrema and points of inflection.

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 0Any 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: