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Prove whether f(x) = x/(1+x^2) with domain & codomain = reals is one-to-one, onto, or both

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove whether the function f(x) = x/(1+x^2) with domain & codomain = reals is one-to-one, onto, or both.

    2. Relevant equations


    3. The attempt at a solution
    I know to show if it's one-to-one I have to show a/(1+a^2) = b/(1+b^2), ultimately that a = b, I don't know how to simplify them to that.
     
    Last edited: Oct 18, 2014
  2. jcsd
  3. Oct 18, 2014 #2

    PeroK

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    Have you tried drawing a graph of the function?
     
  4. Oct 18, 2014 #3

    RUber

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    A good trick is to assume that b = a + h and then show that h must be equal to zero.
     
  5. Oct 18, 2014 #4

    RUber

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    To show onto, pick an arbitrary b in the reals and show that there is an a such that f(a) = b. if there aren't any conditions on b, then it is onto.
     
  6. Oct 18, 2014 #5

    Mark44

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    What is the formula for the function you're working with. In the problem statement you wrote f(x) = x(1 + x2). In your attempt, the formula appears to be f(x) = ##\frac{x}{1 + x^2}##.

    Which is it?
     
  7. Oct 18, 2014 #6
    Oh whoops, thanks Mark44. It is the latter. Can't change the post title unfortunately but I fixed it in the body.
     
  8. Oct 18, 2014 #7
    Thanks RUber but I still don't know how to execute that...
     
  9. Oct 18, 2014 #8
    How do you simplify a/(1+a^2) = (a+h)/(1+(a+h)^2) to a = a+h ?
     
  10. Oct 18, 2014 #9

    LCKurtz

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    As has already been suggested, draw the graph. That will at least tell you what you what the answers are. Then you can think about how to prove them analytically.
     
  11. Oct 18, 2014 #10

    Mark44

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    Set a/(1 + a2) = b, and then solve for a.
     
  12. Oct 19, 2014 #11

    pasmith

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    Let [itex]f(x) = y[/itex]. Then [itex](1 + x^2)y = x[/itex] so [itex]yx^2 - x + y = 0[/itex]. Hence, [itex]f(x) = y[/itex] if and only if [itex]x \in \mathbb{R}[/itex] is a solution of [tex]
    yx^2 - x + y = 0.[/tex] Now consider:

    (1) For what values of [itex]y \in \mathbb{R}[/itex] does that quadratic have real roots?
    (2) For what values of [itex]y \in \mathbb{R}[/itex] does that quadratic have a repeated root?
     
  13. Oct 19, 2014 #12
    Do you know differential Calculus?
    You can use the intermediate value theorem to show that there are distinct x, y such that f(x) = f(y)
    (Note this won't tell you what the x,y are, but that's not important)
     
    Last edited: Oct 19, 2014
  14. Oct 20, 2014 #13
    GFauxPas: The most I know of calculus is pre-, and that I took years ago (my most recent math class prior to this one). So as you might guess I'm very much out of practice.

    I found a counterexample for f being injective: f(1/2) = 2/5 = f(2). Thus the function is not one-to-one.

    As for f being surjective:
    Assume y = 1 = x/(1+x^2). Then
    1+x^2 = x
    x^2 - x + 1 = 0
    x = (1 [+/-] sqrt(1 - 4*1)) / (2*1) ∉ ℝ
    which is a contradiction since ℝ is the domain, x must be an element contained within it.

    Thanks for the hints, though I didn't see any of these since my last reply until now... Only spent 5.5 hours in tutoring yesterday... =\
     
  15. Oct 20, 2014 #14

    Mark44

    Staff: Mentor

    A better approach that doesn't assume anything about y, is to let y = b and then solve for x.

    $$\frac{x}{x^2 + 1} = b$$
    $$\Rightarrow x = bx^2 + b$$
    $$\Rightarrow bx^2 - x + b = 0$$
    $$\Rightarrow x = \frac{1 \pm \sqrt{1 - 4b^2}}{2b}$$
    For x to be real, 1 - 4b2 has to be nonnegative. From this it's easy to see what the range of this function is, and that the function is not onto the reals.
     
  16. Oct 20, 2014 #15
    I see your point thanks a lot.
     
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