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Math Help - help me solve this y=lnx/x x>0

  1. #1
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    help me solve this y=lnx/x x>0

    Given that y=lnx/x for x>0, find the set of values of x for which y is an increasing function of x
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  2. #2
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    Quote Originally Posted by Joelyks View Post
    Given that y=lnx/x for x>0, find the set of values of x for which y is an increasing function of x
    y=\frac{\ln(x)}{x}

    y'=\frac{x\frac{1}{x}-\ln(x)}{x^2}=\frac{1-\ln(x)}{x^2}

    A function is increasing when f'(x) > 0

    Just solve from here.
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  3. #3
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    Using Calculus, you don't even have to think to solve this problem. Simply take the derivative and find the critical point to get the extrema, which happens to be 1 and e (over the given interval).

    This can be solved without calculus by simply thinking about the function. The function is Lnx/x, so the denominator is dominant. Compare this to the function x/x which is always 1.

    We can know that the starting value (minimum) is going to be 1, since the Ln1 = 0. Less than 1 will get you negative numbers.

    The maximum will occur when the numerator becomes greater than 1, when Lnx > 1. This is because once you get over one in the numerator, the ratio will switch over and your divisions will get smaller and smaller. So what value for x will get you Lnx = 1? Lne of course.

    So the set is from 1 to e the function is increasing, and then greater than e the function is decreasing. There are no positive values for the function when x < 1.

    Patrick

    EDIT: I originally misread the question. I thought the question wanted where the function values are > 0, rather than x > 0. The answer can be changed because the function is still increasing when x < 1. The function has a vertical asymptote at zero, since that would be undefined. So the answer would be (0, e].
    Last edited by PatrickFoster; October 29th 2009 at 07:20 PM. Reason: Correction
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  4. #4
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    oh thanks
    So is the solution : lnx<1
    x<e^1
    x<2.72 ( 3s.f) ?
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  5. #5
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    Yep, just don't forget the lower bound of zero (exclusive).
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