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Maximum And Minimum of a function

  1. Jul 2, 2011 #1
    Function:x+x2/3
    I got 0 and -(8/27) as the criticical numbers.

    Am I right?

    I also want to know if this function as a max and a min.

    Note:2/3 is an exponent.
     
  2. jcsd
  3. Jul 2, 2011 #2

    micromass

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    Hi realism877! :smile:

    Use the x2 and the x2 buttons for displaying exponents.

    Your critical points are correct. So can you tell me when the function is increasing and decreasing?? This will give you information about possible maxima and minima.
     
  4. Jul 2, 2011 #3

    SammyS

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    Hello realism877.

    (Use the ' X2 ' button above the "Advanced Message" box for exponents (superscripts)).

    How did you get the critical numbers?

    What did you get for a first derivative?

    x+x2/3 has relative extrema.
     
  5. Jul 2, 2011 #4
    The function is increasing (-infinity, -8/27)u(0,+infinity)

    Decreasing (-8/27)

    Absolute max= (-8/27, .1481)

    Absolute min= (0,0)

    Are my findings correct?
     
  6. Jul 2, 2011 #5

    micromass

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    You mean that it's decreasing in (-8/27,0), right?

    These certainly are relative minima and maxima, but that doesn't make them absolute. the function increases after 0, so it might get very big. The function increases before -8/27, so it might get really small.

    The only way to know is by calculating the limits in [itex]\pm \infty[/itex] and see to where the function increases...
     
  7. Jul 2, 2011 #6
    Yes, (-8/27, 0) decreasing

    Are they local maxima and minima? The values I posted.
     
  8. Jul 2, 2011 #7

    micromass

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    The function increases before -8/27 and decreases after, so it's a local maximum. The same with local minimum.

    But you'll need to do some more work to determine whether they are global maxima/minima.
     
  9. Jul 2, 2011 #8
    There are none.

    The function is (-infinity, +infinity)

    Am I on to something?
     
  10. Jul 2, 2011 #9

    micromass

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    Yes, you are entirely correct! You might want to prove that the range is [itex](-\infty,+\infty)[/itex]...
     
  11. Jul 2, 2011 #10

    SammyS

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    Since the critical point and thus the rel. min. at x=0 comes from the non-existence of the derivative at x=0, you might want to show (or state) that the function is continuous at x = 0 . --- just for completeness.
     
  12. Jul 2, 2011 #11
    Thanks

    I did the second derivative check, and I tried to get the critical numbers. it resulted to 1=0.

    From my calculator it looks like there is concavity down. But how can I determine concavity algebraically?
     
  13. Jul 2, 2011 #12
    How do I show?
     
  14. Jul 2, 2011 #13

    SammyS

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    To be continuous at x=0:
    limx→0+ f(x) = limx→0- f(x) = f(0)​

    For concavity:
    Isn't the second derivative negative everywhere, except at x = 0, where it does not exist ?​
     
  15. Jul 2, 2011 #14
    No,,when I tried to get the 0s from the second derivative, I got this 0=1

    How do I know where thereis convavity, if I don't have 0s to do a number line to check where is positive or negative?
     
  16. Jul 2, 2011 #15
    I don't know why you got 0=1, you shouldn't have. To determine concavity, use the second derivative and choose a point in each interval between critical points, including a point further left than your smallest critical point and a point further right than your largest critical point. Positive values mean concave up, negative concave down. If you're having trouble with the second derivative, try finding it again.

    If you can't get it, what do you have as the second derivative?

    I just have to say I think this is a squirrely function...bends and corners... icky.
     
  17. Jul 2, 2011 #16
    I'm on a mobile device, and I'm not near the paper where I did the work.
    I do remember not being able to solve for x.

    Can someone verify for me that there are no 0s for the second derivative test?
     
  18. Jul 2, 2011 #17

    SammyS

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    There are no zeros for the second derivative, but there is a critical point.

    Look at the second derivative (when you get back to your working paper). It should be obvious that it is negative wherever it's defined.
     
  19. Jul 3, 2011 #18
    I don't understand. How can I get a criticial point if I can't get a 0?
     
  20. Jul 3, 2011 #19
    The second derivative test actually favors non-zero values. You're looking for positivity or negativity. When the second derivative = zero, it's a possible inflection point. Everywhere else, it's indicative of concavity. In this situation, it's all the same concavity everywhere.
     
  21. Jul 3, 2011 #20
    Where do I plug in values in?
     
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