Results 1 to 13 of 13

Math Help - confused about domain and range

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    40
    Thanks
    1

    confused about domain and range

    How do you find these out algebraically given a function?

    It'd be appreciated if you could do an example with an exponential (2^x) function, a root function, a reciprocal function, a cubic function, a semi-circle function, a quadratic function, absolute value function, conics (ellipse, hyperbola, parabola of the form y=1/4px^2, etc.) and whatever else.

    Or, if you could just explain how it's done in general. My textbook has no information on this and my teacher didn't explain any of this. I googled around, but they all use different methods. Could I please get one method that works for all cases?

    Here's a few examples that I just came up with that I have no idea how to solve.

    y = 9x^2 / (25-x^2)

    y = sqrt(4x^2-9)

    y = (4x^2-9)/x^2

    y = 5 / sqrt(3-x)

    y = 4x^2+6x+6

    y = |x-3|+6
    Last edited by iragequit; March 31st 2011 at 04:14 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member rtblue's Avatar
    Joined
    Mar 2009
    From
    Birmingham, Alabama.
    Posts
    221
    I don't know of any set way to find the domain; it depends on the function.

    For example, look at y = 5/ sqrt(3-x)

    The factors limiting the domain are the square root, and the denominator. The root cannot be less than 0 (that would be imaginary), and the denominator cannot equal zero (that's undefined).

    Therefore, we have the domain as (x : x < 3)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    40
    Thanks
    1
    Quote Originally Posted by rtblue View Post
    I don't know of any set way to find the domain; it depends on the function.

    For example, look at y = 5/ sqrt(3-x)

    The factors limiting the domain are the square root, and the denominator. The root cannot be less than 0 (that would be imaginary), and the denominator cannot equal zero (that's undefined).

    Therefore, we have the domain as (x : x < 3)
    Could you do range.?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member rtblue's Avatar
    Joined
    Mar 2009
    From
    Birmingham, Alabama.
    Posts
    221
    (0,infinity)

    You can't get a negative obviously. You can't include zero, because you will never reach zero. As the square root in the denominator gets smaller, y approaches infinity. As x gets more negative, y approaches 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2011
    Posts
    40
    Thanks
    1
    y=x/(x-4)

    i don't get this. why is the range =/= 1.. when i re-arrange for x i get x = x/y + 4 so its y =/= 0 -_-
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member rtblue's Avatar
    Joined
    Mar 2009
    From
    Birmingham, Alabama.
    Posts
    221
    Well, I'm not sure what "the range =/= 1...." is supposed to mean. If you're going to ask for help, at least be clear.

    As for the range of y = x / (x-4):

    The range will be (-infinity,infinity). See that as x approaches 4 from negative infinity (3.999 and so on), y nears negative infinity. As x approaches 4 from infinity (4.00001 and so on), y nears positive infinity. It is also worthwhile to note that as x approaches positive or negative infinity, the limit of y is one.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2011
    Posts
    40
    Thanks
    1
    Quote Originally Posted by rtblue View Post
    Well, I'm not sure what "the range =/= 1...." is supposed to mean. If you're going to ask for help, at least be clear.

    As for the range of y = x / (x-4):

    The range will be (-infinity,infinity). See that as x approaches 4 from negative infinity (3.999 and so on), y nears negative infinity. As x approaches 4 from infinity (4.00001 and so on), y nears positive infinity. It is also worthwhile to note that as x approaches positive or negative infinity, the limit of y is one.
    after 1 hour of thinking i finally understand that one.

    but then i came upon y=9x^2/(25-x^2).

    when i rearranged for x, i got x=sqrt(-9x^2/y +25). how do i find the range from that? and please don't use all this "limit" language, i didn't learn it that way :s. we don't use infinity either, we just use > < =/= signs?

    and y=x^2 / (x^2 - 1 )
    range, i have no idea. when re-arranged to solve for x, i get x = sqrt(x^2/y+1), so its y =/= 0, but it's not
    Last edited by iragequit; March 31st 2011 at 06:18 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Quote Originally Posted by rtblue View Post
    Well, I'm not sure what "the range =/= 1...." is supposed to mean. If you're going to ask for help, at least be clear. ..
    It means "not equal".
    ragequitter, you should take a gander at Online LaTeX Equation Editor
    and add a little Latex markup to your posts.
    \neq is achieved by "\neq" between tex tags.

    There is a general way for determining the domain of real-valued functions such as the ones you listed.

    1. Start with all real numbers.
    2. Exclude any values that would lead to zero in the denominator, or a negative under an even radical (including square roots, 4th roots, etc). logarithms must have positive arguments also, but we're looking at rational functions (adjoin sqrt) for the moment.

    y=\frac{9x^2}{25-x^2}

    The bottom cannot be zero, so solve
    25 - x^2 \neq 0 to get
    x \neq \pm 5

    That. Is. It! Sure, you can write that in interval notation, but the result is the same.
    --------------------
    For the range, it's not as cut-and-dry. You can't always solve for x. More advanced techniques might come into play.
    It's a shame that you aren't familiar (comfortable) with end behavior (i.e. "limits at infinity") and the like.
    Have you heard about even/odd functions? You can use symmetry in the case of y as given above.
    You can use partial fraction decomposition... actually, that might be best in this case. I suspect the range is (almost?) every real number.
    Last edited by TheChaz; March 31st 2011 at 06:52 PM. Reason: Added range discussion
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2011
    Posts
    40
    Thanks
    1
    Quote Originally Posted by TheChaz View Post
    It means "not equal".
    ragequitter, you should take a gander at Online LaTeX Equation Editor
    and add a little Latex markup to your posts.
    \neq is achieved by "\neq" between tex tags.

