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 =(x^2+3)/(2x))
for 
. Define a sequence of real numbers 
by
)
for 
, 
.
A) Show that if 
, then >\sqrt{3})
.
B) Show that if , then )
.
C) Conclude that the sequence converges."
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  = \sqrt{3})
. Furthermore,  = \frac{2x(2x) - 2(x^2 + 3)}{4x^2} = \frac{2x^2 - 6}{4x^2})
, which is always positive when 
. Hence, the minimal value of )
is )
, or 
.
Therefore  > \sqrt{3})
for all .
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*
