# 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. ## 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."

2. ## 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 $x > \sqrt{3}$.

3. ## 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*