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Math Help - Range of y=3/(x^2+1)?

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    Range of y=3/(x^2+1)?

    I want to know how to find the range of the equation y=3/(x^2+1) and y= (x-4)/x without using the graphical method?

    Detailed explanation would be appreciated.
    Please and thank you.
    Last edited by Basketball; September 20th 2010 at 05:42 PM.
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    Quote Originally Posted by Basketball View Post
    I want to know how to find the range of the equation y=3/(x^2+1)
    For any value of x y will be positive. Do you know why?

    Quote Originally Posted by Basketball View Post
    y= (x-4)/4
    This is a linear function. The range should be obvious!
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  3. #3
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    y = \frac{x - 4}{4} = \frac{1}{4}x - 1.

    This is a linear function. Linear functions have domain \mathbf{R} and range \mathbf{R} unless the domain is restricted.



    y = \frac{3}{x^2 + 1}.

    It should be clear that the denominator is never negative. In fact, the denominator is always \geq 1.

    You should know that as a denominator gets larger, the fraction gets smaller. So that means that the largest possible value y can take is \frac{3}{1} = 3, and the fraction will get closer to 0 as x increases/decreases.

    So the range is y \in (0, 3].
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    Sorry I meant y=(x-4)/x, thank you for explaining the other equation to me.
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    Quote Originally Posted by Basketball View Post
    Sorry I meant y=(x-4)/x, thank you for explaining the other equation to me.
    y = \frac{x - 4}{x} = 1 - \frac{4}{x}.

    This is a hyperbola. You should know that the standard hyperbola of the form y = \frac{a}{x} has a horizontal asymptote at y = 0. So the standard hyperbolda's range would be y \in \mathbf{R}\backslash\{0\}.

    In this case, the hyperbola has a vertical translation of 1 unit up.

    So what do you think its range is?
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    Quote Originally Posted by Prove It View Post
    y = \frac{x - 4}{x} = 1 - \frac{4}{x}.

    This is a hyperbola. You should know that the standard hyperbola of the form y = \frac{a}{x} has a horizontal asymptote at y = 0. So the standard hyperbolda's range would be y \in \mathbf{R}\backslash\{0\}.

    In this case, the hyperbola has a vertical translation of 1 unit up.

    So what do you think its range is?
    y cannot be 1?
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    If I were to rewrite y=1-(4/x) in the format of y=(a/x), it would be y=-(4/x)+1. What does the negative in front of (4/x) do to the graph or what does it mean? Our teacher really didn't teach how the graphs would look like, I only know what y=(a/x) looks like. I don't really know how the numbers would affect the outcome of this graph.
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    Correct, the range is y \in \mathbf{R}\backslash\{1\}.

    The negative gives a reflection in the x axis. It doesn't affect the position of the asymptotes.
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    Quote Originally Posted by Basketball View Post
    I want to know how to find the range of the equation y=3/(x^2+1) and y= (x-4)/x without using the graphical method?

    Detailed explanation would be appreciated.
    Please and thank you.
    Personally, I think that generally speaking you are asking for trouble by trying to find the range of a function without using assistance from a graph. But that's just my opinion.
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    Personally, I think that generally speaking you are asking for trouble by trying to find the range of a function without using assistance from a graph. But that's just my opinion.
    Unless of course you use your knowledge of functions to try to graph the function yourself
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  11. #11
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    Another way to find the range of y= \frac{x- 4}{x}- find its inverse (solve for x): xy= x- 4, x- xy= 4, x(1- y)= 4, x= \frac{4}{1- y}.

    If we think of that as a function of y, it is obvious that its domain is " y\ne 1. Since a function and its inverse "swap" domain and range, the range of the original function is "all y except 1".
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