Results 1 to 9 of 9

Math Help - range of function?

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    55

    range of function?

    I'm having trouble figuring the range of this function:

    y = [square root] (x-2) / (6+x) [/square root]


    I know that the horizontal asymptote is y = 1, so there would be some restrictions there. I am just not sure how to use that info to find the range.

    Thank you!

    The answers are y > 0 and y < 1 .. I graphed it and I can see how it works out, just need some explanation please
    Last edited by finalfantasy; January 7th 2008 at 04:02 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by finalfantasy View Post
    I'm having trouble figuring the range of this function:

    m(u) = [square root] (u-2) / (6+u) [/square root]


    I know that the horizontal asymptote is y = 1, so there would be some restrictions there. I am just not sure how to use that info to find the range.

    Thank you!
    do you mean m(u) = \sqrt{\frac {u - 2}{6 + u}} ?

    remember, the range is all the possible outputs we have. what are the possible outputs if u \in (- \infty, 6-) \cup [2, \infty)? (assuming of course, we are dealing with real valued functions here.)

    EDIT: if the above is your function, then we do not require that y < 1. and by the way, there is no y here. there is u and there is m
    Last edited by Jhevon; January 7th 2008 at 04:25 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    55
    Yes, I mean that equation
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by finalfantasy View Post
    Yes, I mean that equation
    well, the range is [0, \infty) graph it again, or try and tell me why through explanation

    maybe you did somethign wrong when graphing by hand. graph it with this
    Follow Math Help Forum on Facebook and Google+

  5. #5
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    well, the range is [0, \infty) graph it again, or try and tell me why through explanation

    maybe you did somethign wrong when graphing by hand. graph it with this
    Wouldn't the range be: [0,1] because the function \frac{x-2}{x+6} has an infinite limit of one giving it a horizontal asymptote of y=1.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by colby2152 View Post
    Wouldn't the range be: [0,1] because the function \frac{x-2}{x+6} has an infinite limit of one giving it a horizontal asymptote of y=1.
    horizontal asymptotes tell how the graph behaves at the "ends", that is, as x \to \infty and x \to - \infty. what about in the middle? how big can the function get as x \to -6, for instance?

    see the graph below. indeed, the horizontal asymptote is 1, but as x \to -6 (from the left, of course), \sqrt{ \frac {x - 2}{x + 6}} \to \infty
    Attached Thumbnails Attached Thumbnails range of function?-graph.jpg  
    Follow Math Help Forum on Facebook and Google+

  7. #7
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    True, I had a mental block there from looking at the negative values - it's that whole square root thing!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by colby2152 View Post
    Wouldn't the range be: [0,1] because the function \frac{x-2}{x+6} has an infinite limit of one giving it a horizontal asymptote of y=1.
    Let g(x) = \frac{x - 2}{x + 6} = \frac{x + 6 - 8}{x + 6} = 1 - \frac{8}{x + 6}. g has a range of (-oo, 1) U (1, +oo).

    Let f(x) = \sqrt{x}. The maximal domain of f is [0, +oo).

    Therefore  y = \sqrt{\frac{x-2}{x+6}} = f(g(x)) is only defined when the range of g is resticted to lie in [0, oo).
    Under this restriction, the 'maximal input' for f will be [0, 1) U (1, +oo) and so the 'maximal output' of f will be [0, 1) U (1, +oo).

    Therefore the range of  y = \sqrt{x-2}{x+6} is [0, 1) U (1, +oo).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mr fantastic View Post
    Let g(x) = \frac{x - 2}{x + 6} = \frac{x + 6 - 8}{x + 6} = 1 - \frac{8}{x + 6}. g has a range of (-oo, 1) U (1, +oo).

    Let f(x) = \sqrt{x}. The maximal domain of f is [0, +oo).

    Therefore  y = \sqrt{\frac{x-2}{x+6}} = f(g(x)) is only defined when the range of g is resticted to lie in [0, oo).
    Under this restriction, the 'maximal input' for f will be [0, 1) U (1, +oo) and so the 'maximal output' of f will be [0, 1) U (1, +oo).

    Therefore the range of  y = \sqrt{x-2}{x+6} is [0, 1) U (1, +oo).
    yes, 1 is not in the range. thanks for pointing out that oversight
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. range of the function
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 30th 2010, 06:19 AM
  2. The range of this function?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 17th 2010, 12:42 PM
  3. Range of a function?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 18th 2009, 01:00 PM
  4. Function Range.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 28th 2009, 07:26 PM
  5. range of function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 18th 2009, 12:45 PM

Search Tags


/mathhelpforum @mathhelpforum