Saturday, March 12, 2016

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 01/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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