# Detecting Convergence, Esimating Limits

• March 31st 2008, 11:45 PM
larson
Detecting Convergence, Esimating Limits
Let $f(x) = \frac {e^x}{1 + e^x}$

a. Show that $f(x) > \frac {1}{2}$ for all $x >= 0$
• April 1st 2008, 12:39 AM
angel.white
Quote:

Originally Posted by larson
Let $f(x) = \frac {e^x}{1 + e^x}$

a. Show that $f(x) > \frac {1}{2}$ for all $x >= 0$

$\frac {e^x}{1 + e^x} > \frac 12$

cross multiply

$2e^x > 1 + e^x$

combine like terms

$e^x > 1$

take natural log

$x > 0$

Note that this is not what you posted, you said when x = 0, f(x) > 1/2. But we have that this is not true. It can be verified by plugging in 0 and seeing that your initial criteria are not correct:

$f(0) = \frac {e^0}{1 + e^0}$

$f(0) = \frac {1}{1 + 1}$

$f(0) = \frac 12$

and since 1/2 is not greater than 1/2, your initial criteria should have been either
a. Show that $f(x) \geq \frac {1}{2}$ for all $x \geq 0$

or

a. Show that $f(x) > \frac {1}{2}$ for all $x > 0$
• April 1st 2008, 02:27 AM
mr fantastic
Quote:

Originally Posted by larson
Let $f(x) = \frac {e^x}{1 + e^x}$

a. Show that $f(x) > \frac {1}{2}$ for all $x >= 0$

The result follows from:

1. $f'(x) = \frac{e^x}{(1 + e^x)^2} > 0$ for all x therefore f(x) is strictly increasing.

2. f(0) = 1/2.