find the largest possible domain of this function

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

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- Jun 3rd 2011, 02:42 AMdomenfrandolicfind 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 AMmr fantastic
- Jun 3rd 2011, 04:03 AMdomenfrandolic
the problem is i dont know how to do it

- Jun 3rd 2011, 04:20 AMHallsofIvy
The square root is only defined for non-negative numbers so the domain is $\displaystyle \frac{\pi}{4}- arccos(x)> 0$ which is the same as saying that $\displaystyle arccos(x)< \frac{\pi}{4}$. One boundary of that will be where $\displaystyle arccos(x)= \frac{\pi}{4}$ or $\displaystyle x= cos(\pi/4)$. Remember that cosine is a

**decreasing**function. - Jun 3rd 2011, 04:38 AMdomenfrandolic
so the answer is cos(pi/4)?

- Jun 3rd 2011, 04:55 AMHallsofIvy
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 AMdomenfrandolic
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 AMskeeter
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

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

then determine the interval where $\displaystyle y \ge 0$ - Jun 3rd 2011, 01:44 PMmr fantastic
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.