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Math Help - Showing a function is not surjective

  1. #1
    Super Member craig's Avatar
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    Showing a function is not surjective

    Hi, I've got a question here and I've got to show that a function is not surjective.

    To do this I've got to find a value that the function cannot equal, but I'm not sure how to do this, is trail and error the right way or is there a more formal approach?

    Here's the actual function:

    f : \mathbb{R} \ \frac{1}{3} \to \mathbb{R} given by f(x) = \frac{5x+1}{3x-1}

    Thanks in advance
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  2. #2
    Member Black's Avatar
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    There's no real x such that f(x)=5/3, which is the horizontal asymptote.
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    Super Member craig's Avatar
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    Quote Originally Posted by Black View Post
    There's no real x such that f(x)=5/3, which is the horizontal asymptote.
    Ohh yeh thank you!
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  4. #4
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    A direct way to do it is to set up the equation \frac{5x+1}{3x-1}= y and try to solve for x.

    5x+1= (3x+1)y= 3xy+ y. 5x- 3xy= y - 1. (5- 3y)x= y- 1.

    x= \frac{y-1}{5- 3y} as long as the denominator is not 0.

    If 5- 3y= 0, that is, if y= \frac{3}{5}, there is no such x.
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    A direct way to do it is to set up the equation \frac{5x+1}{3x-1}= y and try to solve for x.

    5x+1= (3x+1)y= 3xy+ y. 5x- 3xy= y - 1. (5- 3y)x= y- 1.

    x= \frac{y-1}{5- 3y} as long as the denominator is not 0.

    If 5- 3y= 0, that is, if y= \frac{3}{5}, there is no such x.
    Thanks for the reply, that's a good way of thinking of it.
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