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Math Help - domain and range

  1. #1
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    Question domain and range

    am i right?
    y=-\sqrt{x^2+2}

    domain:  x|x\ge-2
    range;?

    how can i graph this?
    thanks.
    Last edited by princess_21; March 30th 2009 at 07:30 AM.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by princess_21 View Post
    am i right?
    y=-\sqrt{x^2+2}

    domain:  x|x\ge-2
    range;?

    how can i graph this?
    thanks.
    ---------------------------
    domain:  x|x\ge-2


    Not exactly

    This will give real values if

    x^2 + 2 > 0

    Since Square of every real number is >0
    And +2 will not make it negative

    Hence its true for all real x
    ------------------------------------------------
    Range:



    Thus
    x^2 = y^2 - 2

    x= \pm \sqrt{y^2 -2}

    For rage you need to find all y that satisfy above values

    thus

    y^2 \ge 2

    So all y EXCEPT ( -2,2)
    But y should be negative because of the question's negative sign
    so range is
    [2,infinity)


    ------------------------------------

    Graph:
    Note that y will always be negative

    y^2 = x^2 + 2

    y^2 -x^2 = 2

    This is a 90degree rotated hyperbola (Vertical hyperbola)

    the upper part of vertical hyperbola should not be drawn

    The graph below is for






    Again Don't draw the upper leaf of this type of hyperbola
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  3. #3
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    Quote Originally Posted by ADARSH View Post
    ---------------------------
    domain:  x|x\ge-2


    Not exactly

    This will give real values if

    x^2 + 2 > 0

    Since Square of every real number is >0
    And +2 will not make it negative

    Hence its true for all real x
    ------------------------------------------------
    ok I understand this part. thanks


    Range:



    (y)^{2}=(-\sqrt{x^2+2})^{2}

    y^2=-x^{2}-2

    x^2=-y^{2}-2


    how did you get x^2 = y^2 - 2

    Thus
    x^2 = y^2 - 2

    x= \pm \sqrt{y^2 -2}

    For range you need to find all y that satisfy above values

    thus

    y^2 \ge 2

    So all y EXCEPT ( -2,2)
    But y should be negative because of the question's negative sign
    so range is
    [2,infinity)
    I didn't understand the range can you explain further? thanks so much
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by princess_21 View Post
    ok I understand this part. thanks




    I didn't understand the range can you explain further? thanks so much
    First of all there was a TYPO at the last step, corrected here (written y as negative and gave you +ve range)



    <br />
\implies y^2 = (-1)^2 \times (\sqrt{x^2 +2})^2

    y^2 = 1 \times (x^2 +2)

    Now
    add (-2) on both sides

    y^2 - 2 = x^2

    And




    Now this represents x in terms of y

    Domain of x is R

    what are the values that y can take to give all x in R

    This is done by finding all the values in this function

    in other words we find the domain(meaning all values that y can take) of this function


    Hence I followed the steps below

    -----------------------------------------------------

    thus



    So all y EXCEPT ( -2,2)
    -----------------------------------------------------
    But y should be negative because of the question's negative sign
    this happens because



    has LHS negative (-1 x square root= negative number)

    so range is

    (-infinity,-2]
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