Results 1 to 4 of 4

Math Help - Domains?

  1. #1
    Junior Member
    Joined
    Jan 2007
    Posts
    35

    Domains?

    Hi, I just did these two problems and the domain was the same for both problems, am I overlooking something?

    h(x) = 4x/Square Root(x^2 - 16)
    and
    j(x) = Square Root[4x/(x^2 - 16)]

    Maybe she's trying to trick me?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2007
    Posts
    35

    Oops!

    One more quick thing. I didn't quite know how to approach this one:

    The function y=k(x) is a polynomial with x-intercepts 1,2,5 and 8. The solution set to k(x) is (1, 2)U(5,8).

    Use this information to find the domain of y= Square Root[k(x)]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by leviathanwave View Post
    Hi, I just did these two problems and the domain was the same for both problems, am I overlooking something?

    h(x) = 4x/Square Root(x^2 - 16)
    and
    The denominator cannot be zero. And the radical must be positive. Thus,
    x^2-16>0
    x^2>16
    x>4 \mbox{ or }x<-4
    j(x) = Square Root[4x/(x^2 - 16)]
    Over here we need term inside the radical to be positive.
    First x^2-16\not = 0 thus, x\not = \pm 4.
    And,
    \frac{4x}{x^2-16}>0
    Both positive or both negative

    1) 4x\geq 0 \to x\geq 0
    And,
    x^2-16 >0 thus, x>4 \mbox{ or }x<-4
    Combination of those two leads us to,
    x>4

    2) 4x\leq 0 \to x\leq 0
    x^2-16 >0 thus, x>4 \mbox{ or }x<-4.
    Combination of those two leads us to,
    x<-4.

    Even though they are different functions they have the same solution set.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,664
    Thanks
    298
    Awards
    1
    Quote Originally Posted by leviathanwave View Post
    One more quick thing. I didn't quite know how to approach this one:

    The function y=k(x) is a polynomial with x-intercepts 1,2,5 and 8. The solution set to k(x) is (1, 2)U(5,8).

    Use this information to find the domain of y= Square Root[k(x)]
    Typically if you want a polynomial with specific x-intercepts you can simply write them out term by term:

    If you have the intercepts 1, 2, 5, and 8 a polynomial that fits this is:
    f(x) = (x - 1)(x - 2)(x - 5)(x - 8)

    Now we want the function y = \sqrt{k(x)} to have the domain (1, 2) \cup (5, 8) which means that this is the part of the (real) x-axis for which k(x) is positive. So let's graph f(x). (See the attachment below.)

    Note that in the graph the set (1, 2) \cup (5, 8) is the region where f(x) is negative. So we need to "flip the function over" to make this the region where the function is positive. This is easy. Just define k(x) to be the negative of f(x):
    k(x) = -f(x) = -(x - 1)(x - 2)(x - 5)(x - 8).

    Then y = \sqrt{k(x)} has the correct domain.

    Note: Technically the domain is  [1, 2] \cup [5, 8] . I couldn't think of a way to define a finite polynomial that excludes the end-points of the sets. (Unless I'm missing the obvious.) You might wish to point this out to your teacher.

    -Dan
    Attached Thumbnails Attached Thumbnails Domains?-polynomial.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prufer Domains and Valuation Domains
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 27th 2009, 07:15 AM
  2. Domains
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 22nd 2009, 07:39 AM
  3. Domains
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 9th 2009, 08:27 PM
  4. Domains
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 28th 2009, 08:56 AM
  5. Domains
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 23rd 2008, 01:49 PM

Search Tags


/mathhelpforum @mathhelpforum