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Thread: Absolute Max and Min on an Interval? =[

  1. #1
    Jan 2010

    Absolute Max and Min on an Interval? =[

    Find the absolute maximum and absolute minimum of f on the interval (0,3] when f(x)=(x^3-4x^2+7x)/x
    A. max:none, min: (3,4)
    B. max(0,7), min3,4)
    C. max: none, min: (2,3)
    D. max (0,7), min: (2,3)
    E. none of the above

    I found the derivative, 2x-4
    the critical number, 2
    and plugged in these values into f, to see which gives the highest and lowest value.
    we dont plug in 0 because its (0,3] and not [0,3].
    f(2) = 3
    f(3) = 4
    However, the correct answer is C max:none, and min2,3).
    I dont get why there is no maximum, when clearly, when (3,4) and also (1,4) are values in the interval
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009
    the limit as x goes to 0 is 7 which is bigger than 4. Since 0 is not in the interval there is no max . Note there are an inifinity of values of x near 0 which are greater than 4.
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