Results 1 to 3 of 3

Math Help - Finding range of fuctions.

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    17

    Exclamation Finding range of fuctions.

    hello.. Could someone please show me how to find the range of these equations. Thanks ~~

    Find domain for the following relation:
    a) (x + 2)^2 + (y-3)^2 = 49

    Find the range of the following functions:
    a) m(x) = (x^2 + 2x +5)
    b) (x – 2)^2 + (y+1)^2 = 64
    c) h(x) = √(x+4) + √(5-x)

    Just some homework qus i m stuck with and cant work out... Thanks for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by tom_asian View Post
    hello.. Could someone please show me how to find the range of these equations. Thanks ~~

    Find domain for the following relation:
    a) (x + 2)^2 + (y-3)^2 = 49

    Find the range of the following functions:
    a) m(x) = (x^2 + 2x +5)
    b) (x – 2)^2 + (y+1)^2 = 64
    c) h(x) = √(x+4) + √(5-x)

    Just some homework qus i m stuck with and cant work out... Thanks for your help
    The simplest way to find the range is to graph the relation/function.
    a) (x + 2)^2 + (y - 3)^2 = 49
    This is a circle of radius 7, with center (-2, 3), so the y value is between 3 - 7 = -4 and 3 + 7 = 10. Thus the range is  [-4, 10 ] .

    a) m(x) = \sqrt{x^2 + 2x +5}
    This can be modified to
    m(x) = \sqrt{(x + 1)^2 + 4}
    Thus the minimum value of the quadratic function is 4, so the minimum m value is 2. There is no upper limit on the radicand, so the upper "value" of m is \infty. Thus the range is  [2, \infty ).

    b) (x - 2)^2 + (y + 1)^2 = 64
    This is a circle of radius 8, with center (2, -1). You do this one.

    c) h(x) = \sqrt{x + 4} + \sqrt{5 - x}
    This is a bit trickier to see what to do, so I graphed it. (See below.)
    The minimum value of the range is apparently 3. The peak of the function is at x = 1/2 (we can prove this with Calculus, but you can also see it on the graph), so the maximum y value is h \left ( \frac{1}{2} \right ) = \sqrt{\frac{1}{2} + 4} + \sqrt{5 - \frac{1}{2}} = 3 \sqrt{2}
    So the range is  [ 3,  3\sqrt{2} ]

    -Dan
    Attached Thumbnails Attached Thumbnails Finding range of fuctions.-graph.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tom_asian View Post
    c) h(x) = √(x+4) + √(5-x)
    Here is a purely algebraic way of doing this one.

    The domain of this function is -4\leq x\leq 5.

    Consider the equation for y:
    y=\sqrt{x+4}+\sqrt{5-x}
    The question is for what y does this have at least one real solution for x\mbox{ on }[-4,5].

    Thus,
    y^2 = x+4 + 5 - x + 2\sqrt{(x+4)(5-x)}
    y^2 - 9 = 2\sqrt{20+x-x^2}
    (y^2-9)^2 = 80+4x-4x^2
    -4x^2+4x+[80-(y^2-9)^2]=0
    In order to have a real solution we need,
    16 + 16[80-(y^2-9)^2]\geq 0
    81 - (y^2-9)^2 \geq 0
    Thus,
    (y^2-9)^2 \leq 81
    With the solution,
    x=\frac{-4\pm \sqrt{16+16[80-(y^2-9)^2]}}{-8}
    x=\frac{1\pm \sqrt{81-(y^2-9)^2}}{2}
    But we want,
    -4\leq 1-\sqrt{81-(y^2-9)^2} \mbox{ and }1+\sqrt{81-(y^2-9)^2} \leq 5
    If you solve these inequalities system (which is not that hard) you get what topsquark got.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. finding range
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 2nd 2010, 06:33 AM
  2. Finding the Range using f '(x)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 17th 2010, 11:28 AM
  3. Finding Range
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 6th 2009, 12:57 PM
  4. Finding The Range Help!
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 22nd 2008, 08:33 PM
  5. SAT HELP: finding range
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 28th 2007, 11:34 AM

Search Tags


/mathhelpforum @mathhelpforum