Examining the injectivity and surjectivity of $f(x) = x/(x^2+1)$ I'm beginning with my self study in math from the $11$th grade, so bear with the simplicity.
I've got to test the objectivity of $x/(x^2+1)$, function is from $\Bbb R$ to $\Bbb R$.
First issue is, I think it is injective, because my algebra after equating the output of two inputs leads to both inputs coming equal to each other. The book answer says I'm wrong (without explanation).
With surjectivity, I can't find a simple expression for the inverse of the function, so I thought of this:
Let $g(x) = x, t(x) = x^2+1$
Then $f(x) = [g(x)/t(x)]$. Now $g(x)$ is surjective and $t(x)$ isn't. Can I argue that the ratio however, will be surjective because the numerator is surjective, and thus even if it is being divided only by numbers greater than or equal to $1$, the numerator's surjectivity will make the whole function surjective?
If I can, is there a rigorous way of arguing this? If I can't, a counterexample or counter-proof would help.
 A: Another way, taking some shortcuts:

*

*$\displaystyle f(x) = \frac{1}{x+\frac{1}{x}}\,$ for $\,x \ne 0\,$, so $\,f(x)=f\left(\frac{1}{x}\right)\,$, and therefore $\,f\,$ cannot be injective;


*$\displaystyle |f(x)| = \frac{1}{|x|+\frac{1}{|x|}} \le \frac{1}{2}\,$ for $\,x \ne 0\,$, where the inequality follows from $\,|x|+\frac{1}{|x|} \ge 2\,$ by the AM-GM inequality, and therefore $\,f\,$ cannot be surjective.
A: In this case, I think there is a simpler way to check injectivity/surjectivity. Let $f(x)=c$. Surjective means that for any $c$ there is a solution. Injective means that for any solution, the solution is unique. $$\frac x{x^2+1}=c\\cx^2-x+c=0$$
This is a simple quadratic equation if $c\ne 0$. Then you have a real solution if the discriminant is positive. $$\Delta=1-4c^2$$
If the discriminant is negative you have no solution. In this case it means $|c|>\frac12$. So the function is not surjective. Also if $|c|<\frac12$ and $c\ne 0$ the quadratic has two distinct solution, so the function is not injective either.
A: For injectivity, note that you can "test" this graphically. Look at the graph: if any horizontal line crosses the graph twice or more, then it is not injective.
Here our graph is this:

To debunk it, you can simply use that, if $x,y$ are distinct yet $f(x),f(y)$ are equal, then it is not injective. Pick a point (e.g. $1/2$ or $1/4$) and find $f$ of that point. For instance,
$$f \left( \frac 1 2 \right) = \frac 2 5$$
The graph suggests there is another point $x$ with that value. Question is, what? We can solve for it by simply solving $f(x) = 1/4$; we can get the other value this way (you would get two solutions here), and you can ensure it is not some sort of extraneous solution by finding $f(x)$ after finding $x$.
You can also use a more analytical argument. Notice the following:

*

*$f'(x) > 0$ on $(-1,1)$

*It is negative elsewhere

*$f$ achieves global extreme values of $\pm 1/2$

*$\displaystyle \lim_{x \to \pm \infty} f(x) = 0$

*Hence, $f$ maps both $(-1/2,1/2) \setminus \{0\}$ and $(-\infty,-1/2)\cup(1/2,\infty)$ onto (-1/2,1/2) \setminus {0}$, clearly violating injectivity

I'll leave justifying those arguments up to you.

As for surjectivity, the graph makes that quite obvious. You can argue against it a few ways:

*

*As before, we can find the global extrema. Then by continuity and the intermediate value theorem, we can't achieve values outside that interval.


*Suppose we want to argue nonsurjectivity from a more basic perspective. Suppose otherwise; then there is a $x$ such that $f(x) = 2$. Try solving
$$\frac{x}{x^2+1} = 2$$
to find applicable values of $x$. An issue will arise. (The discriminant in the resulting quadratic will be negative, so such an $x$ cannot be a real number.)


With surjectivity, I can't find a simple expression for the inverse of the function, so I thought of this:
Let $g(x) = x, t(x) = x^2+1$
Then $f(x) = [g(x)/t(x)]$. Now $g(x)$ is surjective and $t(x)$ isn't. Can I argue that the ratio however, will be surjective because the numerator is surjective, and thus even if it is being divided only by numbers greater than or equal to $1$, the numerator's surjectivity will make the whole function surjective?

Surjectivity or not of the denominator and numerator determine nothing really.
Half of the reason is simply because the choice of codomain is somewhat arbitrary of a decision. I can make (for $f(x) = x/(x^2+1)$) $f$ have codomain $\Bbb R, \Bbb C, \Bbb Q \cup (-1/2,1/2), [-10,10]$, or whatever. So long as the range of $f$ - $f(\Bbb R)$ in this case - is contained in that codomain, it's a valid function.
A: $\displaystyle \because f(x)=\frac{x}{x^{2}+1}=\frac{1}{x+\frac{1}{x}}, \quad $
$\therefore f(2)=f\left(\frac{1}{2}\right) \Rightarrow f$ is not injective.
$\forall x \in \mathbb{R} $, letting $ \displaystyle y=\frac{x}{x^{2}+1}$ yields
$$
y x^{2}-x+y=0 \cdots(*)
$$
$$\Delta_{(\star)}  \geqslant 0 \quad \Leftrightarrow \quad 1-4 y^{2} \geqslant 0 \quad \Leftrightarrow \quad |y| \leqslant \frac{1}{2}.$$
Therefore $\displaystyle \operatorname{Ran} (f)=\left[-\frac{1}{2}, \frac{1}{2}\right]$ and hence $f$ is NOT surjective.
A: Honestly I find the current answers overly complicated. The graph hits some horizontal lines more than once. For example, it looks like $f(x)=1/4$ has two solutions, and you can find that they are when $x$ is $2\pm\sqrt 3$.
Since the graph doesn't hit every horizontal line, the function is not surjective. For example, $f(x)=1$ implies $x=x^2+1$ has a solution, which is false (over the real numbers).
You only need the quadratic formula for both these claims.
