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Math Help - Maxima and Minima

  1. #1
    Junior Member
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    Maxima and Minima

    Q) At what integral value of x will the function

    (x^2 + 3x + 1)/(x^2 -3x+1) attains its maximum value?

    Any help would be greatly appreciated.

    Thanks,
    Ashish
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  2. #2
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    simplify the function-> 1+(6/(x-3+1/x))
    you will see that when x=1,it attains max and when x=-1,it attains min.
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  3. #3
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    How do you simplify the above expression to 1+(6/(x-3+1/x)) ?
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  4. #4
    MHF Contributor Amer's Avatar
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    the critical points (3+5^1/2)/2 and (3-5^1/2)/2

    since

    (x^2 +3x+1)/(x^2 -3x+1)=(x^2-3x+1 +6x)/(x^2-3x+1) = 1+6x/(x^2-3x+1)

    = 1+6(x-3+1/x)^-1= f(x)
    f`(x) = -1 (1 -1/x^2)(x-3+1/x)^-2 = 0

    \frac{-1+\frac{1}{x^2}}{(x-3+\frac{1}{x})^2}

    the root of the numerator {-1,1} and the root denominator

     \frac{3-+\sqrt{5}}{2}

    but the denominator is always positive
    look at this

    Maxima and Minima-photo2.jpg
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