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

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Answer 1 · 2018-03-21T12:12:06

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]}