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Show that is not onto (##\frac{x}{x^2+1}##)

  1. Feb 7, 2015 #1
    1. The problem statement, all variables and given/known data
    I need to show that $$\frac{x}{x^2+1}$$ is either onto or not.
    My domain is $$R-{0}$$ and range is $$R$$

    2. Relevant equations
    I have learn to do this to show that a function is surjective
    y = $$\frac{x}{x^2+1}$$ and solve for x, but I am not sure how to proceed here.

    3. The attempt at a solution
    By trial and error and a calculator I have seen that the range is $$\frac{1}{2} \leq x \leq \frac{1}{2}$$
    It means that it's not onto.

    Any help on how to proceed?
    Thank you.
     
    Last edited: Feb 7, 2015
  2. jcsd
  3. Feb 7, 2015 #2

    Mark44

    Staff: Mentor

    That is not the domain, and the range is not all real numbers.
    Solve the equation above for x (which you said). First thing to do is multiply both sides by x2 + 1.
     
  4. Feb 7, 2015 #3
    The question says that A= R-{0} and B =R. Then, that f:A ->B and I need to show whether they 1-1 and whether they are onto. Prove.
    Thanks for the hint.
     
  5. Feb 7, 2015 #4
    Once, I have ##y(x^2+1) = x## what else can I do?
    ## y x^2 + y = x## divide all over x^2 wouldn't work.
     
  6. Feb 7, 2015 #5

    Mark44

    Staff: Mentor

    Move the x term over to the left side -- you have a quadratic in x, which you should know how to solve.

    The reason for doing this (i.e., solving for x from the original equation) is to determine whether there are any restrictions on y. If there are, your function is not onto the entire real line (the y-axis). If there are no restrictions, the function is onto the reals. IOW, for any y value whatsoever, there is an x value that maps to it.
     
  7. Feb 7, 2015 #6

    Mark44

    Staff: Mentor

    For the function in this problem, there is no inherent reason for the domain to be R - {0}. The function is defined for all real numbers.
     
  8. Feb 7, 2015 #7

    vela

    User Avatar
    Staff Emeritus
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    With the given domain, though, it becomes obvious that the function isn't surjective.
     
  9. Feb 8, 2015 #8

    Mark44

    Staff: Mentor

    It wasn't clear to me that that restriction was actually part of this problem.
     
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