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Math Help - Find the number of extremum

  1. #1
    Senior Member pankaj's Avatar
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    Find the number of extremum

    Consider the function f:R\rightarrow R

     <br />
f(x)=(x-1)(x-2)^2(x-3)^3(x-4)^4(x-5)^5......(x-n)^n<br />

    Find the number of points at which the function has an extremum(i.e points of local maxima/minima).
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Both Captain Black and Noncomalg are correct.

    I used the 2nd derivative test incorrectly.
    Last edited by TheEmptySet; October 30th 2009 at 08:50 AM.
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  3. #3
    Senior Member pankaj's Avatar
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    Is x=2 the only extremum
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by TheEmptySet View Post

    Now for any m >2 notice thatwhen x=m that f'(m)=f''(m)=0

    So m is a saddle point for every m>2 and x=2 is either a max or a min.
    Would your argument not mean that y=x^4 does not have a minima but only a saddle point at x=0?

    (you have misinterpreted the second derivative test for classification of extrema)

    CB
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  5. #5
    MHF Contributor

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    Quote Originally Posted by pankaj View Post
    Consider the function f:R\rightarrow R

     <br />
f(x)=(x-1)(x-2)^2(x-3)^3(x-4)^4(x-5)^5......(x-n)^n<br />

    Find the number of points at which the function has an extremum(i.e points of local maxima/minima).
    considering the fact that the graph of a polynomial at the roots with even (odd) multiplicity touches (crosses) the x-axis, and hoping that nothing weird happens in between, i believe that the

    number of required points is: n + \left \lfloor \frac{n}{2} \right \rfloor -1.
    Thanks from pankaj
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