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

Question

Aurora Alvarado · Accepted Answer

#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.

William Abdalla · Answer

undefined To find the critical numbers of $ f(x) = x^{-2}\ln(x) $, you need to first find its derivative, then set it equal to zero and solve for $ x $. After finding the critical numbers, you can determine whether they correspond to maximum or minimum points by using the first or second derivative test.

The derivative of $ f(x) = x^{-2}\ln(x) $ can be found using the product rule and the chain rule. The derivative is $ f'(x) = -2x^{-3}\ln(x) + x^{-3} $.

Setting $ f'(x) $ equal to zero and solving for $ x $, we have $ -2x^{-3}\ln(x) + x^{-3} = 0 $. Multiplying through by $ x^3 $, we get $ -2\ln(x) + 1 = 0 $.

Solving $ -2\ln(x) + 1 = 0 $ for $ x $, we find $ x = e^{1/2} $ or approximately $ x = 1.64872 $.

To determine whether $ x = e^{1/2} $ corresponds to a maximum or minimum, you can use the second derivative test or analyze the behavior of the function around that point.