# find the largest possible domain of this function

• Jun 3rd 2011, 02:42 AM
domenfrandolic
find the largest possible domain of this function
find the largest possible domain of this function

square root of pi/4-arcosx...everything is under the square root
• Jun 3rd 2011, 03:42 AM
mr fantastic
Quote:

Originally Posted by domenfrandolic
f:x

square root of pi/4-arcosx...everything is under the square root

Draw the graph of $w = \frac{\pi}{4} - arccos(x)$. The implied domain will be the values of x for which $w \geq 0$.
• Jun 3rd 2011, 04:03 AM
domenfrandolic
the problem is i dont know how to do it
• Jun 3rd 2011, 04:20 AM
HallsofIvy
The square root is only defined for non-negative numbers so the domain is $\frac{\pi}{4}- arccos(x)> 0$ which is the same as saying that $arccos(x)< \frac{\pi}{4}$. One boundary of that will be where $arccos(x)= \frac{\pi}{4}$ or $x= cos(\pi/4)$. Remember that cosine is a decreasing function.
• Jun 3rd 2011, 04:38 AM
domenfrandolic
• Jun 3rd 2011, 04:55 AM
HallsofIvy
Do you not understand what a "domain" is? The domain of a functions is a set of numbers, not a single number.
• Jun 3rd 2011, 06:48 AM
domenfrandolic
yeah i knew that but like which others are the answers?sorry about that but i am quit stuck with the given function
• Jun 3rd 2011, 07:56 AM
skeeter
you should already know what the basic arccos function looks like ...

http://www.mathamazement.com/images/.../arccosine.JPG

perform the necessary transformations to obtain the graph of

$y = \frac{\pi}{4} - \arccos{x}$

then determine the interval where $y \ge 0$
• Jun 3rd 2011, 01:44 PM
mr fantastic
Quote:

Originally Posted by domenfrandolic
yeah i knew that but like which others are the answers?sorry about that but i am quit stuck with the given function

Your problem is that you do not have a sufficient understanding of the pre-requisite mathematical knowledge assumed for this question. I suggest you go back to your class notes or textbook and review inverse trigonometric functions before making further attempts on this question.