- #1

Cosmophile

- 111

- 2

## Homework Statement

Identify the open intervals on which the function ##f(x) = \cos \frac {x}{2}, 0 < x < 2 \pi## is increasing or decreasing.

## Homework Equations

##f(x) = \cos \frac {x}{2}##

##f'(x) = -\frac {1}{2} \sin \frac {x}{2}##

## The Attempt at a Solution

I'm attempting to do these types of problems using inequalities, as Serge Lang does in his text, and some problems are easier than others (as one would expect). Here's my attempt so far:

##f## is increasing/decreasing when ##f' > 0## and ##f' < 0## respectively.

##-\frac {1}{2} \sin \frac {x}{2} > 0##

##\sin \frac{x}{2} < 0##

Here, I am torn. If I take ##\arcsin \frac {x}{2} < \arcsin 0## I end up with ##x < 0##. This essentailly means that ##f## is increasing when ##x## is negative. There are no ##x < 0## on the interval, so ##f(x)## is never increasing on the interval ##0 < x < 2 \pi##. This happens to be the case, but I don't think this is the reason. For instance, if I look for the interval on which ##f## is decreasing, using the same method, I get:##\sin \frac{x}{2} < 0##

##-\frac {1}{2} \sin \frac {x}{2} < 0 \to \sin \frac {x}{2} > 0 \to x > 0##

Which says that ##f## is decreasing when ##x>0##, which is true on this interval, but fails beyond the interval.

I have tried doing this without taking the ##\arcsin## of both sides:

##-\frac {1}{2} \sin \frac {x}{2} < 0 \to \sin \frac {x}{2} > 0##

This is less upsetting for me, as I can see this and see which values of ##x## hold. Ultimately, I do not know which is the best method, or if there is another method entirely that I should be using.

Thanks in advance!