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Thread: 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 $\displaystyle f(x)=(x^2+3)/(2x)$ for $\displaystyle x \neq 0$. Define a sequence of real numbers $\displaystyle x_n$ by

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

    A) Show that if $\displaystyle x>\sqrt{3}$, then $\displaystyle f(x)>\sqrt{3}$.
    B) Show that if $\displaystyle x>\sqrt{3}$, then $\displaystyle x>f(x)$.
    C) Conclude that the sequence $\displaystyle x_n$ 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 $\displaystyle f(\sqrt{3}) = \sqrt{3}$. Furthermore, $\displaystyle f'(x) = \frac{2x(2x) - 2(x^2 + 3)}{4x^2} = \frac{2x^2 - 6}{4x^2}$, which is always positive when $\displaystyle x > \sqrt{3}$. Hence, the minimal value of $\displaystyle f(x)$ is $\displaystyle f(\sqrt{3})$, or $\displaystyle \sqrt{3}$.

    Therefore $\displaystyle f(x) > \sqrt{3}$ for all $\displaystyle x > \sqrt{3}$.
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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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