# domain

• Jul 9th 2013, 05:15 AM
mazen
domain
 Will someone help me with this problem? thank you.
Attachment 28764
• Jul 9th 2013, 05:29 AM
HallsofIvy
Re: domain
What sort of "help" do you want or need? Do you know what "domain of a function" means? Do you know what $\sqrt{x}$ and |x| mean? It is impossible to give suggestions or hints without know what you do understand.
• Jul 9th 2013, 05:35 AM
Prove It
Re: domain
Quote:

Originally Posted by mazen
 Will someone help me with this problem? thank you.
Attachment 28764

In order for there to be a composition of functions \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 AM
mazen
Re: domain
I tried to solve 3abs(sinx)-sinx-2>=0 but i cannot Continue
• Jul 9th 2013, 05:47 AM
Prove It
Re: 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 AM
mazen
Re: domain
• Jul 9th 2013, 06:14 AM
Prove It
Re: 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 AM
HallsofIvy
Re: domain
I don't think I would square both sides. Rather, look at $sin(x)\ge 0$ and $sin(x)< 0$ separately. Of course, $sin(x)\ge 0$ for $2n\pi\le x\le (2n+1)\pi$ for any integer n, $sin(x)< 0$ for $(2n-1)\pi< x< 2n\pi$ for any integer n.

If $sin(x)\ge 0$, |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 $\pi$.

If $sin(x)< 0$, |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 $(2n-1)\pi\le x\le 2n\pi$.
• Jul 9th 2013, 12:01 PM
Plato
Re: domain
Quote:

Originally Posted by mazen
 Will someone help me with this problem? thank you.
Attachment 28764

@ All, Have a look at this webpage.
• Jul 9th 2013, 12:16 PM
mazen
Re: domain
it is so help full 4 me