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

1. Feb 7, 2015

### knowLittle

1. The problem statement, all variables and given/known data
I need to show that $$\frac{x}{x^2+1}$$ is either onto or not.
My domain is $$R-{0}$$ and range is $$R$$

2. Relevant 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.

3. 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: Feb 7, 2015
2. Feb 7, 2015

### Staff: Mentor

That is not the domain, and the range is not all real numbers.
Solve the equation above for x (which you said). First thing to do is multiply both sides by x2 + 1.

3. Feb 7, 2015

### knowLittle

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. Feb 7, 2015

### knowLittle

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. Feb 7, 2015

### Staff: Mentor

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. Feb 7, 2015

### Staff: Mentor

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. Feb 7, 2015

### vela

Staff Emeritus
With the given domain, though, it becomes obvious that the function isn't surjective.

8. Feb 8, 2015

### Staff: Mentor

It wasn't clear to me that that restriction was actually part of this problem.