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

[knowLittle]

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.

 

[Mark44]

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.

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.

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.

 

[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.

 

[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.

 

[Mark44]

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.

 

[Mark44]

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.

 

[vela]

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.

 

[Mark44]

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.

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