## Question

###### What are the extrema and saddle points of #f(x)=2x^2 lnx#?

What are the extrema and saddle points of #f(x)=2x^2 lnx#?

## Answers

The domain of definition of:

#f(x) = 2x^2lnx# is the interval

#x in (0,+oo)# .Evaluate the first and second derivatives of the function:

#(df)/dx = 4xlnx +2x^2/x = 2x(1+2lnx)#

#(d^2f)/dx^2 = 2(1+2lnx)+2x*2/x = 2+4lnx+4 = 6+lnx# The critical points are the solutions of:

#f'(x) = 0#

#2x(1+2lnx) = 0# and as

#x > 0# :

#1+2lnx =0#

#lnx = -1/2#

#x =1/sqrt(e)# In this point:

#f''(1/sqrte) = 6-1/2 = 11/2 > 0# so the critical point is a local minimum.

The saddle points are the solutions of:

#f''(x) = 0#

#6+lnx =0#

#lnx = -6#

#x= 1/e^6# and as

#f''(x)# is monotone increasing we can conclude that#f(x)# is concave down for#x < 1/e^6# and concave up for#x > 1/e^6# graph{2x^2lnx [-0.2943, 0.9557, -0.4625, 0.1625]}

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