What are the local extrema of #f(x)=x^2/lnx#?

1 Answer
Nov 3, 2016

Answer:

#f(x)# has local extrema at #x=0# and #x=sqrt(e)#

Explanation:

#f(x) = x^2/lnx#

#f'(x) = (lnx*2x - x^2*1/x)/(lnx)^2#

#f(x)# will have local extrema where #f'(x)=0#

#(lnx*2x - x^2*1/x)/(lnx)^2 = 0#

#2xlnx-x=0#

#x(2lnx-1)=0#

#x=0# or #lnx=1/2 -> x=e^(1/2) = sqrt(e)#

Hence #f(x)# has local extrema at #x=0, (0,0)# and #x=sqrt(e), (sqrt(e), 2e)#

This can be seen from the graph of #f(x)# below:

graph{x^2/lnx [-11.91, 33.68, -8.58, 14.24]}