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!

Finding domain of a function.

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the domain of following functions : Note: [.] denotes greatest integer function. {.} denotes fractional part of a function.

    1. f(x)=sin-1([x]/{x})

    2. √(2{x}2-3{x}+1)

    3. Solve the equation :

    y=[sinx+[sinx+[sinx]]]/3
    [y+[y]]=2cosx

    2. Relevant equations



    3. The attempt at a solution

    1. For this function to exist,

    -1≤[x]/{x}≤1
    As 0≤{x}<1

    -{x}≤[x]≤{x}
    Now [x]=x-{x}

    So

    0≤x≤2{x}

    Now ??

    2. For this to exist,

    2{x}2-3{x}+1≥0

    From here I found values of {x} but to no avail.

    3. I do not know how to proceed. :(

    Please help !!

    Thanks in advance... :smile:
     
  2. jcsd
  3. Oct 26, 2013 #2
    For the first question, it is obvious that it is not defined for integers.

    I am unsure what would be the best approach. I would begin with considering different intervals for x. Check what happens to the ratio when 0<x<1, check similarly for 1<x<2 and others.
     
  4. Oct 26, 2013 #3
    As per you the domain should be all fractions. But this is not true for negatives, as for the the greatest integer function "back-shifts".

    So what is wrong in my approach ?

    -1≤[x]/{x}≤1
    As 0≤{x}<1

    -{x}≤[x]≤{x}
    Now [x]=x-{x}

    So

    0≤x≤2{x}

    and

    0≤2{x}<2
     
  5. Oct 26, 2013 #4
    Nope, I never said that. I asked you to check for 0<x<1. I already know what happens in the case when -1<x<0.
    Not sure but it is usually a good idea to check specific intervals when you have to deal with these kind of functions.
     
  6. Oct 26, 2013 #5
    Case I: Does not work for integers.

    Case II: Non integers.

    Ok I considered 2 intervals.

    at 0<x<1

    at 1<x<2

    And I got the answers. But I still can not get the actual mathematical procedure except plotting graph. I must wait till morning for more replies regarding first question.

    Anyone else ?
     
  7. Oct 26, 2013 #6

    pasmith

    User Avatar
    Homework Helper

    Therefore
    [tex]
    -1 < -\{x\} \leq [x] \leq \{x\} < 1
    [/tex]
     
  8. Oct 26, 2013 #7
    And hence from here, taking [x]=x-{x}

    −1+{x}<0≤x≤2{x}<1+{x}

    And so,

    How can I get the values of x from here ? Thanks till now.

    Edit: From this:

    1+{x} is from [1,2). If x is always less than this, then x<1. Now x>{x}-1. {x}-1 is from [-1,0). If x be always greater than this, it has to be x>0. Hence combining the two, we get,

    0<x<1

    Which is the correct answer. Thanks to both the posters !! :)

    Now for the second question ?
     
    Last edited: Oct 26, 2013
  9. Oct 28, 2013 #8
    I give my attempt of second question a step forward :

    2{x}2-3{x}+1≥0
    2{x}2-2{x}-{x}+1≥0
    (2{x}-1)({x}-1)≥0

    So by method of intervals,
    {x}≥1 OR {x}≤1/2
    Also we know that,

    0≤{x}<1

    So using these two conditions, how to find for the values of x ? Please help !!
     
  10. Oct 28, 2013 #9
    So what is the range of {x}?
     
  11. Oct 28, 2013 #10
    Why, for this question it is, 0≤{x}≤1/2....

    So what will be the values of x corresponding to this ?
     
  12. Oct 28, 2013 #11
    Yes. :)
    {x} is periodic with period of 1, does that help?

    Sketch a graph if that helps.
     
  13. Oct 29, 2013 #12
    Well good to know.. :)


    Well I already know that. For all xεR, 0≤{x}<1
    Now here we have 0≤{x}≤0.5. Making use of the knowledge that {x} has period 1, we can say that domain of x is,

    xε[0,0.5]U[1,1.5]U[2,2.5],...... it will keep moving on.....U(0,-0.5]U[-1,-1.5]U......and so on...
    But this is not the answer !!! :(

    ....Does not give answer.... :(
     
  14. Oct 29, 2013 #13
    A much better way to write the above is
    $$x \in \left[ m,m+\frac{1}{2} \right] \,\, \forall \,\, m \in \mathbb{Z}$$

    EDIT: How do you get [-0.5,0]? What is the value of, say, {-0.1}?
    What is the answer then?
     
    Last edited: Oct 29, 2013
  15. Nov 1, 2013 #14
    Sorry for my haste. The question explicitly stated that we have to restrict domain in [-1,1].
    So we have,

    x∈[m,m+1/2]∀m∈Z

    putting, m=-1 and 0, m=1 lies outside [-1,1].

    x∈[-1,-1/2]U[0,1/2]U{1}∀m∈Z , as 1 alone also satisfies the function.

    Now can you give me a start for the third question ?
     
  16. Nov 1, 2013 #15
    I am not entirely sure if there is a "proper" way to do it but currently, I cannot find a better method than checking the cases.

    Look at the second equation, the left side is an integer so the right side must be also an integer.

    The possible values the right side can have is -2,-1,0,1,2. When 2cosx=0, then sin(x)=1. Substitute this in the first equation and you get y=1. Check if this y satisfies the second equation. We find that when y=1, [y+[y]]=2 but right side is zero so there is no value of x satisfying this condition.

    Similarly check for other cases.

    Does the question restricts the domain? Can you re-check with the source?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted