One-to-one function determination

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ehrenfest
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Homework Statement


Without a graphing calculator, how can you tell that the function

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


Homework Equations





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.
 
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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.
 
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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.
 
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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
 
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?
 
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.
 
What is f(0)?

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

What does this tell you?