Attachment 28764

Will someone help me with this problem? thank you.

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- Jul 9th 2013, 05:15 AMmazendomain
Attachment 28764*Will someone help me with this problem? thank you.* - Jul 9th 2013, 05:29 AMHallsofIvyRe: domain
What sort of "help" do you want or need? Do you know what "domain of a function"

**means**? Do you know what $\displaystyle \sqrt{x}$ and |x| mean? It is impossible to give suggestions or hints without know what you**do**understand. - Jul 9th 2013, 05:35 AMProve ItRe: domain
In order for there to be a composition of functions $\displaystyle \displaystyle \begin{align*} f \circ g(x) = f \left( g(x) \right) \end{align*}$, the range of the second function needs to be completely contained in the implied domain of the first function (so that all these values can be inserted into the first function).

If this is the case, then the domain of your composition will be the same as the domain as the second function. If not, you will need to determine the largest possible domain for your second function so that the range WILL be completely contained in the implied domain of the first function. - Jul 9th 2013, 05:41 AMmazenRe: domain
I tried to solve 3abs(sinx)-sinx-2>=0 but i cannot Continue

- Jul 9th 2013, 05:47 AMProve ItRe: domain
I have made a mistake in my original post, I'm editing now. But what you have done is a good start. Why not move everything that's not in the absolute value to the other side, then square both sides to remove the absolute value?

- Jul 9th 2013, 06:01 AMmazenRe: domain
- Jul 9th 2013, 06:14 AMProve ItRe: domain
That looks good so far. Can you determine where this equation equals 0? Can you use this to determine where the function is positive?

- Jul 9th 2013, 09:00 AMHallsofIvyRe: domain
I don't think I would square both sides. Rather, look at $\displaystyle sin(x)\ge 0$ and $\displaystyle sin(x)< 0$ separately. Of course, $\displaystyle sin(x)\ge 0$ for $\displaystyle 2n\pi\le x\le (2n+1)\pi$ for any integer n, $\displaystyle sin(x)< 0$ for $\displaystyle (2n-1)\pi< x< 2n\pi$ for any integer n.

If $\displaystyle sin(x)\ge 0$, |sin(x)|= sin(x) so the argument of the square root is 3sin(x)- sin(x)- 2= 2 sin(x)- 2 which will be less than 0 if and only if sin(x)< 1 which is true for all such x except multiples of $\displaystyle \pi$.

If $\displaystyle sin(x)< 0$, |sin(x)|= -sin(x) so the argument of the square root is 3sin(x)+ sin(x)- 2= 4sin(x)- 2 which will be less than 0 if and only if sin(x)< 1/2 which is true for**all**x such that sin(x)< 0[/tex].

Notice that I have looked at values of x such that the argument of the square root is NEGATIVE- where the function does NOT exist. That is, the domain of this function is all real numbers**except**$\displaystyle (2n-1)\pi\le x\le 2n\pi$. - Jul 9th 2013, 12:01 PMPlatoRe: domain
@ All, Have a look at this webpage.

- Jul 9th 2013, 12:16 PMmazenRe: domain
it is so help full 4 me