What is #(lnx)^2#?

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Answer 1 · 2017-06-30T10:39:30

See below.

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

#lnx# is defined for #x>0# hence, #f(x)# is defined #x>0#

#lim_(x-> 0) f(x) = +oo# and #lim_(x->oo) f(x) =+oo#

#f'(x) = 2lnx*(1/x)# {Chain rule]

For a turning point #f'(x) = 0#

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

Since #1/x !=0-> 2lnx=0#

#2lnx=0 -> x=1#

Hence #f(x)# has a turning point at #x=1#

#f(1) = (ln1)^2 = 0#

Since #f(x) >= 0 -> f(1) = f_"min"#

#:. f_"min" = 0# at #x=1#

We can see these results on the graph of #f(x) # below.

graph{(lnx)^2 [-10, 10, -5, 5]}