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

Answer 1

There is a local minimum of #0# at #1#. (Which is also global.) and a local maximum of #4/e^2# at #e^2#.

For #f (x) =(lnx)^2/x#, note first that the domain of #f# is the positive real numbers, #(0,oo)#.

Then find

#f'(x) = ([2(lnx)(1/x)]*x - (lnx)^2[1])/x^2#
# = (lnx(2-lnx))/x^2#.
#f'# is undefined at #x=0# which is not in the domain of #f#, so it is not a critical number for #f#.
#f'(x)=0# where
#lnx=0# # # or # # #2-lnx=0#
#x=1# # # or # # #x=e^2#
Test the intervals #(0,1)#, #(1,e^2)#, and #(e^2,oo)#.
(For test numbers, I suggest #e^-1, e^1, e^3# -- recall #1=e^0# and #e^x# is increasing.)
We find that #f'# changes from negative to positive as we pass #1#, so #f(1)=0# is a local minimum,
and that #f'# changes from positive to negative as we pass #e^2#, so #f(e^2)=4/e^2# is a local maximum.
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Answer 2

To find the local extrema of ( f(x) = \frac{(\ln x)^2}{x} ), we first need to find the critical points by setting the derivative equal to zero and solving for ( x ).

The derivative of ( f(x) ) can be found using the quotient rule:

[ f'(x) = \frac{d}{dx} \left( \frac{(\ln x)^2}{x} \right) = \frac{(2\ln x \cdot \frac{1}{x}) - ((\ln x)^2 \cdot 1)}{x^2} ] [ = \frac{2\ln x - (\ln x)^2}{x^2} ]

To find critical points, set ( f'(x) = 0 ):

[ 2\ln x - (\ln x)^2 = 0 ]

This equation can be a bit complex to solve directly. Instead, we can simplify it by introducing a substitution. Let ( u = \ln x ), so ( u^2 = (\ln x)^2 ). Then we have:

[ 2u - u^2 = 0 ] [ u(2 - u) = 0 ]

This gives us two possible values for ( u ): ( u = 0 ) or ( u = 2 ). Now we need to solve for ( x ) using ( u = \ln x ):

If ( u = 0 ), then ( \ln x = 0 ), which implies ( x = e^0 = 1 ).

If ( u = 2 ), then ( \ln x = 2 ), which implies ( x = e^2 ).

Now, we need to check whether these points are local extrema. We can use the second derivative test to determine this.

The second derivative of ( f(x) ) can be found by differentiating ( f'(x) ):

[ f''(x) = \frac{d}{dx} \left( \frac{2\ln x - (\ln x)^2}{x^2} \right) ]

This requires the quotient rule and the chain rule. After the calculation, we find:

[ f''(x) = \frac{2(\ln x - 1)}{x^3} ]

Now, we evaluate ( f''(x) ) at our critical points:

At ( x = 1 ): ( f''(1) = 0 ) At ( x = e^2 ): ( f''(e^2) > 0 )

Since ( f''(e^2) > 0 ), and ( f''(1) = 0 ), we can conclude that ( x = 1 ) corresponds to a point of inflection, while ( x = e^2 ) corresponds to a local minimum.

Therefore, the local minimum of ( f(x) = \frac{(\ln x)^2}{x} ) is at ( x = e^2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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