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Math Help - Absolute extrema

  1. #1
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    Absolute extrema

    For business calculus. I just want to check my answer ...

    3x^2-6x+5

    x is a member of [0,4] find the absolute extrema
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  2. #2
    Super Member wingless's Avatar
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    To find the global extrema, check the local extrema and the limit of the function approaching the bounds.

    Local extrema:
    y=3x^2-6x + 5

    y' = 6x - 6 = 0

    x = 1

    We have a local extremum at x = 1. What's f(x) of the point?
    y=3x^2-6x + 5
    y=2


    Now check the bounds.
    \lim_{x\to 0^{+}}3x^2-6x+5 = 5

    \lim_{x\to 4^{-}}3x^2-6x+5 = 29


    We have 3 possible values to be the global extrema.
    y=2 -- (the local extrema, x=1)
    y=5 -- (left bound)
    y=29 -- (right bound)

    The greatest of these values, y=29 is the global maximum, and the least, y=2 is the global minimum.
    Last edited by wingless; February 18th 2008 at 03:01 PM.
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  3. #3
    Super Member wingless's Avatar
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    The answer above is enough to solve. In addition, there are a few more things that may come in handy

    y= 3x^2-6x + 5

    Firstly, this is a quadratic (second degree) function.
    Quadratic functions look like,


    (position of the vertex and the functions closeness may vary)

    Quadratic functions are of the form y=a~x^2 + b~x + c
    The first coefficient a indicates the side of the function (upwards or downwards) and its narrowness. If a>0, the function is upwards. If a<0, the function is downwards.

    In this example, a=3, which is greater than zero. This makes the function upwards. So, the vertex is the local and global minimum, and the global maximum is at the bounds.
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