Find the domain of the function

[says]

Homework Statement

Find domain of f(x, y) = ln ((x²+2x+y²) / (x²−2x+y²))

Homework Equations

The Attempt at a Solution

Because the function is a natural ln, the restriction on the function is that everything inside the brackets has to >0, otherwise the function is undefined.

I worked on the problem for a while but wasn't getting anywhere so i plugged the equation into wolfram and got the domain { (x,y) ∈ R2 | 4x² < (x² + y²)²}

I'm not sure how to get there though...

[(x²+2x+y²) / (x²−2x+y²)] > 0

 

[Mark44]

Homework Statement

Find domain of f(x, y) = ln ((x²+2x+y²) / (x²−2x+y²))

Homework Equations

The Attempt at a Solution

Because the function is a natural ln, the restriction on the function is that everything inside the brackets has to >0, otherwise the function is undefined.

I worked on the problem for a while but wasn't getting anywhere so i plugged the equation into wolfram and got the domain { (x,y) ∈ R2 | 4x² < (x² + y²)²}

I'm not sure how to get there though...

[(x²+2x+y²) / (x²−2x+y²)] > 0

Because of the natural log, $\frac{x^2 + 2x + y^2}{x^2 - 2x + y^2}$ has to be positive, but you're forgetting that there is a restriction on the denominator.

 

[PeroK]

Homework Statement

Find domain of f(x, y) = ln ((x²+2x+y²) / (x²−2x+y²))

Homework Equations

The Attempt at a Solution

Because the function is a natural ln, the restriction on the function is that everything inside the brackets has to >0, otherwise the function is undefined.

I worked on the problem for a while but wasn't getting anywhere so i plugged the equation into wolfram and got the domain { (x,y) ∈ R2 | 4x² < (x² + y²)²}

I'm not sure how to get there though...

[(x²+2x+y²) / (x²−2x+y²)] > 0

First, note that for a fixed $y$ you have two quadratics in $x$. Perhaps start by analysing those quadratics.

 

[Ray Vickson]

Homework Statement

Find domain of f(x, y) = ln ((x²+2x+y²) / (x²−2x+y²))

Homework Equations

The Attempt at a Solution

Because the function is a natural ln, the restriction on the function is that everything inside the brackets has to >0, otherwise the function is undefined.

I worked on the problem for a while but wasn't getting anywhere so i plugged the equation into wolfram and got the domain { (x,y) ∈ R2 | 4x² < (x² + y²)²}

I'm not sure how to get there though...

[(x²+2x+y²) / (x²−2x+y²)] > 0

So, if $(x^2+2x+y^2)/(x^2-2x+y^2)>0$, then the numerator and denominator must both be > 0 or both be < 0. It might help to look at those two cases separately.