Determine the Set of Points at Which the Function is Continu

[cosmos42]

Homework Statement

"Determine the set of points at which the function is continuous"
ƒ(x,y,z) = √y/(√(x²-y²-z²))

Homework Equations

ƒ(x,y,z) = √y/(√(x²-y²-z²))

The Attempt at a Solution

If y is zero this function would be discontinuous or if the denominator became zero.
As much detail as you could provide in your explanation would be greatly appreciated, thank you.

 

[RUber]

Why would you have a discontinuous function at y = 0?
First, you need to define the domain for the function. The denominator being zero would be inadmissible, values of x, y, z such that $x^2 -y^2 -z^2 =0$ would be outside the domain.
Secondly, if you are only dealing with real numbers, you would also want to remove any negatives under the square roots.
Other than that, rational expressions like this are generally continuous everywhere they are defined.

 

[cosmos42]

Why would you have a discontinuous function at y = 0?
First, you need to define the domain for the function. The denominator being zero would be inadmissible, values of x, y, z such that $x^2 -y^2 -z^2 =0$ would be outside the domain.
Secondly, if you are only dealing with real numbers, you would also want to remove any negatives under the square roots.
Other than that, rational expressions like this are generally continuous everywhere they are defined.

You're right, the numerator could be zero but the denominator should not ever be zero so, would the domain then be {(x,y,z) ∈ ℝ³ : x²+y²+z²>0}?

 

[RUber]

Double check your signs.
And there is also a requirement for y to be *edit* non-negative (from the numerator).
If you draw the region, it should look like a half-cone.

 

[HallsofIvy]

Why would you have a discontinuous function at y = 0?
First, you need to define the domain for the function. The denominator being zero would be inadmissible, values of x, y, z such that $x^2 -y^2 -z^2 =0$ would be outside the domain.
Secondly, if you are only dealing with real numbers, you would also want to remove any negatives under the square roots.
Other than that, rational expressions like this are generally continuous everywhere they are defined.

This is not a "rational expression".

 

[RUber]

This is not a "rational expression".

Thank you, I misused the term. Is there a general term for functions of this type?

What I meant to say was: f(x,y,z), as a ratio of continuous functions, g(y) and h(x,y,z) will be continuous anywhere in the intersection of the domains of g(y) and h(x,y,z) with the exception of those points where h(x,y,z) = 0.