What are the critical values of #f(x) = xlnx# ?

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Answer 1 · 2017-03-19T22:53:43

#f(x)# is defined for #x>0#
#f(x)# has a zero at #x=1#
#f(x)# has an absolute minimum of#-1/e# at #x=1/e#

#f(x) =xlnx#

Since #lnx# is defined for #x>0 -> f(x)# is defined for #x>0#

To find the zero: #f(x) = 0#

#xlnx = 0 -> lnx =0# Since #x>0#

#lnx = 0 -> x=1#

#f'(x) = x*1/x + lnx# (product rule and standard differential)

#f'(x) = 0 -> lnx = -1#

#x=e^-1#

#x=1/e ~= 0.3679#

#f''(x) = 1/x# which is #>0 forall x>0#

Hence: #f_min = f(e^-1) = -e^-1 ~= -0.3679#

These points can be seen on the graph of #f(x)# below:

graph{xlnx [-10, 10, -5, 5]}