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Math Help - min/max with a twist

  1. #1
    Junior Member winterwyrm's Avatar
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    min/max with a twist

    Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = ln(x2 + 3x + 5)
    [-2, 2]

    Thanks in advance
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  2. #2
    MHF Contributor
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    I don't know how to use any graphing calculator so I cannot see now the graph of f(x).

    Let us see if there are maximun or minimum between [-2,2].
    f(x) = ln(x^2 +3x +5)
    Differentiate both sides with respect to x,
    f'(x) = [1/(x^2 +3x +5)](2x +3)
    Set that to zero,
    0 = 2x +3
    x = -1.5 ------a critical point, maximum or minimum.

    Now let us just see the values of f(x) at the ends of the interval, at x = -1.5, and at x=0.

    f(-2) = ln[(-2)^2 +3(-2) +5] = ln(3)
    f(-1.5) = ln[(-1.5)^2 +3(-1.5) +5] = ln(2.75)
    f(0) = ln[(0)^2 +3(0) +5] = ln(5)
    f(2) = ln[(2)^2 +3(2) +5] = ln(15)

    So, in that interval, f(-1.5) is the lowest, while f(2) is the highest.
    Therefore, in that interval, the absolute maximum is ln(15), and the absolute minimum is ln(2.75). --------------answer.
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  3. #3
    Junior Member winterwyrm's Avatar
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    wow, I kept thinking the derivative was 2x+5 for who knows what reason, lol..... I kept checking everything and wondering why it just wouldn't work out
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