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    saywhat234's Avatar
    saywhat234 Posts: 4, Reputation: 1
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    #1

    Apr 27, 2010, 04:32 AM
    Curve Sketching & Optimization
    f(x)=ln²x/x
    Find local maxima & minima and then find all inflection points. Find asymptotes(vertical or horizontal) after stating the domain of the given function
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Apr 27, 2010, 10:10 AM

    I assume you know that the local minima and maxima can be determined by solving the equation f'(x) =0, and that the inflection points are at f''(x) = 0. I suggest you try to solve this and then post back with your results - we'll be happy to check it for you.
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    saywhat235 Posts: 5, Reputation: 1
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    #3

    Apr 27, 2010, 05:20 PM
    OK I got the local minimum at(1,0) and no local max.. the inflection pt at(2.61,0.35), the domain of the function is f(x)>0.. there is a vertical and horizontal assypmtote but I'm having trouble finding the limits.. can u please tell me if my answers are correct and if not tell me what the right answers are and the shape of the graph.. I have been working on this problem for an entire week and still didn't come up with the right answers.. thanks a lot
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    saywhat235 Posts: 5, Reputation: 1
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    #4

    Apr 27, 2010, 05:20 PM
    OK I got the local minimum at(1,0) and no local max.. the inflection pt at(2.61,0.35), the domain of the function is f(x)>0.. there is a vertical and horizontal assypmtote but I'm having trouble finding the limits.. can u please tell me if my answers are correct and if not tell me what the right answers are and the shape of the graph.. I have been working on this problem for an entire week and still didn't come up with the right answers.. thanks a lot
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    saywhat234 Posts: 4, Reputation: 1
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    #5

    Apr 28, 2010, 03:03 AM

    Please can you help me with this as soon as possible? I'm trying so hard to find the correct answers... thank you
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Apr 28, 2010, 06:07 AM

    In addition to the minimum that you found, there is also a maximum. From the dereivative:



    you can see that this is equal to zero at ln(x) = 0 and ln(x) = 2.

    The domain is x>0 (because ln(x) is limited to only positjve values of x), and the range is y>=0 (since ln^2(x) can't be negative). There are two asymtotes - one is pretty obviously the vertical at x = 0. To find the other, consider what happens in the limit as x goes to infinity. Use l'Hospital's rule:

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    saywhat234 Posts: 4, Reputation: 1
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    #7

    Apr 28, 2010, 06:18 AM

    Thanks a lot... I managed to figure out the whole problem after a lot of hard work.. I just want to make sure that my graph is correct.. how does the graph look like.. thanks again
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #8

    Apr 28, 2010, 06:48 AM

    Do you have a graphing calculator? Or you can do what I do - set up a spreadsheet.
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