Proving f(x): One-to-One, Onto, or Both?

[cilla]

Homework Statement

Prove whether the function f(x) = x/(1+x^2) with domain & codomain = reals is one-to-one, onto, or both.

Homework Equations

The Attempt at a Solution

I know to show if it's one-to-one I have to show a/(1+a^2) = b/(1+b^2), ultimately that a = b, I don't know how to simplify them to that.

 

[PeroK]

Homework Statement

Prove whether the function f(x) = x(1+x^2) with domain & codomain = reals is one-to-one, onto, or both.

Homework Equations

The Attempt at a Solution

I know to show if it's one-to-one I have to show a/(1+a^2) = b/(1+b^2), ultimately that a = b, I don't know how to simplify them to that.

Have you tried drawing a graph of the function?

 

[RUber]

A good trick is to assume that b = a + h and then show that h must be equal to zero.

 

[RUber]

To show onto, pick an arbitrary b in the reals and show that there is an a such that f(a) = b. if there aren't any conditions on b, then it is onto.

 

[Mark44]

Homework Statement

Prove whether the function f(x) = x(1+x^2) with domain & codomain = reals is one-to-one, onto, or both.

Homework Equations

The Attempt at a Solution

I know to show if it's one-to-one I have to show a/(1+a^2) = b/(1+b^2), ultimately that a = b, I don't know how to simplify them to that.

What is the formula for the function you're working with. In the problem statement you wrote f(x) = x(1 + x²). In your attempt, the formula appears to be f(x) = ##\frac{x}{1 + x^2}##.

Which is it?

 

[cilla]

Oh whoops, thanks Mark44. It is the latter. Can't change the post title unfortunately but I fixed it in the body.

 

[cilla]

Thanks RUber but I still don't know how to execute that...

 

[cilla]

How do you simplify a/(1+a^2) = (a+h)/(1+(a+h)^2) to a = a+h ?

 

[LCKurtz]

As has already been suggested, draw the graph. That will at least tell you what you what the answers are. Then you can think about how to prove them analytically.

 

[Mark44]

Thanks RUber but I still don't know how to execute that...

Set a/(1 + a²) = b, and then solve for a.

 

[pasmith]

Homework Statement

Prove whether the function f(x) = x/(1+x^2) with domain & codomain = reals is one-to-one, onto, or both.

Homework Equations

The Attempt at a Solution

I know to show if it's one-to-one I have to show a/(1+a^2) = b/(1+b^2), ultimately that a = b, I don't know how to simplify them to that.

Let [itex]f(x) = y[/itex]. Then [itex](1 + x^2)y = x[/itex] so [itex]yx^2 - x + y = 0[/itex]. Hence, [itex]f(x) = y[/itex] if and only if [itex]x \in \mathbb{R}[/itex] is a solution of [tex]
yx^2 - x + y = 0.[/tex] Now consider:

(1) For what values of [itex]y \in \mathbb{R}[/itex] does that quadratic have real roots?
(2) For what values of [itex]y \in \mathbb{R}[/itex] does that quadratic have a repeated root?

 

[GFauxPas]

Do you know differential Calculus?
You can use the intermediate value theorem to show that there are distinct x, y such that f(x) = f(y)
(Note this won't tell you what the x,y are, but that's not important)

 

[cilla]

GFauxPas: The most I know of calculus is pre-, and that I took years ago (my most recent math class prior to this one). So as you might guess I'm very much out of practice.

I found a counterexample for f being injective: f(1/2) = 2/5 = f(2). Thus the function is not one-to-one.

As for f being surjective:
Assume y = 1 = x/(1+x^2). Then
1+x^2 = x
x^2 - x + 1 = 0
x = (1 [+/-] sqrt(1 - 4*1)) / (2*1) ∉ ℝ
which is a contradiction since ℝ is the domain, x must be an element contained within it.

Thanks for the hints, though I didn't see any of these since my last reply until now... Only spent 5.5 hours in tutoring yesterday... =\

 

[Mark44]

As for f being surjective:
Assume y = 1 = x/(1+x^2). Then
1+x^2 = x
x^2 - x + 1 = 0
x = (1 [+/-] sqrt(1 - 4*1)) / (2*1) ∉ ℝ
which is a contradiction since ℝ is the domain, x must be an element contained within it.

A better approach that doesn't assume anything about y, is to let y = b and then solve for x.

$$\frac{x}{x^2 + 1} = b$$
$$\Rightarrow x = bx^2 + b$$
$$\Rightarrow bx^2 - x + b = 0$$
$$\Rightarrow x = \frac{1 \pm \sqrt{1 - 4b^2}}{2b}$$
For x to be real, 1 - 4b² has to be nonnegative. From this it's easy to see what the range of this function is, and that the function is not onto the reals.

 

[cilla]

I see your point thanks a lot.