    There is a general way for determining the domain of real-valued functions such as the ones you listed.

    1. Start with all real numbers.
    2. Exclude any values that would lead to zero in the denominator, or a negative under an even radical (including square roots, 4th roots, etc). logarithms must have positive arguments also, but we're looking at rational functions (adjoin sqrt) for the moment.

    y=\frac{9x^2}{25-x^2}

    The bottom cannot be zero, so solve
    25 - x^2 \neq 0 to get
    x \neq \pm 5

    That. Is. It! Sure, you can write that in interval notation, but the result is the same.
    --------------------
    For the range, it's not as cut-and-dry. You can't always solve for x. More advanced techniques might come into play.
    It's a shame that you aren't familiar (comfortable) with end behavior (i.e. "limits at infinity") and the like.
    Have you heard about even/odd functions? You can use symmetry in the case of y as given above.
    You can use partial fraction decomposition... actually, that might be best in this case. I suspect the range is (almost?) every real number.
    Okay, first of all, thanks a lot.

    I'm having trouble with problems where there is more than 1 variable under the square root when solved for x, like the one I mentioned. Like you said, domain isn't a problem for me.. I'm having trouble with range. Could you PLEASE go in depth about your methods? I'm willing to try anything, because I searched this everywhere on the internet and I can't find anything.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Quote Originally Posted by iragequit View Post
    How do you find these out algebraically given a function?

    It'd be appreciated if you could do an example with an exponential (2^x) function, a root function, a reciprocal function, a cubic function, a semi-circle function, a quadratic function, absolute value function, conics (ellipse, hyperbola, parabola of the form y=1/4px^2, etc.) and whatever else.

    ...
    a) y = \frac{9x^2}{ 25-x^2}

    b) y = \sqrt{4x^2-9}

    c) y = \frac{4x^2-9}{x^2}

    d) y = \frac{5}{\sqrt{3-x}}

    e) y = 4x^2+6x+6

    f) y = |x-3|+6
    Maybe a better man than I can come along and give you a universal way to work these!

    In the order or least resistance.
    e) y = f(x) has minimum (since the leading coefficient is positive) value when x = -b/(2a). So the range is y >= (whatever it is when x = -b/(2a)

    f) The absolute value achieves all non-negative values.
    So |x - 3| >= 0
    Add six to both sides to get (y =) |x - 3| + 6 >= 6. Range.

    c) Can be rewritten as 4 - 9/x^2. Since the part "9/x^2" is always positive, you will always be subtracting a positive value from 4, so you can never get your function to EQUAL 4. And it will never be higher. As x gets close to zero, the fraction 9/x^2 gets big, so you can achieve any negative value for your function. Range : y < 4
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2011
    Posts
    40
    Thanks
    1
    Quote Originally Posted by TheChaz View Post
    Maybe a better man than I can come along and give you a universal way to work these!

    In the order or least resistance.
    e) y = f(x) has minimum (since the leading coefficient is positive) value when x = -b/(2a). So the range is y >= (whatever it is when x = -b/(2a)

    f) The absolute value achieves all non-negative values.
    So |x - 3| >= 0
    Add six to both sides to get (y =) |x - 3| + 6 >= 6. Range.

    c) Can be rewritten as 4 - 9/x^2. Since the part "9/x^2" is always positive, you will always be subtracting a positive value from 4, so you can never get your function to EQUAL 4. And it will never be higher. As x gets close to zero, the fraction 9/x^2 gets big, so you can achieve any negative value for your function. Range : y < 4
    i can see how you did c), but what if the fraction was in the form a / b + c instead of a + b / c? how do you break it up?

    for example, x^2 / (x^2-1)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by iragequit View Post
    i can see how you did c), but what if the fraction was in the form a / b + c instead of a + b / c? how do you break it up?

    for example, x^2 / (x^2-1)
    \displaystyle \frac{x^2}{x^2 - 1} = \frac{(x^2 - 1) + 1}{x^2 -1} = 1 + \frac{1}{x^2 - 1}.

    The problem is that you want a single magic bullet that will solve every question. There is none. Each question has to taken on its own merits, you must apply your own mathematical knowledge and experience to solve it.

    As a general rule, you should first draw a graph when trying to find the range of a function. (As to how to draw graphs, you should already have been taught that).
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    Thanks, Mr. F! I started typing a similar response, but it was past my bedtime
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 19th 2011, 12:07 PM
  2. Confused about domain and range
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 21st 2010, 10:08 AM
  3. Replies: 2
    Last Post: February 19th 2010, 05:50 AM
  4. Replies: 6
    Last Post: September 16th 2009, 06:25 AM
  5. Replies: 4
    Last Post: November 17th 2006, 11:57 PM

Search Tags


/mathhelpforum @mathhelpforum