One-to-one function determination

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Homework Help Overview

The problem involves determining whether the function f(x) = x/(x^2+1) is one-to-one without the aid of a graphing calculator. The discussion centers around the definition of one-to-one functions and various approaches to prove this property.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of one-to-one functions and suggest proving that f(x) satisfies this definition. Some explore the contrapositive approach, while others consider analyzing the function through its reciprocal or by examining specific values and limits.

Discussion Status

The discussion is active, with participants offering different perspectives and methods for analyzing the function. Some have provided insights into the implications of the quadratic form derived from the function, while others are exploring the behavior of the function at specific points and limits.

Contextual Notes

There is an emphasis on not using graphing calculators, and participants are encouraged to explore various mathematical approaches to the problem. Some constraints regarding the methods of proof and the nature of the function are being discussed.

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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The contrapositive is often easier to work with; if f(x)=f(y), then x=y.
 
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
 
Find what value/values of x belong to a certain value of y. ehild
 
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?
 
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.
 
You can factorise, my quadratic equation.

Mat
 
  • #10
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.
 
  • #11
I know, I was only winding you up.
 
  • #12
What is f(0)?

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

What does this tell you?
 
  • #13
Or even(coming from my quadratic equation) what is f(a) and f(1/a)?
 

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