Question

Find the domain of function f(x) = sqrt(sin(sqrt(x)))

Answer 1

f(x) = \[\sqrt{\sin{\sqrt{x}}}\]

Answer 2

Huh... interesting... for one thing, x cannot be negative...

Answer 3

Yeah

Answer 4

x >= 0

Answer 5

Also, \(\sin\sqrt{x}\ge 0\)

Answer 6

ok yeah

Answer 7

So, for instance, we have \(\Large \sin(\theta) \ge 0\) At which intervals does \(\theta\) fall so that \(\large \sin(\theta) \) is not negative?

Answer 8

0 to pi

Answer 9

True... the interval
\[\Large\left[0, \pi\right]\]Anything else?
Remember, we are dealing with the entirety of the real numbers here.

Answer 10

aah,
\[[2n \pi , (2n+1) \pi]\]

Answer 11

Yes, where n ranges from...?

Answer 12

n is an integer.... good enough...

Answer 13

yeah -infinity < n < infinity... n belongs to Z

Answer 14

So that means
\[\huge \sqrt x \in [2n\pi \quad , \quad (2n+1)\pi]\quad\quad n\in\mathbb{Z}?\]

Answer 15

yeah

Answer 16

I think we are close..

Answer 17

Are you sure it's not...
\[\huge n \in [0,\infty)\cap\mathbb{Z}\]?
After all, x can only be positive, right?

Answer 18

Wow..
\[x \in [(2n \pi)^2,{(2n+1) \pi}^2]\]

Answer 19

Am I right ?

Answer 20

Yup :)

Answer 21

Yay Love u bro :D

Answer 22

But you forgot to square this part... close enough...
\[x \in [(2n \pi)^2,{\color{green}{(2n+1)} \pi}^2]\]
And I think the proper notation is this
\[\huge \bigcup_{n=0}^{\infty}[(2n\pi)^2\quad , \quad (2n+1)^2\pi^2]\]