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Finding the local extrema of this function

  1. Jan 26, 2009 #1
    F(x)=x^m * (1-x)^n where m and n is greater or equal to 2?
    find the x coordinates of local extrema?

    would my answer be with respect to m and n? or do i plug in a number for these variables
     
  2. jcsd
  3. Jan 26, 2009 #2
    Well how do you find extrema in general?
     
  4. Jan 26, 2009 #3
    take the derivative of the function set equal to zero and solve for x?
    then if the function increases at a point less than the critical number and decreses after the critlcal point then its a max, and otherway if its min

    but this function has the variables m and n and i cant solve for ciritcal numbers
     
  5. Jan 26, 2009 #4
    You can still differentiate and group things, hopefully you can factor them into a product of a few things?
     
  6. Jan 27, 2009 #5

    Chalnoth

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    Right, so your answer will be in terms of m and n. Just figure out the derivative in terms of these two unknown numbers, and set it equal to zero.

    A little trick that should help:

    [tex]x^m = x x^{m-1}[/tex]
     
  7. Jan 27, 2009 #6
    ok i got up to here:

    m/(n+m) = x

    would that be the only possible x coordinate for a local extrema?
     
  8. Jan 27, 2009 #7
    Well let's let [tex] n = m = 2[/tex] you then have

    [tex] f(x) = x^2(1-x)^2 = x^4 - 2x^3 + x^2 \Rightarrow f'(x) = 4x^3 - 6x^2 + 2x [/tex]

    If we set that equal to 0 we get

    [tex] f'(x) = 2x(2x^2 - 3x + 1) = 0 \Rightarrow x = 0, \frac{1}{2}, 1 [/tex]

    This gives one of your responses

    [tex] \frac{m}{n+m} [/tex]

    Since if we are doing my case we would get

    [tex] m = n = 2 \Rightarrow \frac{2}{2+2} = \frac{1}{2} [/tex]

    Unless I'm misunderstanding you?
     
    Last edited: Jan 28, 2009
  9. Jan 27, 2009 #8
    so i should assume m=n=2? why is that
    also would the answer be 1/2 ?
     
  10. Jan 27, 2009 #9
    I was picking an example, if your answer is right it should work for ALL (m,n). You said the only restriction was that m, n has to be greater than 2 and so I picked the simplest example. Using basic calculus I got one set of answers, using YOUR answer I got a different one, that shouldn't happen. This is why it's important to show your work instead of just giving an answer you think is right.
     
  11. Jan 27, 2009 #10
    ok using your method i got 3+- root 1/ 4

    how did you get i(?)root 7?
     
  12. Jan 27, 2009 #11
    I made a mistake the roots are x = 0, 1, 1/2, so your original solution might be partially right
     
  13. Jan 28, 2009 #12

    Chalnoth

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    You definitely made a mistake here.

    First, you should have three different roots. So you appear to have taken the one interesting-looking root and just ignored the other two. Make sure to get all of them.

    Second, this root you have here looks very similar to one of the three roots, but it isn't quite right. I'm not quite sure where your error is, but it's probably a very, very simple one.
     
  14. Jan 28, 2009 #13
    m/(m+n) is local extrema. 0 and 1 are points of inflection I believe. Unless you define your function on a closed subset of R... you need to look at limit behavior too.
     
  15. Jan 28, 2009 #14

    Chalnoth

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    m/(m+n) isn't quite correct, though. And I'm pretty sure that whether or not any of the roots is a point of inflection depends upon m and n, though I could be mistaken there.

    Edit: Yes, I went back and checked, and whether or not m or n are points of inflection depends upon m and n. I also verified that m/(m+n) is not an extremum (but it's close).
     
    Last edited: Jan 28, 2009
  16. Jan 28, 2009 #15
    Are you sure, maple gives it as one, I'm curious why you don't think it is.
     
  17. Jan 28, 2009 #16

    Chalnoth

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    Ah, my bad. I wrote down the problem wrong (transposed m and n). Sorry about any confusion.
     
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