Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ . 
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2  -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2  -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
 A: We need to prove that$$\left|\frac{2x}{x^2+1}\right|\leq1$$ or
$$x^2-2|x|+1\geq0$$ or
$$\left(|x|-1\right)^2\geq0.$$
The equality occurs for $|x|=1$, which gives:
$(1,1)$ is a maximum point and $(-1-1)$ is a minimum point.
Done!
A: 
For what values of $a$ does the equation
  $$
\frac{2x}{1+x^2}=a
$$
  admit a solution?

The equation can be rewritten
$$
ax^2-2x+a=0
$$
and its discriminant is
$$
4(1-a^2)
$$
which is $\ge0$ if and only if $-1\le a\le 1$. For $a=0$ it's not a degree $2$ equation, but of course $f(0)=0$, so also $a=0$ is a value attained by the function.
A: HINT:
If calculus is not mandatory,
Method$\#1:$
Set $x=\tan A$
Method$\#2:$
$$\implies x^2y-2x+y=0$$
As $x$ is real, the discriminant must be $\ge0$
Method$\#3:$
If $x>0,$  $$\dfrac{2x}{x^2+1}=\dfrac2{x+\dfrac1x}$$
Now using AM-GM inequality, $$\dfrac{x+\dfrac1x}2\ge\sqrt{x\cdot\dfrac1x}=?$$
If $x<0,$ set $x=-u$
A: i would write
$$-1\le \frac{2x}{x^2+1}\le 1$$ this is equivalent to
$$-x^2-1\le 2x$$ or $$(x+1)^2\geq 0$$ and $$2x\le x^2+1$$ this is equivalent to $$(x-1)^2\geq 0$$
A: use AM-GM:
If $x>0$ then $$\frac{2x}{1+x^2}\le\frac{2x}{2\sqrt{x^2}}=\frac{2x}{2x}=1$$
Similarly $x<0$
$$\frac{2x}{1+x^2}\ge\frac{2x}{2\sqrt{x^2}}=\frac{2x}{2|x|}=-1$$
A: Your second and third approaches are workable, but they lack the part proving that the fraction otherwise is between $-1$ and $1$.
For the second you use the sign of the derivate between and beyond the stationary points, especially beyond. You then can see that for $x>1$ the function is decreasing and for $x<-1$ too (and for $-1\le x\le 1$ it's increasing.
For the discriminant you consider for a given candidate value of $y$ the equation $2x/(x^2+1) = y$ which becomes
$$yx^2 - 2x + y = 0$$
which has the discriminant $(-2)^2 - 4y^2 = 4(1-y^2)$ and the equation only have solutions if $1-y^2=(1-y)(1+y)\ge 0$ which means that the factors are both positive or both negative (but they can't be both negative). So $1-y \ge 0$ and $1+y\ge 0$ that is $-1\le y\le 1$.
A: Squares are non-negative so 
$$-(x+1)^2\le 0\le (x-1)^2$$
so
$$-x^2-2x-1\le 0\le x^2-2x+1$$
so
$$-x^2-1\le {2x}\le x^2+1$$
so
$$-1\le \frac{2x}{x^2+1}\le 1$$
since this is division by $x^2+1>0$
A: All this is usually summarized in a good old variation table.
$$f(x)=\frac{2x}{x^2+1}\quad,\quad f'(x)=\frac{-2(x-1)(x+1)}{(x^2+1)^2}$$
$\begin{array}{|l|ccccccc|}\hline
x & -\infty& & -1  && 1 && +\infty\\
\hline
f'&& - & 0 & + & 0 & -\\ 
\hline
f&0&\searrow&-1&\nearrow&+1&\searrow&0\\
\hline
\end{array}$
Adding that $x^2+1$ does not annulate, thus $f$ is continuous and so range is given by the closed interval made of miminum and maximum values in the table since they are reached : $[-1,1]$.
A: Since
$$\frac{2x}{x^2+1}=\frac{(x+1)^2}{x^2+1}-1$$
it suffices to show that $\frac{(x+1)^2}{x^2+1}\le2$, as this expression is clearly positive. This is obvious, since
$$\frac{(x+1)^2}{x^2+1}-2=-\frac{(x-1)^2}{x^2+1}\le 0$$
