# 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)

• Jul 3rd 2012, 01:02 PM
juanma101285
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."
• Jul 3rd 2012, 01:09 PM
richard1234
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 $\displaystyle x > \sqrt{3}$.
• Jul 3rd 2012, 01:37 PM
juanma101285
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*