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

1 Answer
Jul 21, 2017

Answer:

#f_min =f(1) =0#
#f_max =f(e^(-2)) approx 0.541#

Explanation:

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

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

#= x(lnx)^2#

Applying the product rule

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

#= (lnx)^2 + 2lnx#

For local maxima or minima: #f'(x) = 0#

Let #z= lnx#

#:. z^2 +2z = 0#

#z(z+2) =0 -> z=0 or z=-2#

Hence for local maximum or minimum:

#lnx = 0 or lnx = -2#

#:.x= 1 or x=e^-2 approx 0.135#

Now examine the graph of #x(lnx)^2# below.

graph{x(lnx)^2 [-2.566, 5.23, -1.028, 2.87]}

We can observe that simplified #f(x)# has a local minimum at #x=1# and a local maximum at #x in (0, 0.25)#

Hence: #f_min =f(1) =0# and #f_max =f(e^(-2)) approx 0.541#