# Detecting Convergence, Esimating Limits

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

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

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

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

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

cross multiply

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

combine like terms

$\displaystyle e^x > 1$

take natural log

$\displaystyle x > 0$

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:

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

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

$\displaystyle f(0) = \frac 12$

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

or

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

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

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

The result follows from:

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

2. f(0) = 1/2.