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

  • Thread starter knowLittle
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  • #1
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Homework Statement


I need to show that $$\frac{x}{x^2+1}$$ is either onto or not.
My domain is $$R-{0}$$ and range is $$R$$

Homework 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.

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:

Answers and Replies

  • #2
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Homework Statement


I need to show that $$\frac{x}{x^2+1}$$ is either onto or not.
My domain is $$R-{0}$$ and range is $$R$$
That is not the domain, and the range is not all real numbers.
knowLittle said:

Homework 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.
Solve the equation above for x (which you said). First thing to do is multiply both sides by x2 + 1.
knowLittle said:

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.
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
34,678
6,387
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.
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.
 
  • #6
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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.
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.
 
  • #7
vela
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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.
With the given domain, though, it becomes obvious that the function isn't surjective.
 
  • #8
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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.
vela said:
With the given domain, though, it becomes obvious that the function isn't surjective.
It wasn't clear to me that that restriction was actually part of this problem.
 

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