1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Explanation as regards range and domains

  1. Apr 3, 2014 #1
    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. jcsd
  3. Apr 3, 2014 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##.
     
  4. Apr 3, 2014 #3
    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.
     
  5. Apr 3, 2014 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##?
     
  6. Apr 3, 2014 #5
    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.
     
  7. Apr 3, 2014 #6
    I am unsure. Could you point me in the right direction?
     
  8. Apr 3, 2014 #7

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What is the range of possible values of ##\cos(\theta)##, where ##\theta## can be any real number?
     
  9. Apr 3, 2014 #8
    How is my answer to part(a)?
    Cos(Θ) has a range of [-1,1]
     
  10. Apr 3, 2014 #9

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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)##?
     
  11. Apr 3, 2014 #10

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
  12. Apr 3, 2014 #11
    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
     
  13. Apr 3, 2014 #12
    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.
     
  14. Apr 3, 2014 #13

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Explanation as regards range and domains
  1. ? Domain & Range ? (Replies: 24)

  2. Domain and range (Replies: 3)

  3. Domain and Range (Replies: 6)

Loading...