How do you find the critical numbers for #f(x) = x^(-2)ln(x)# to determine the maximum and minimum?

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Answer 1 · 2017-04-15T20:13:18

#0# is the minimum

#e^(1/2)# is the maximum

Take the derivative of #f(x)#. You will need to use the product rule. You also need to know that the derivative of #ln(x)# is #1/x#:

#f'(x)=x^-2(1/x) + ln(x)(-2x^-3)#

#f'(x)=x^-3-2ln(x)x^-3#

Factor out a #x^-3#

#f'(x)=x^-3(1-2ln(x))#

Solve for #x#:

#x=0,e^(1/2)#

Plug in these numbers into the initial equation:

#f(0)=0ln(0)=DNE#.

We'll need to use limits for this:

#lim_(xrarr0) x^-2ln(x)=-oo#. This is definitely a minimum

#f(e^(1/2))=(e^-1)(ln(e^(1/2)))=1/(2e)#. This is a maximum because plugging in anything before or after this will give a value less than this.