Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)
-   -   Curve Sketching & Optimization (https://www.askmehelpdesk.com/showthread.php?t=467378)

  • Apr 27, 2010, 04:32 AM
    saywhat234
    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
  • Apr 27, 2010, 10:10 AM
    ebaines

    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.
  • Apr 27, 2010, 05:20 PM
    saywhat235
    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
  • Apr 27, 2010, 05:20 PM
    saywhat235
    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
  • Apr 28, 2010, 03:03 AM
    saywhat234

    Please can you help me with this as soon as possible? I'm trying so hard to find the correct answers... thank you
  • Apr 28, 2010, 06:07 AM
    ebaines

    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:

  • Apr 28, 2010, 06:18 AM
    saywhat234

    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
  • Apr 28, 2010, 06:48 AM
    ebaines
    1 Attachment(s)

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

  • All times are GMT -7. The time now is 05:32 AM.