find extreme values of $\frac{2x}{x²+4}$ I am doing my homework like a good little boy. I know that when I want to find the extreme values of a function I have to put the derivative equal to zero so I can find the x values. I've done it before with easier functions where I only have to use the power rule.
But with the function 
$\dfrac{2x}{x²+4}$
I think I have to use the quotient rule:
$f'(x)= \dfrac {2x²+4-(2x)^2} {(x²+4)^2} \implies f'(x)= \dfrac{-2x²+4}{(x²+4)^2}$
Is this correct? If it is, I think the following step would be
$\dfrac{-2x²+4}{x^4+8x²+16}=0$
I don't know how I can find the solutions to this... Any tips?
 A: You employed the quotient rule correctly, but because you omitted a necessary pair of parentheses in the numerator, you made an algebra error in simplifying it. You should have
$$f\,'(x)=\frac{2(x^2+4)-(2x)(2x)}{(x^2+4)^2}=\frac{2x^2+8-4x^2}{(x^2+4)^2}=\frac{8-2x^2}{(x^2+4)^2}\;.$$
Setting this to $0$ and solving is easier than it looks: assuming that the denominator is not $0$, a fraction is $\mathbf0$ if and only if its numerator is $\mathbf0$. Clearly $(x^2+4)^2$ is never $0$, since it’s always at least $16$, so
$$\frac{8-2x^2}{(x^2+4)^2}=0\quad\text{if and only if}\quad 8-2x^2=0\;,$$
and that’s an easy equation to solve.
A: Here is an algebraic way without using calculus 
$$\text{Let }y=\frac{2x}{x^2+4}\iff x^2y-2x+4y=0$$ which is a Quadratic equation in $x$
As $x$ is real, the discriminant of the above equation must   be $\ge0$
i.e, $(-2)^2\ge 4\cdot y\cdot 4y\iff y^2\le \frac14$
We know, $x^2\le a^2\iff -a\le x\le a$

Alternatively,  let $x=2\tan\theta$ which is legal as $x,\tan\theta$ can assume any real value
$$\implies\frac{2x}{x^2+4}=\frac{2\cdot 2\tan\theta }{(2\tan\theta)^2+4}=2\frac{ 2\tan\theta }{4(1+\tan^2\theta)}=\frac{\sin2\theta}2$$
Now, we know the range of $\sin2\theta$ for real $\theta$
