find extreme values of $\frac{2x}{x²+4}$

Solution 1:

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.

Solution 2:

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 $\theta$