Region of continuity - Geogebra

[mathmari]

Hey!

I am looking the following:

Describe and draw with Geogebra the region where the following functions are continuous:

1. $f(x,y)=\ln (x-3y)$
2. $f(x,y)=\cos^{-1}(xy)$

Do we have to find the region manually or is it possible to find it also using Geogebra? (Wondering)

 

[I like Serena]

Hey mathmari!

'Describe' seems to imply that we should do some analysis.
Drawing a region with Geogebra should then show what we found and/or give us some clues to what we should perhaps have described. And we might include additional objects in Geogebra to show what we found in an analysis. (Thinking)

Usual items for analysis are:

• Identify domain.
• Identify range.
• Find points of discontinuity and/or singular points.
• EDIT (added): Find zeroes, or more generally intersections with the coordinate planes and/or axes.
• Find extrema.
• Find inflection points and identify where the the function is convex, concave, or has saddle points.
• Find asymptotes.

(Thinking)

 

[mathmari]

'Describe' seems to imply that we should do some analysis.
Drawing a region with Geogebra should then show what we found and/or give us some clues to what we should perhaps have described. And we might include additional objects in Geogebra to show what we found in an analysis. (Thinking)

Ahh ok! Let's consider the first function $f(x,y)=\ln (x-3y)$.

Identify domain.

It must hold $x-3y>0 \Rightarrow x>3y$. Therefore the domain is $\{(x,y)\in \mathbb{R}^2\mid x>3y\}$.

Identify range.

The range f a logarithmic function is $\mathbb{R}$, therefore the range of $f$ is $\mathbb{R}$.

Find points of discontinuity and/or singular points.

Aren't these the points that don't belong to the domain? (Wondering)

Find extrema.

We have that $\nabla f=(0,0)\Rightarrow \left (\frac{1}{x-3y}, -\frac{3}{x-3y}\right )=(0,0)$. Since this cannot hold, there are no extrema.

Find inflection points and identify where the the function is convex, concave, or has saddle points.

The Hessian matrix is $\begin{pmatrix}-\frac{1}{(x-3y)^2} & \frac{3}{(x-3y)^2} \\ \frac{3}{(x-3y)^2} & -\frac{9}{(x-3y)^2}\end{pmatrix}$. The eigenvalues are $-\frac{10}{(x-3y)^2}$ and $0$. These are non-positive, and so the matrix is negative semi-definite and so the function is concave. Is that correct? (Wondering)

Find asymptotes.

For that we have to calculate the limits as $x\rightarrow \infty$, $y\rightarrow \infty$, $x\rightarrow 3y$, or not? (Wondering)

 

[mathmari]

Why isn't the region where a function is continuous not equal to the domain of the function? (Wondering)

Why isn't t as follows? Let's consider the function $f(x,y)=\ln (x-3y)$. So that it is defined it must hold that $x-3y>0\Rightarrow x>3y$. Therefore the domain is $\{(x,y)\in \mathbb{R}^2\mid x>3y\}$.

Since the function $\ln$ and the map $(x,y)\mapsto x-3y$ are continuous, the function $f(x,y)$ is also continuous. Therefore the region where the function is continuous is $\{(x,y)\in \mathbb{R}^2\mid x>3y\}$.
Let's consider the function $f(x,y)=\cos^{-1} (xy)$. The range of $\cos$ is $[-1,1]$ and so the domain of $\cos^{-1}$ is $[-1,1]$. Therefore the domain of $\cos^{-1} (xy)$ is $\{(x,y)\in\mathbb R^2\mid xy\in[-1,1]\}$.

Since the function $\cos^{-1}$ and the map $(x,y)\mapsto xy$ are continuous, the function $f(x,y)$ is also continuous. Therefore the region where the function is continuous is $\{(x,y)\in\mathbb R^2\mid xy\in[-1,1]\}$. (Wondering) Is that is correct, what do we have to draw in Geogebra? (Wondering)

I did the following:

View attachment 9605View attachment 9606 Is this what we have to draw? Or is it meant to darw something/somehow else? (Wondering)

 

[I like Serena]

Aren't these the points that don't belong to the domain?

Points of discontinuity are usually defined as points in the domain, such that a limit to them either does not exist, or is different from it.
In most cases we can tell because the derivative does not exist in that point then. (Thinking)

A point of singularity is if all partial derivatives of a parametrization are zero.
In our case of $(x,y)\mapsto (x,y,f(x,y))$ that is not possible because the partial derivative with respect to $x$ is $1\ne 0$. (Nerd)

We have that $\nabla f=(0,0)\Rightarrow \left (\frac{1}{x-3y}, -\frac{3}{x-3y}\right )=(0,0)$. Since this cannot hold, there are no extrema.

The Hessian matrix is $\begin{pmatrix}-\frac{1}{(x-3y)^2} & \frac{3}{(x-3y)^2} \\ \frac{3}{(x-3y)^2} & -\frac{9}{(x-3y)^2}\end{pmatrix}$. The eigenvalues are $-\frac{10}{(x-3y)^2}$ and $0$. These are non-positive, and so the matrix is negative semi-definite and so the function is concave. Is that correct?

For that we have to calculate the limits as $x\rightarrow \infty$, $y\rightarrow \infty$, $x\rightarrow 3y$, or not?

All correct.

To be fair, finding asymptotes on a surface is a bit out of the way.Actually, I forgot one: find zeroes, or more generally the intersections with each of the coordinate planes and/or axes. (Thinking)

Why isn't the region where a function is continuous not equal to the domain of the function?

Consider $h(x,y)=\operatorname{Step}(x+y)$. The function is discontinuous at all points where $x+y=0$. Those points are in the domain though. (Thinking)

Why isn't t as follows? Let's consider the function $f(x,y)=\ln (x-3y)$. So that it is defined it must hold that $x-3y>0\Rightarrow x>3y$. Therefore the domain is $\{(x,y)\in \mathbb{R}^2\mid x>3y\}$.

Since the function $\ln$ and the map $(x,y)\mapsto x-3y$ are continuous, the function $f(x,y)$ is also continuous. Therefore the region where the function is continuous is $\{(x,y)\in \mathbb{R}^2\mid x>3y\}$.
Let's consider the function $f(x,y)=\cos^{-1} (xy)$. The range of $\cos$ is $[-1,1]$ and so the domain of $\cos^{-1}$ is $[-1,1]$. Therefore the domain of $\cos^{-1} (xy)$ is $\{(x,y)\in\mathbb R^2\mid xy\in[-1,1]\}$.

Since the function $\cos^{-1}$ and the map $(x,y)\mapsto xy$ are continuous, the function $f(x,y)$ is also continuous. Therefore the region where the function is continuous is $\{(x,y)\in\mathbb R^2\mid xy\in[-1,1]\}$.

That is correct. So these functions do not have discontinuities in their domain, and so there is nothing special to draw or to see for discontinuities.

Is that is correct, what do we have to draw in Geogebra?

I did the following:

Is this what we have to draw? Or is it meant to draw something/somehow else?

Seems fine to me.
The function has been drawn where it is continuous. Geogebra actually already does that for us.
Additionally you have drawn the domain where the function is continuous, which highlights that. (Nod)