Explanation as regards range and domains

[negation]

Homework Statement

a) f(x,y)=√ [1−(x²+y²)]

b) f(x,y)=2cos(4x+y²)

The Attempt at a Solution

a) The domain is such that x² + y² must not be greater than 1
In other words, this is expressed (as stated as an option on the answer sheet) as "xy-plane without the line x=y"
Why is this so?

b) the equation f(x,y)=2cos(4x+y²) has a range of [-2,2]
As for the domain, for f(x,y)=2cos(4x+y2) to have a range of [-2,2], (4x +y² must be 0 or π. How do I express the domain in terms of x and y?
The domain for (b) as provided by the answer is R²(xy-plane).

 

[jbunniii]

Homework Statement

a) f(x,y)=√ [1−(x²+y²)]

b) f(x,y)=2cos(4x+y²)

The Attempt at a Solution

a) The domain is such that x² + y² must not be greater than 1

OK. Can it equal 1?

In other words, this is expressed (as stated as an option on the answer sheet) as "xy-plane without the line x=y"
Why is this so?

No, this is not equivalent at all. What is the region defined by $x^2 + y^2 < 1$? What is its shape?

b) the equation f(x,y)=2cos(4x+y²) has a range of [-2,2]
As for the domain, for f(x,y)=2cos(4x+y2) to have a range of [-2,2], (4x +y² must be 0 or π. How do I express the domain in terms of x and y?
The domain for (b) as provided by the answer is R²(xy-plane).

The notation $[-2,2]$ means the entire interval between $-2$ and $2$, including the endpoints. It does not mean just the two points $-2$ and $2$.

 

[negation]

OK. Can it equal 1?

Yes it can.

No, this is not equivalent at all. What is the region defined by $x^2 + y^2 < 1$? What is its shape?

Strange, but the answer states it is...

It's a "dashed circle" with radius smaller than 1

The notation $[-2,2]$ means the entire interval between $-2$ and $2$, including the endpoints. It does not mean just the two points $-2$ and $2$.[/QUOTE]

I know.

 

[jbunniii]

Yes it can. Strange, but the answer states it is...

It's a "dashed circle" with radius smaller than 1

Well, that's almost correct, but as you said, $x^2 + y^2 = 1$ is allowed, so the circle itself is included in the domain.

But how does that relate to the answer you gave previously?

In other words, this is expressed (as stated as an option on the answer sheet) as "xy-plane without the line x=y"
Why is this so?

Regarding the second part, I assumed you were interpreting $[-2,2]$ to mean the set containing just the points $-2$ and $2$, because you said that the argument must be either $0$ or $\pi$. Since you know what $[-2,2]$ means, I'm not sure what you are trying to do. Is there any value of $4x+y^2$ which results in $2\cos(4x+y^2)$ being greater than $2$ or less than $-2$?

 

[negation]

Well, that's almost correct, but as you said, $x^2 + y^2 = 1$ is allowed, so the circle itself is included in the domain.

But how does that relate to the answer you gave previously?Regarding the second part, I assumed you were interpreting $[-2,2]$ to mean the set containing just the points $-2$ and $2$, because you said that the argument must be either $0$ or $\pi$. Since you know what $[-2,2]$ means, I'm not sure what you are trying to do. Is there any value of $4x+y^2$ which results in $2\cos(4x+y^2)$ being greater than $2$ or less than $-2$?

I'm not exactly sure cause the system is giving me erratic answer.

When I did the same question using the answer it reflected originally, it reported that my answer was wrong.

In response to part(a):
my answer is "The disk with radius 1 and center (0,0)" for the domain and "(−∞,0]" for the range.

 

[negation]

Regarding the second part, I assumed you were interpreting $[-2,2]$ to mean the set containing just the points $-2$ and $2$, because you said that the argument must be either $0$ or $\pi$. Since you know what $[-2,2]$ means, I'm not sure what you are trying to do. Is there any value of $4x+y^2$ which results in $2\cos(4x+y^2)$ being greater than $2$ or less than $-2$?

I am unsure. Could you point me in the right direction?

 

[jbunniii]

I am unsure. Could you point me in the right direction?

What is the range of possible values of $\cos(\theta)$, where $\theta$ can be any real number?

 

[negation]

What is the range of possible values of $\cos(\theta)$, where $\theta$ can be any real number?

How is my answer to part(a)?
Cos(Θ) has a range of [-1,1]

 

[jbunniii]

How is my answer to part(a)?

What is your final answer for (a)?

Cos(Θ) has a range of [-1,1]

Good, so $2\cos(\theta)$ has a range of $[-2,2]$, and $\theta$ can be anything. So what are the domain and range of $2\cos(4x + y^2)$?

 

[jbunniii]

Sorry, just noticed your answer for (a)

In response to part(a):
my answer is "The disk with radius 1 and center (0,0)" for the domain and "(−∞,0]" for the range.

For the domain, you may want to explicitly state that it's the closed disk of radius 1 and center (0,0). The range is wrong: does the square root give you a negative result?

 

[negation]

What is your final answer for (a)?


Good, so $2\cos(\theta)$ has a range of $[-2,2]$, and $\theta$ can be anything. So what are the domain and range of $2\cos(4x + y^2)$?

my answer for (a) is "The disk with radius 1 and center (0,0)" for the domain and "(−∞,0]" for the range.

In response to (b): The domain for cos(4x +y²) is any real number on the xy-plane.
And it's range is [-1,1]-min bound of -1 with max bound of 1.
Since the question asked for the domain and range of 2 cos(4x +y²),
the domain is any real number on the xy-plane. The range is the interval [-2,2]-min at -2, max at 2

 

[negation]

Sorry, just noticed your answer for (a)


For the domain, you may want to explicitly state that it's the closed disk of radius 1 and center (0,0). The range is wrong: does the square root give you a negative result?

Apparently, there's no option for a CLOSED disk. Yes the range is wrong. It should be [0,+ve infinity]. Brain isn't really working well at 2am.

 

[jbunniii]

Apparently, there's no option for a CLOSED disk.

In some contexts, a domain is required to be an open set, so if $D$ is the set of all allowable values, we take the domain to be the largest open set contained in $D$. This is called the interior of $D$ and is usually denoted $\text{int}(D)$ or $D^o$. Perhaps that is the intent in your situation? Or maybe there's just an error in the set of possible answers.