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Math Help - Global max and Minimum

  1. #1
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    Global max and Minimum

    I have a few of these I'm practicing out of my book. Can you explain this to me a little better then the book. Here is their practice question:

    f(x)=x^3+3x^2+1; -3<=x<=2


    I would like to be able to solve their practice problems before pop quiz on Monday.
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  2. #2
    Senior Member Twig's Avatar
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    Hi

    What you typically do in these cases are:

    You differentiate the function and find max and minima.
    But here you also need to test the endpoints of the interval.

    You know how to perform a second derivative test to test for minima or maxima?

    Some functions may not have a global maxima, if say a function is defined on  (-2,2) , that is, the endpoints are not included, and the function in strictly growing, there might now exist a global maxima. Because whenever you pick a number close to 2, say 1.999 , you can always choose insted 1.999999 , and the function attains a higher value.

    So keep a distinction between closed and open intervals.
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  3. #3
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    so for t(x)=x^3+3x^2+1 -3<=x<=2

    I found the critical points x=0 and x=-2

    I plugged -3,-2,0,2 into the original function and came up with

    absolute max (2,21)
    absolute min (-3,-53)
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by lisa1984wilson View Post
    so for t(x)=x^3+3x^2+1 -3<=x<=2

    I found the critical points x=0 and x=-2

    I plugged -3,-2,0,2 into the original function and came up with

    absolute max (2,21)
    absolute min (-3,-53)
    as I can see f(0)=f(-3) right
    so (0,1) and (-3,1) is local min

    absolute max (2,21) it is correct as I see
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