How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #F(x)= (x^2)(lnx)#?

1 Answer
Jan 3, 2017

By analysing #F'(x)# over the domain of #F(x): x>0#
#F(x)# has an absolute minimum at #x=1/sqrt(e)#
#F(x)# is decreasing for #0< x<1/sqrt(e)# and increasing for #x>1/sqrt(e)#

Explanation:

#F(x) = x^2lnx#

#F'(x) = x^2.!/x + lnx*2x# (Product Rule)

#= x+2x*lnx#

For local extrema #F'(x) =0#

Thus: #x+2xlnx = 0#

#x(1+2lnx)=0#

#x=0# or #lnx=-1/2#

Since #F(x)# is defined for #x>0# our local extrema will be at:

#lnx=-1/2 -> x=e^(-1/2) = 1/sqrt(e) ~= 0.60653#

Hence:
#F(x)# has an absolute minimum at #x=1/sqrt(e)#
#F(x)# is decreasing for #x<1/sqrt(e)# and increasing for #x>1/sqrt(e)#

This is shown by the graph of #F(x)# below:

graph{x^2*lnx [-0.542, 2.877, -0.449, 1.26]}