Hi everyone! If you are trying to find the domain of a function sin(xy)/(x^2+y^2-4), I know that x^2+y^2 can't = 4 but can someone tell me the proper way to express the domain? Thanks!

Question

abb0t · Answer

Domain is all values for which the function is undefined. I think that your domain for the given function is all real, where (x,y) ≠ 0. But, if you use the (x,y) = (0,0) for your example, that makes the function: $\sf \color{}{\frac{(sin(0))}{0}}$ = 0 and so domain is $\mathbb{R}^2$ I believe

abb0t · Answer

I hope that makes sense. I am not good at explaining this type of domain stuff. I just kind of learned how to find it. Haha.

anonymous · Answer

well if x=2 or -2 while y=0 or vice versa, then that's a no go, so since I clearly can't have those conditions, do you still think my domain is really All Real Numbers?

abb0t · Answer

I didn't see the -4 at the end. Sorry. In that case, your domain is $\sf \color{}{\mathbb{R}^2: x ≠ \pm ~2}$  
I am not sure if notation is right.

anonymous · Answer

so since y isn't on the x-axis, when we speak about finding the domain, do we kinda just exclude it from mention?

anonymous · Answer

so maybe R2: X can't = + or - 2, y can't = 0 ???

anonymous · Answer

or would you just leave the y out of the conversation?

amistre64 · Answer

the limit of a function does not depend on the value of the function at the point in question .... therefore a function can be discontinuous and still have a limit.  But the domain is still restricted.

abb0t · Answer

@amistre64 always swaggin' it up :)

amistre64 · Answer

x^2+y^2-4 $\ne$ 0

x^2+y^2 $\ne$ 4

x^2+2xy+y^2 $\ne$ 4+2xy

(x+y)^2 $\ne$ 4+2xy

x+y $\ne$ $\pm\sqrt{4+2xy}$

prolly a bad approach :)

y^2 $\ne$ 4-x^2

y $\ne \pm\sqrt{4-x^2}$

amistre64 · Answer

:)

amistre64 · Answer

$$D=\{x,y:x\in R~and~y\ne\pm\sqrt{4-x^2}\}$$
might be a good notation

anonymous · Answer

wow! thanks! :o)

amistre64 · Answer

hope it makes sense :)

anonymous · Answer

kinda confused about why I have to mention "y" because the domain is along the x-axis after all...kinda confused about that

amistre64 · Answer

to domain is in the xy plane

amistre64 · Answer

a function of 2 variables; neither one is dependant on the other ... but there are certain combinations that provide bad math spots.

anonymous · Answer

oh yeah...i see that...i'm so used to playing with lines that I forgot I'm now playing with partials :o)

amistre64 · Answer

could be:  x,y in R : y ne ...

anonymous · Answer

could you re-post your final notation with how you would say it exatcly in words so I understand how to tell someone what that notation means?

amistre64 · Answer

the domain is the set of all ordered pairs (x,y) in R^2, such that y does not equal (this) of (that).

amistre64 · Answer

edit ...

*(this) or (that)

anonymous · Answer

that's what that fancy notation means from up above with the big bold "D" ?

amistre64 · Answer

essentially, yes.  But i was using my "final" setup, not my initial idea

anonymous · Answer

D={x,y:x∈R and y≠±4−x2−−−−−√}

anonymous · Answer

thanks amistre!

amistre64 · Answer

D = {......}  ; the domain is the set of

D = { (x,y) $\in$ R^2 :  .... }  all ordered pairs (x,y) in R^2 such that .....

D = { (x,y) $\in$ R^2 : y $\ne$ _____ }  y is not equal to what we found