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Math Help - Detecting Convergence, Esimating Limits

  1. #1
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    Detecting Convergence, Esimating Limits

    Let  f(x) = \frac {e^x}{1 + e^x}

    a. Show that f(x) > \frac {1}{2} for all x >= 0
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by larson View Post
    Let  f(x) = \frac {e^x}{1 + e^x}

    a. Show that f(x) > \frac {1}{2} for all x >= 0
    \frac {e^x}{1 + e^x} > \frac 12

    cross multiply

    2e^x > 1 + e^x

    combine like terms

    e^x > 1

    take natural log

    x > 0

    Note that this is not what you posted, you said when x = 0, f(x) > 1/2. But we have that this is not true. It can be verified by plugging in 0 and seeing that your initial criteria are not correct:

    f(0) = \frac {e^0}{1 + e^0}

    f(0) = \frac {1}{1 + 1}

    f(0) = \frac 12

    and since 1/2 is not greater than 1/2, your initial criteria should have been either
    a. Show that f(x) \geq \frac {1}{2} for all x \geq 0

    or

    a. Show that f(x) > \frac {1}{2} for all x > 0
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  3. #3
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    Quote Originally Posted by larson View Post
    Let  f(x) = \frac {e^x}{1 + e^x}

    a. Show that f(x) > \frac {1}{2} for all x >= 0
    The result follows from:

    1. f'(x) = \frac{e^x}{(1 + e^x)^2} > 0 for all x therefore f(x) is strictly increasing.

    2. f(0) = 1/2.
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