# Explanation as regards range and domains

1. Apr 3, 2014

### negation

1. The problem statement, all variables and given/known data

a) f(x,y)=√ [1−(x2+y2)]

b) f(x,y)=2cos(4x+y2)

3. 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).

Last edited: Apr 3, 2014
2. Apr 3, 2014

### jbunniii

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

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 $2$.

3. Apr 3, 2014

### negation

Yes it can.
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.

4. Apr 3, 2014

### jbunniii

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 $-2$?

5. Apr 3, 2014

### negation

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.

6. Apr 3, 2014

### negation

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

7. Apr 3, 2014

### jbunniii

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

8. Apr 3, 2014

### negation

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

9. Apr 3, 2014

### jbunniii

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

10. Apr 3, 2014

### jbunniii

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?

11. Apr 3, 2014

### negation

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

12. Apr 3, 2014

### negation

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.

13. Apr 3, 2014

### jbunniii

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.