# find the domain of the function

• Oct 26th 2010, 11:32 AM
jayshizwiz
find the domain of the function
$\displaystyle f(x) = \sqrt{4-x^2}$ +$\displaystyle \sqrt{sinx}$

$\displaystyle \sqrt{4-x^2}$ is defined if $\displaystyle -2 \leq x \leq 2$

$\displaystyle \sqrt{sinx}$ is defined if $\displaystyle 2\pi k \leq x \leq \pi + 2 \pi k$

So, to find the domain of the function, am I supposed to determine their union or intersection?

(I think it's the intersection)

$\displaystyle (-2 \leq x \leq 2) \cap (2\pi k \leq x \leq \pi + 2 \pi k) = 0 \leq x \leq 2$

Is that the correct way to find the domain or am I supposed to find their union?

Thanks!
• Oct 26th 2010, 01:09 PM
pickslides
Quote:

Originally Posted by jayshizwiz
$\displaystyle = 0 \leq x \leq 2$

I agree.
• Oct 26th 2010, 01:44 PM
HallsofIvy
In order to be able to find a value for f(x), you must be able to calculate both $\displaystyle \sqrt{4- x^2}$ and $\displaystyle \sqrt{sin(x)}$. That is why both "$\displaystyle -2\le x\le 2$" and "$\displaystyle 2k\pi\le x\le \pi+ 2k\pi$ must be true.

I think it is amusing that Pickslides says "I agree with you" and just below has his "If I agreed with you we'd both be wrong"!
• Oct 26th 2010, 11:21 PM
jayshizwiz
Quote:

Originally Posted by HallsofIvy
In order to be able to find a value for f(x), you must be able to calculate both $\displaystyle \sqrt{4- x^2}$ and $\displaystyle \sqrt{sin(x)}$. That is why both "$\displaystyle -2\le x\le 2$" and "$\displaystyle 2k\pi\le x\le \pi+ 2k\pi$ must be true.

I think it is amusing that Pickslides says "I agree with you" and just below has his "If I agreed with you we'd both be wrong"!

Haha!!