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27 February, 23:42

Find the absolute minimum and absolute maximum values of f on the given interval. f (x) = x - ln 8x, [1/2, 2]

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  1. 28 February, 01:35
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    The given function is

    f (x) = x - ln (8x), on the interval [1/2, 2].

    The derivative of f is

    f' (x) = 1 - 1/x

    The second derivative is

    f'' (x) = 1/x²

    A local maximum or minimum occurs when f' (x) = 0.

    That is,

    1 - 1/x = 0 = > 1/x = 1 = > x = 1.

    When x = 1, f'' = 1 (positive).

    Therefore f (x) is minimum when x=1.

    The minimum value is

    f (1) = 1 - ln (8) = - 1.079

    The maximum value of f occurs either at x = 1/2 or at x = 2.

    f (1/2) = 1/2 - ln (4) = - 0.886

    f (2) = 2 - ln (16) = - 0.773

    The maximum value of f is

    f (2) = 2 - ln (16) = - 0.773

    A graph of f (x) confirms the results.

    Answer:

    Minimum value = 1 - ln (8)

    Maximum value = 2 - ln (16)
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