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Math Help - Let f(x)=(x^2+3)/(2x) for x not equal to 0. Show that if x>3^(1/2), then f(x)>3^(1/2)

  1. #1
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    Unhappy Let f(x)=(x^2+3)/(2x) for x not equal to 0. Show that if x>3^(1/2), then f(x)>3^(1/2)

    Hi, this exercise is giving headaches . I only need help with Part A because I know part B and have an idea of how to solve other exercises similar to part C. Thanks a lot!

    The whole exercise says:

    "Let f(x)=(x^2+3)/(2x) for x \neq 0. Define a sequence of real numbers x_n by

    x_{n+1}=f(x_n) for n \ge 1, x_1=2.

    A) Show that if x>\sqrt{3}, then f(x)>\sqrt{3}.
    B) Show that if x>\sqrt{3}, then x>f(x).
    C) Conclude that the sequence x_n converges."
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  2. #2
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    Re: Let f(x)=(x^2+3)/(2x) for x not equal to 0. Show that if x>3^(1/2), then f(x)>3^(

    A) Note that f(\sqrt{3}) = \sqrt{3}. Furthermore, f'(x) = \frac{2x(2x) - 2(x^2 + 3)}{4x^2} = \frac{2x^2 - 6}{4x^2}, which is always positive when x > \sqrt{3}. Hence, the minimal value of f(x) is f(\sqrt{3}), or \sqrt{3}.

    Therefore f(x) > \sqrt{3} for all x > \sqrt{3}.
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  3. #3
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    Re: Let f(x)=(x^2+3)/(2x) for x not equal to 0. Show that if x>3^(1/2), then f(x)>3^(

    Thanks! I didn't even think it could be solved so easily *sigh*
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