Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - implied domain of a function

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    16

    implied domain of a function

    Got another question for you guys - i hope someone is up and around here to help me out, i am totally stuck...

    i am to find the implied domain of this function...



    f(x)= [the square root of (x+6)] +3



    i am understanding the domain and range theory and what not, but i just don't get it with this particular problem...

    thanks for any help!

    ryan
    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 shipwreck View Post
    Got another question for you guys - i hope someone is up and around here to help me out, i am totally stuck...

    i am to find the implied domain of this function...



    f(x)= [the square root of (x+6)] +3



    i am understanding the domain and range theory and what not, but i just don't get it with this particular problem...

    thanks for any help!

    ryan
    just find the domain. the implied domain would be the largest possible domain you can have in the real numbers. that is, there are otherwise no restrictions than those the function itself imposes on us
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    16
    so does that mean its basically just negative infinity to positive infinity?
    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 shipwreck View Post
    so does that mean its basically just negative infinity to positive infinity?
    no

    the function itself gives us restrictions, correct? the square root function does not work everywhere. the domain will be restricted based on that
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    16
    thats true, i wasn't thinking of that.

    so because of the square root, x cannot equal -6...so we would exclude this, no?

    (-infinity, -6), (-6, infinity)....does that look right, or am i just falling asleep while trying to learn this?
    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 shipwreck View Post
    thats true, i wasn't thinking of that.

    so because of the square root, x cannot equal -6...so we would exclude this, no?

    (-infinity, -6), (-6, infinity)....does that look right, or am i just falling asleep while trying to learn this?
    no. say x = -7, which is in the domain you gave. does the square root work?

    the domain in the real numbers for the function \sqrt{x} is x \ge 0
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    16
    i am still lost i guess, i'm not understanding what you're telling me unfortunatley.

    it has to be written in interval notation, and maybe that is what is throwing me off?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    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 shipwreck View Post
    i am still lost i guess, i'm not understanding what you're telling me unfortunatley.

    it has to be written in interval notation, and maybe that is what is throwing me off?
    first get to the answer, then we worry about expressing it in the right way

    we need what is being square rooted to be zero or bigger

    by the way, as an interval x \ge 0 = [0, \infty)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Oct 2008
    Posts
    16
    ok, so x+6 is greater than or equal to 0,
    so x is greater than or equal to -6...

    this would give me [-6, infinity)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    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 shipwreck View Post
    ok, so x+6 is greater than or equal to 0,
    so x is greater than or equal to -6...

    this would give me [-6, infinity)
    yes
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Oct 2008
    Posts
    16
    phew! thank you SO much!! it just took some talking it out to get it through my head...thank you again!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    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 shipwreck View Post
    phew! thank you SO much!! it just took some talking it out to get it through my head...thank you again!
    no problem.

    thanks for being a good sport and not one of those just-tell-me-the-answer-for-crying-out-loud! types
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Oct 2008
    Posts
    16
    no believe me, i need to KNOW how to do it rather than just be told what the answer is. i have a test over all of this and more tomorrow, and i won't be able to have someone give me answers then, so my little brain has to do the thinking on its own.

    i am already stuck on the next problem though - any chance I could get some assistance on how to start it at least?

    g(x)= 3x^2 -x -1

    for the following function, determine g(x+a)-g(x)

    any ideas? i have NO clue how to even start this. there is absolutely no example of this in my book, nor any practice problems that look like this...
    Follow Math Help Forum on Facebook and Google+

  14. #14
    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 shipwreck View Post
    no believe me, i need to KNOW how to do it rather than just be told what the answer is. i have a test over all of this and more tomorrow, and i won't be able to have someone give me answers then, so my little brain has to do the thinking on its own.

    i am already stuck on the next problem though - any chance I could get some assistance on how to start it at least?

    g(x)= 3x^2 -x -1

    for the following function, determine g(x+a)-g(x)

    any ideas? i have NO clue how to even start this. there is absolutely no example of this in my book, nor any practice problems that look like this...
    you should post new questions in new threads.

    as for composite functions, which is what these are, here's a hint.

    function notation works as follows, saying something like f(x) means, the function f takes any value x in its domain and plugs it in the order given. so whatever is in the brackets, takes the spot of x.

    some examples are in order:

    using f(x) = x^2 + 2x + 1

    f(a) = a^2 + 2a + 1

    f(1) = 1^2 + 2(1) + 1 = 4

    f(c) = c^2 + 2c + 1

    f(bc + d) = (bc + d)^2 + 2(bc + d) + 1

    f( ) = ( )^2 + 2( ) + 1

    so whatever shows up in the brackets, we replace x with it, provided that thing is in the domain. ( is in the domain of all functions, in case you were wondering )

    now, how about your problem?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Oct 2008
    Posts
    16
    ok - so for my problem then...

    (3x^2 - x - 1 + a) - (3x^2 - x - 1),

    and then combine all like terms? is that right though? that doesn't feel right, as everything cancels out obviously, leaving only a behind.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Implied Domain
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 2nd 2010, 01:24 PM
  2. Implied Domain
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 7th 2010, 11:15 AM
  3. Implied domain
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 7th 2009, 02:11 AM
  4. Implied Domain
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: March 6th 2008, 10:47 PM
  5. implied domain
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: February 10th 2008, 08:40 AM

Search Tags


/mathhelpforum @mathhelpforum