Results 1 to 3 of 3

Math Help - Ranges without the aid of a calculator Help

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    1

    Ranges without the aid of a calculator Help

    This is a pre-calc review for Calculus (I think this is the right area)

    I can do domains completely fine but I have trouble with ranges. Like for instance y=sqrt(x+1), y=1/(sqrt(x+1)) and y= sqrt((1/x)+1). For the first one, I know its a radical function and y >= 0. But Mathematically how do I calculate this? I heard you put the equation into the inverse function but I'm not really sure what to do afterward.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    You don't need to calculate it for the first square root. If you need a justification you can start by looking at y^2 = x. This is a parabola centered at the origin and it opens to the right (towards positive x-values). So x can only have positive values. This is equivalent to y = \pm \sqrt{x}. In your case you are taking the positive square root, which will be the part of the parabola above the x-axis, that is the square root evaluates to only value greater or equal to 0.

    In your first case, the range is [0,\infty[. In the second expression you are taking the reciprocal of the first function. The range is ]0,\infty[. Here is how you get it. Lets look at a point y in the first range. To get the range of the second one, we take the reciprocal, i.e. 1/y. When y is close to 0 we get a big number, so on one side the range can be arbitrarily large. However, if we make y really large, we are making 1/y really tiny, so the range is 0 at the other end. The range ought to be [0,\infty[, but since you can never make 1/y zero, it must be an open interval - ]0,\infty[
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32
    Quote Originally Posted by Nekuni View Post
    This is a pre-calc review for Calculus (I think this is the right area)

    I can do domains completely fine but I have trouble with ranges. Like for instance y=sqrt(x+1), y=1/(sqrt(x+1)) and y= sqrt((1/x)+1). For the first one, I know its a radical function and y >= 0. But Mathematically how do I calculate this? I heard you put the equation into the inverse function but I'm not really sure what to do afterward.

    Thanks!
    Personally, I find the study of basic functions and transformations (vertical/horizontal stretch, shift up/down/left/right) to be useful when determining the domain and range of a function. For example, if you know the domain and range of x^2, then you can derive the domain/range list of 2x^2, (x-2)^2, and x^2 + 2 by realizing that these functions are simply products of transformations + their effect.

    Also, here's a trick, (although it may be limited). you should be careful with it as some inverse functions can get quite messy.

    We know that f: x -> y, where x is the domain and y is the range.
    Now consider that the inverse function of f, call it g: y -> x

    The domain of f is the range of g, and the range of f is the domain of g *verify*

    For example, let f(x) = y = x^2
    Domain: the real line
    Range: y>=0

    Now, g(x) = +- sqrt(x)
    Domain: x>=0
    Range: the real line

    EDIT:// as an afterthought, if all else fails: graph it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 6th 2012, 02:16 PM
  2. Ranges of a function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 23rd 2010, 01:35 AM
  3. Nulls and Ranges
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 10th 2009, 12:38 AM
  4. Domains and Ranges
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: March 7th 2008, 06:03 PM
  5. Ranges of function f(x)?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 2nd 2008, 05:39 AM

Search Tags


/mathhelpforum @mathhelpforum