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Math Help - Absoulute max and min

  1. #1
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    Absoulute max and min

    the Absoulute max and min f(x)=x^4 -2x^2 +3 on [-2,3] is at x=?

    I do not know how to do this.

    here are my thoughts.

    1) take derivative of f(x) and set equal to 0

    f'(x)= 4x^3 -4x = 0

    4x(4x^2-1) =0

    x=0

    x= +1/2
    x= -1/2

    f(0) = 3
    f(1/2) = 2.54 ???

    I am confused
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Absoulute max and min

    Quote Originally Posted by NeoSonata View Post
    f'(x)= 4x^3 -4x = 0
    4x(4x^2-1) =0
    4x(x^2-1)=0
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  3. #3
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    Re: Absoulute max and min

    so
    f(0)=3
    f(-1)=2
    f(1)=2

    What am I suppose to find next?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Absoulute max and min

    Quote Originally Posted by NeoSonata View Post
    so f(0)=3 f(-1)=2 f(1)=2
    What am I suppose to find next?
    Compare with f(-2) and f(3) .
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  5. #5
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    Re: Absoulute max and min

    I am sorry but I do not understand. How am I suppose to compare them?
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Absoulute max and min

    The values of the function at the critical points are f(0)=3,f(1)=f(-1)=2 (you needn't verify if these points are local maximum or minimum). The values of f at the endpoints of the closed interval are f(-2)=11,f(3)=66 . Comparing those values we obtain the absolute maximum and minimum of f that is, f_{\max}(3)=66 and f_{\min}(1)=f_{\min}(-1)=2 .
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