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One-to-one function determination

  1. Jul 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Without a graphing calculator, how can you tell that the function

    f(x) = x/(x^2+1) is one-to-one?


    2. Relevant equations



    3. The attempt at a solution

    You can sketch both x and 1/(x^2+1) separately but I did not think it was obvious that when you multiplied them togethor the result was not one-to-one.
     
    Last edited: Jul 29, 2007
  2. jcsd
  3. Jul 29, 2007 #2

    Hurkyl

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    Is your homework problem actually to show that that function is one-to-one? One way to show that a function is one-to-one is to start by stating the definition of one-to-one, and then prove that this function satisfies the definition.
     
    Last edited: Jul 29, 2007
  4. Jul 29, 2007 #3
    A function is one-to-one if whenever s1 and s2 are two different elements in the domain, f(s1) is not equal to s2.
     
    Last edited: Jul 29, 2007
  5. Jul 29, 2007 #4

    Hurkyl

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    The contrapositive is often easier to work with; if f(x)=f(y), then x=y.
     
  6. Aug 22, 2010 #5
    It is one-to-one function so that assuming that
    f(a)=f(b)
    we find easily that a=b.
    you can follow the link to see the graph

    http://www.4shared.com/file/d8TFg6zC/emad.html
     
  7. Aug 23, 2010 #6

    ehild

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    Find what value/values of x belong to a certain value of y.


    ehild
     
  8. Aug 23, 2010 #7

    hunt_mat

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    You are asked to show:
    [tex]
    \frac{a}{1+a^{2}}=\frac{b}{1+b^{2}}
    [/tex]
    So expand and write as a quadratic:
    [tex]
    b^{2}-\Bigg( a+\frac{1}{a}\Bigg) b+1=0
    [/tex]
    If f(x) is one to one, the above quadratic should have one and only one solution, does it?
     
  9. Aug 23, 2010 #8

    vela

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    If you flip the function over, you get

    [tex]y=\frac{x}{1+x^2} \Rightarrow \frac{1}{y} = x+\frac{1}{x}[/tex]

    That might be a bit easier to analyze.
     
  10. Aug 23, 2010 #9

    hunt_mat

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    You can factorise, my quadratic equation.

    Mat
     
  11. Aug 23, 2010 #10

    vela

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    Yeah, I know. I was just offering yet another way to look at the problem. I hadn't really thought about the problem until I saw the linear term in your quadratic and realized you could easily deduce the answer looking at the reciprocal of the function.
     
  12. Aug 23, 2010 #11

    hunt_mat

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    I know, I was only winding you up.
     
  13. Aug 23, 2010 #12
    What is f(0)?

    What is [tex]\lim_{x \to \infty} f(x)[/tex]?

    What does this tell you?
     
  14. Aug 23, 2010 #13

    hunt_mat

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    Or even(coming from my quadratic equation) what is f(a) and f(1/a)?
     
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