Yes! This question is most certainly asking about domain! You need to figure out for what values x, y, and z is f(x,y,z) not defined. I copied the question as it was written, but since we have "z" in the denominator, I figured now we simply have a function of...

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three variables. But don't worry! Even if that wasn't intended (like if z were some sort of constant), the analysis remains the same.

There are two cases in which `f(x,y,z)` would be undefined:

1) `2x-y+1<=0` because `ln` cannot be calculated for nonpositive inputs

2) `x^2+z^2 - 1 = 0` because the denominator of a function cannot be 0.

Let's consider the **first case** first.

`2x-y+1<=0`

Let's solve for y:

`2x-y+1<=0`

`2x+1<=y`

`y >= 2x+1`

This tells us that our domain is `y<2x+1` (because we calculated where the function is not defined) and that in the shaded region, our function is defined. Keep in mind! This is only a projection on the x,y axes! Because our function is a function of 3 variables, we need to consider the 3-D case. In our case, then, this inequality tells us that Our defined region includes the space to the right of the dotted line extending infinitely in both z directions. In other words, the dotted line is a plane cutting into the space dividing our domain from impermissible values for the function.

Alright, there's our domain for the first case.

Our **second case** considers when the denominator is
equivalent to zero:

`x^2 + z^2 -1 = 0`

Well, we do some rearranging and we see something interesting. The equation above looks a lot like the equation for a circle with a center of (0,0) and a radius of 1:

`x^2 + z^2 = 1`

However, again, we forgot that we're looking at the 3-D case! The above
equation is **actually a cylinder in space**! We can see the
cross-section on the x-z plane here:

However, looking at the cross-section on the x-y plane is much different!

Combining the two regions on the x-y plane gives us a decent idea of what our domain looks like visually:

In 3-D we have to imagine the dashed line as a plane, and the red lines as a cylinder going through the plane. Using this idea, we have separated the domain into two regions to the right of the plane: one inside the cylinder, and one outside the cylinder. Mathematically, they can be expressed as:

Region 1: `{x,y,z|2x+1>y and x^2 + z^2 < 1}`

Region 2: `{x,y,z|2x+1>y and x^2 + z^2 > 1}`

Now, it'd be easy to figure that the region inside of the cylinder looks a
lot smaller than the region outside the cylinder. **However, it all
depends on how you define "smaller."**

If you just think of it intuitively, then you could say "Well, that small cylinder can certainly be seen as smaller than the outside of the cylinder."

On the other hand, if we're considering the size (cardinality) of the two domains, then they would be the same size (`aleph`)! In fact, if you ever take Analysis, you'll learn that you can prove that the cardinality of the set of points contained within Region 1 is equal to the cardinality of the set of points contained within Region 2 by constructing a bijective function between them. In other words, I can map all points in Region 1 to all points in Region 2 and have no repeats in either set!

So what is the largest region in which f(x,y,z) is continuous?

Well, **either Region 2 or Both Regions**, depending on what
you mean by largest!

I hope that helps!