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

1 Answer
Jan 6, 2016

Answer:

There is a local minimum of #0# at #1#. (Which is also global.) and a local maximum of #4/e^2# at #e^2#.

Explanation:

For #f (x) =(lnx)^2/x#, note first that the domain of #f# is the positive real numbers, #(0,oo)#.

Then find

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

# = (lnx(2-lnx))/x^2#.

#f'# is undefined at #x=0# which is not in the domain of #f#, so it is not a critical number for #f#.

#f'(x)=0# where

#lnx=0# # # or # # #2-lnx=0#

#x=1# # # or # # #x=e^2#

Test the intervals #(0,1)#, #(1,e^2)#, and #(e^2,oo)#.

(For test numbers, I suggest #e^-1, e^1, e^3# -- recall #1=e^0# and #e^x# is increasing.)

We find that #f'# changes from negative to positive as we pass #1#, so #f(1)=0# is a local minimum,

and that #f'# changes from positive to negative as we pass #e^2#, so #f(e^2)=4/e^2# is a local maximum.