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Math Help - Find the domain

  1. #1
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    Find the domain

    Hi, im trying to find the domain of f(x)=arcsin(1/x)

    I know that y=arcsinx is defined on the interval (-1,1),
    so when I look at f(x) then (1/x) should be on the interval:

    -1<=(1/x)<=1

    I know that the solution is x>=1 and x<=-1 (x is bigger or equal to 1, and less or equal to -1) But I dont know how they got there? any help? thanks!
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  2. #2
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    Re: Find the domain

    Hey Duffman.

    Hint: When is |1/x| <= 1 (where |.| is the absolute value)? [Remember that sin(x) always lies in-between -1 and 1 inclusive).
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  3. #3
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    Re: Find the domain

    You know that 0<\frac{1}{x}\le{1} if and only if x \ge 1, and -1\le\frac{1}{x}<0 if and only if x \le -1. There is no real number x that makes \frac{1}{x} zero. So your condition on \frac{1}{x} and the answer have the same meaning.

    - Hollywood
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  4. #4
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    Re: Find the domain

    As I said the correct answer should be x>=1 and  x<=-1 which can also be written as (-infinite,-1]u[1,infinite).


    Im sorry but I dont fully understand You. This is how I try to solve it:


    If:
    -1<=(1/x)<=1

    Then: (multiplying all sides with x)
    -x<=1<=x

    Then I have:

    -x<=1 and 1<=x

    which is equal to (?):

    x>=-1 and 1<=x which can be written as [-1, infinite)]u[1,infinite). (This solution is not OK because it should be (-infinite,-1]u[1,infinite).

    I know that something is wrong here but I dont seem to understand what :/ Thanks for our help
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  5. #5
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    Re: Find the domain

    Rule: |x|<a if -a<x<a
    ---------------------
    If I try to continue:

    -1<=(1/x)<=1

    is equal to:
    |1/x| <= 1

    should be equal to:
    1/|x| <= 1

    multiplying with |x| on both sides, gives:

    1<=|x|

    This expression has "two solutions",
    x>=1
    -x>=1

    Can be written as:
    x>=1
    x<=-1

    Which seems to give the correct answer:
    (-infinite,-1]u[1,infinite).

    My question: Is this method mathematically correct? And can I solve this in any other way?
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  6. #6
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    Re: Find the domain

    Quote Originally Posted by Duffman View Post

    If:
    -1<=(1/x)<=1

    Then: (multiplying all sides with x)
    -x<=1<=x
    When you multiply both sides of an inequality by a negative number, the direction of the inequality is reversed. So you have -x \ge 1 instead of -x \le 1.

    - Hollywood
    Thanks from Duffman
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  7. #7
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    Re: Find the domain

    Im sorry I seem slow

    But when I have
    -1 \le (1/x)\le 1
    ... and multiply both sides with x, isnt x a positive number? So why is the direction of the inequality reversed on the left side? (All I have done is multiplying with positive x).

    How do you make it?
    -x \ge (1)\le x

    Best regards
    /Duffman
    Last edited by Duffman; August 7th 2013 at 07:49 AM.
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  8. #8
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    Re: Find the domain

    The problem is that x can represent a negative number. So when you multiply by x you either reverse or don't reverse the direction of the inequality depending on whether x is positive or negative. In other words, there are two cases: x < 0 and x > 0.

    - Hollywood
    Thanks from Duffman
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  9. #9
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    Re: Find the domain

    Quote Originally Posted by Duffman View Post
    Hi, im trying to find the domain of f(x)=arcsin(1/x)

    I know that y=arcsinx is defined on the interval (-1,1),
    so when I look at f(x) then (1/x) should be on the interval:

    -1<=(1/x)<=1
    The problem with this is that 0 is between -1 and 1 and 1/x is never 0. Better to say -1<= 1/x< 0 and 0< 1/x<= 1. From -1<= 1/x< 0 we have 1<= -1/x, x<= -1. From 0< 1/x<= 1, 1<= x.
    That gives two separate intervals, (-\infty, -1)\cup (1, \infty).

    I know that the solution is x>=1 and x<=-1 (x is bigger or equal to 1, and less or equal to -1) But I dont know how they got there? any help? thanks!
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